contestId int64 0 1.01k | index stringclasses 57 values | name stringlengths 2 58 | type stringclasses 2 values | rating int64 0 3.5k | tags listlengths 0 11 | title stringclasses 522 values | time-limit stringclasses 8 values | memory-limit stringclasses 8 values | problem-description stringlengths 0 7.15k | input-specification stringlengths 0 2.05k | output-specification stringlengths 0 1.5k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 425k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3 values | verdict stringclasses 14 values | testset stringclasses 12 values | passedTestCount int64 0 1k | timeConsumedMillis int64 0 15k | memoryConsumedBytes int64 0 805M | code stringlengths 3 65.5k | prompt stringlengths 262 8.2k | response stringlengths 17 65.5k | score float64 -1 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
721 | A | One-dimensional Japanese Crossword | PROGRAMMING | 800 | [
"implementation"
] | null | null | Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). | The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. | [
"3\nBBW\n",
"5\nBWBWB\n",
"4\nWWWW\n",
"4\nBBBB\n",
"13\nWBBBBWWBWBBBW\n"
] | [
"1\n2 ",
"3\n1 1 1 ",
"0\n",
"1\n4 ",
"3\n4 1 3 "
] | The last sample case correspond to the picture in the statement. | 500 | [
{
"input": "3\nBBW",
"output": "1\n2 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "4\nWWWW",
"output": "0"
},
{
"input": "4\nBBBB",
"output": "1\n4 "
},
{
"input": "13\nWBBBBWWBWBBBW",
"output": "3\n4 1 3 "
},
{
"input": "1\nB",
... | 1,586,460,059 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 124 | 307,200 | from sys import stdin
def read():
return stdin.readline().strip()
n=int(read())
count=0
k=0
decryptList=[]
encryptedList=list((read()))
for i in range(len(encryptedList)):
if(encryptedList[i]=='B'):
count+=1
if(encryptedList[i]=='W' and count!=0):
decryptList.append(count)
count=0
k+=1
if(count>0):
decryptList.append(count)
k+=1
print(k)
print(*decryptList,sep=" ")
| Title: One-dimensional Japanese Crossword
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output Specification:
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Demo Input:
['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n']
Demo Output:
['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 ']
Note:
The last sample case correspond to the picture in the statement. | ```python
from sys import stdin
def read():
return stdin.readline().strip()
n=int(read())
count=0
k=0
decryptList=[]
encryptedList=list((read()))
for i in range(len(encryptedList)):
if(encryptedList[i]=='B'):
count+=1
if(encryptedList[i]=='W' and count!=0):
decryptList.append(count)
count=0
k+=1
if(count>0):
decryptList.append(count)
k+=1
print(k)
print(*decryptList,sep=" ")
``` | 3 | |
270 | A | Fancy Fence | PROGRAMMING | 1,100 | [
"geometry",
"implementation",
"math"
] | null | null | Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? | The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees. | For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. | [
"3\n30\n60\n90\n"
] | [
"NO\nYES\nYES\n"
] | In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | 500 | [
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,697,442,300 | 2,147,483,647 | Python 3 | OK | TESTS | 3 | 62 | 0 | # -*- coding: utf-8 -*-
"""
Created on Mon Oct 16 15:30:39 2023
@author: 刘婉婷 2300012258
"""
t=int(input())
for i in range(t):
a=int(input())
n=360//(180-a)
if a*n==(n-2)*180:
print('YES')
else:
print('NO') | Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | ```python
# -*- coding: utf-8 -*-
"""
Created on Mon Oct 16 15:30:39 2023
@author: 刘婉婷 2300012258
"""
t=int(input())
for i in range(t):
a=int(input())
n=360//(180-a)
if a*n==(n-2)*180:
print('YES')
else:
print('NO')
``` | 3 | |
80 | A | Panoramix's Prediction | PROGRAMMING | 800 | [
"brute force"
] | A. Panoramix's Prediction | 2 | 256 | A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix? | The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4. | Print YES, if *m* is the next prime number after *n*, or NO otherwise. | [
"3 5\n",
"7 11\n",
"7 9\n"
] | [
"YES",
"YES",
"NO"
] | none | 500 | [
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
... | 1,668,853,659 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 124 | 3,584,000 | n, m = map(int, input().split())
def isPrime(num):
for i in range(2, int(sqrt(num)) + 1):
if num % i == 0:
return 1
return 0
for i in range(n+1, n+50):
checkf = isPrime(i)
if checkf == 1:
if m == i:
print("YES")
break
| Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none | ```python
n, m = map(int, input().split())
def isPrime(num):
for i in range(2, int(sqrt(num)) + 1):
if num % i == 0:
return 1
return 0
for i in range(n+1, n+50):
checkf = isPrime(i)
if checkf == 1:
if m == i:
print("YES")
break
``` | -1 |
550 | B | Preparing Olympiad | PROGRAMMING | 1,400 | [
"bitmasks",
"brute force"
] | null | null | You have *n* problems. You have estimated the difficulty of the *i*-th one as integer *c**i*. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least *l* and at most *r*. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least *x*.
Find the number of ways to choose a problemset for the contest. | The first line contains four integers *n*, *l*, *r*, *x* (1<=≤<=*n*<=≤<=15, 1<=≤<=*l*<=≤<=*r*<=≤<=109, 1<=≤<=*x*<=≤<=106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=106) — the difficulty of each problem. | Print the number of ways to choose a suitable problemset for the contest. | [
"3 5 6 1\n1 2 3\n",
"4 40 50 10\n10 20 30 25\n",
"5 25 35 10\n10 10 20 10 20\n"
] | [
"2\n",
"2\n",
"6\n"
] | In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable. | 750 | [
{
"input": "3 5 6 1\n1 2 3",
"output": "2"
},
{
"input": "4 40 50 10\n10 20 30 25",
"output": "2"
},
{
"input": "5 25 35 10\n10 10 20 10 20",
"output": "6"
},
{
"input": "4 15 60 10\n10 20 30 25",
"output": "6"
},
{
"input": "1 10 20 1\n15",
"output": "0"
},... | 1,684,594,066 | 466 | PyPy 3 | OK | TESTS | 30 | 108 | 2,969,600 | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, l, r, x = map(int, input().split())
c = list(map(int, input().split()))
inf = pow(10, 9) + 1
pow2 = [1]
for _ in range(n):
pow2.append(2 * pow2[-1])
ans = 0
for i in range(1, pow2[n]):
u, v, mi, ma = 0, 0, inf, -inf
for j in range(n):
if i & pow2[j]:
u += 1
v += c[j]
mi, ma = min(mi, c[j]), max(ma, c[j])
if u >= 2 and l <= v <= r and ma - mi >= x:
ans += 1
print(ans) | Title: Preparing Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *n* problems. You have estimated the difficulty of the *i*-th one as integer *c**i*. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least *l* and at most *r*. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least *x*.
Find the number of ways to choose a problemset for the contest.
Input Specification:
The first line contains four integers *n*, *l*, *r*, *x* (1<=≤<=*n*<=≤<=15, 1<=≤<=*l*<=≤<=*r*<=≤<=109, 1<=≤<=*x*<=≤<=106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=106) — the difficulty of each problem.
Output Specification:
Print the number of ways to choose a suitable problemset for the contest.
Demo Input:
['3 5 6 1\n1 2 3\n', '4 40 50 10\n10 20 30 25\n', '5 25 35 10\n10 10 20 10 20\n']
Demo Output:
['2\n', '2\n', '6\n']
Note:
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable. | ```python
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, l, r, x = map(int, input().split())
c = list(map(int, input().split()))
inf = pow(10, 9) + 1
pow2 = [1]
for _ in range(n):
pow2.append(2 * pow2[-1])
ans = 0
for i in range(1, pow2[n]):
u, v, mi, ma = 0, 0, inf, -inf
for j in range(n):
if i & pow2[j]:
u += 1
v += c[j]
mi, ma = min(mi, c[j]), max(ma, c[j])
if u >= 2 and l <= v <= r and ma - mi >= x:
ans += 1
print(ans)
``` | 3 | |
340 | A | The Wall | PROGRAMMING | 1,200 | [
"math"
] | null | null | Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question. | The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*). | Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink. | [
"2 3 6 18\n"
] | [
"3"
] | Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | 500 | [
{
"input": "2 3 6 18",
"output": "3"
},
{
"input": "4 6 20 201",
"output": "15"
},
{
"input": "15 27 100 10000",
"output": "74"
},
{
"input": "105 60 3456 78910",
"output": "179"
},
{
"input": "1 1 1000 100000",
"output": "99001"
},
{
"input": "3 2 5 5... | 1,688,458,237 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 92 | 0 | import math
x,y,a,b=map(int,input().split())
g=math.lcm(x,y)
r=(b-a)//g
if (a%g+b%g)==0:
r+=1
print(r) | Title: The Wall
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input Specification:
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
Output Specification:
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
Demo Input:
['2 3 6 18\n']
Demo Output:
['3']
Note:
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | ```python
import math
x,y,a,b=map(int,input().split())
g=math.lcm(x,y)
r=(b-a)//g
if (a%g+b%g)==0:
r+=1
print(r)
``` | 0 | |
10 | A | Power Consumption Calculation | PROGRAMMING | 900 | [
"implementation"
] | A. Power Consumption Calculation | 1 | 256 | Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*]. | The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work. | Output the answer to the problem. | [
"1 3 2 1 5 10\n0 10\n",
"2 8 4 2 5 10\n20 30\n50 100\n"
] | [
"30",
"570"
] | none | 0 | [
{
"input": "1 3 2 1 5 10\n0 10",
"output": "30"
},
{
"input": "2 8 4 2 5 10\n20 30\n50 100",
"output": "570"
},
{
"input": "3 15 9 95 39 19\n873 989\n1003 1137\n1172 1436",
"output": "8445"
},
{
"input": "4 73 2 53 58 16\n51 52\n209 242\n281 407\n904 945",
"output": "5287... | 1,632,583,005 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 92 | 6,963,200 | n,p1,p2,p3,t1,t2 = map(int, input().split())
tp = [] # time periods
for _ in range(n):
tp.append([int(x) for x in input().split()])
res = 0
for i in range(len(tp)-1):
res += p1*(tp[i][1] - tp[i][0])
if (tp[i+1][0] - tp[i][1]) <= t1:
res += p1*t1
continue
else:
res += p1*t1
tmp = tp[i+1][0] - tp[i][1] - t1
if tmp <= t2:
res += p2*t2
continue
else:
res += p2*t2
tmp = tmp - t2
res += p3*tmp
res += p1*(tp[-1][1] - tp[-1][0])
print(res) | Title: Power Consumption Calculation
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*].
Input Specification:
The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work.
Output Specification:
Output the answer to the problem.
Demo Input:
['1 3 2 1 5 10\n0 10\n', '2 8 4 2 5 10\n20 30\n50 100\n']
Demo Output:
['30', '570']
Note:
none | ```python
n,p1,p2,p3,t1,t2 = map(int, input().split())
tp = [] # time periods
for _ in range(n):
tp.append([int(x) for x in input().split()])
res = 0
for i in range(len(tp)-1):
res += p1*(tp[i][1] - tp[i][0])
if (tp[i+1][0] - tp[i][1]) <= t1:
res += p1*t1
continue
else:
res += p1*t1
tmp = tp[i+1][0] - tp[i][1] - t1
if tmp <= t2:
res += p2*t2
continue
else:
res += p2*t2
tmp = tmp - t2
res += p3*tmp
res += p1*(tp[-1][1] - tp[-1][0])
print(res)
``` | 0 |
38 | B | Chess | PROGRAMMING | 1,200 | [
"brute force",
"implementation",
"math"
] | B. Chess | 2 | 256 | Two chess pieces, a rook and a knight, stand on a standard chessboard 8<=×<=8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square. | The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide. | Print a single number which is the required number of ways. | [
"a1\nb2\n",
"a8\nd4\n"
] | [
"44\n",
"38\n"
] | none | 0 | [
{
"input": "a1\nb2",
"output": "44"
},
{
"input": "a8\nd4",
"output": "38"
},
{
"input": "a8\nf1",
"output": "42"
},
{
"input": "f8\nh3",
"output": "42"
},
{
"input": "g8\nb7",
"output": "42"
},
{
"input": "h1\ng5",
"output": "42"
},
{
"inp... | 1,613,134,207 | 2,147,483,647 | PyPy 3 | OK | TESTS | 40 | 280 | 2,457,600 | import string
def solve(x1, y1, x2, y2):
c = [[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0]]
X = [2, 1, -1, -2, -2, -1, 1, 2]
Y = [1, 2, 2, 1, -1, -2, -2, -1]
for i in range(8):
for j in range(8):
if i != x1 and j != y1:
f = True
for k in range(8):
if i == x2 + X[k] and j == y2 + Y[k]:
f = False
for k in range(8):
if i == x1 + X[k] and j == y1 + Y[k]:
f = False
if f:
c[i][j] = 1
c[x2][y2] = 0
# for i in range(8):
# print(c[i])
r = 0
for i in range(8):
for j in c[i]:
if j == 1:
r = r + 1
return r
for _ in range(1):
a = input()
b = input()
y1 = a[0]
x1 = int(a[1]) - 1
y1 = int(string.ascii_lowercase.index(y1))
y2 = b[0]
x2 = int(b[1]) - 1
y2 = int(string.ascii_lowercase.index(y2))
# print(x1, y1, x2, y2)
print(solve(x1, y1, x2, y2))
| Title: Chess
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Two chess pieces, a rook and a knight, stand on a standard chessboard 8<=×<=8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input Specification:
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output Specification:
Print a single number which is the required number of ways.
Demo Input:
['a1\nb2\n', 'a8\nd4\n']
Demo Output:
['44\n', '38\n']
Note:
none | ```python
import string
def solve(x1, y1, x2, y2):
c = [[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0]]
X = [2, 1, -1, -2, -2, -1, 1, 2]
Y = [1, 2, 2, 1, -1, -2, -2, -1]
for i in range(8):
for j in range(8):
if i != x1 and j != y1:
f = True
for k in range(8):
if i == x2 + X[k] and j == y2 + Y[k]:
f = False
for k in range(8):
if i == x1 + X[k] and j == y1 + Y[k]:
f = False
if f:
c[i][j] = 1
c[x2][y2] = 0
# for i in range(8):
# print(c[i])
r = 0
for i in range(8):
for j in c[i]:
if j == 1:
r = r + 1
return r
for _ in range(1):
a = input()
b = input()
y1 = a[0]
x1 = int(a[1]) - 1
y1 = int(string.ascii_lowercase.index(y1))
y2 = b[0]
x2 = int(b[1]) - 1
y2 = int(string.ascii_lowercase.index(y2))
# print(x1, y1, x2, y2)
print(solve(x1, y1, x2, y2))
``` | 3.925422 |
450 | A | Jzzhu and Children | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order? | The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). | Output a single integer, representing the number of the last child. | [
"5 2\n1 3 1 4 2\n",
"6 4\n1 1 2 2 3 3\n"
] | [
"4\n",
"6\n"
] | Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | 500 | [
{
"input": "5 2\n1 3 1 4 2",
"output": "4"
},
{
"input": "6 4\n1 1 2 2 3 3",
"output": "6"
},
{
"input": "7 3\n6 1 5 4 2 3 1",
"output": "4"
},
{
"input": "10 5\n2 7 3 6 2 5 1 3 4 5",
"output": "4"
},
{
"input": "100 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18... | 1,691,769,338 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 49 | 62 | 1,945,600 | s = []
def roll(a):
global s
if all(i <= 0 for i in a):
return s[len(s)-1]+1
else:
for i in range(len(a)):
if a[i] > m:
a[i] = a[i] - m
elif a[i] <= 0:
continue
else:
a[i] = a[i] - m
s.append(i)
return roll(a)
n, m = map(int, input().split(" "))
a = list(map(int, input().split(" ")))
print(roll(a))
| Title: Jzzhu and Children
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output a single integer, representing the number of the last child.
Demo Input:
['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n']
Demo Output:
['4\n', '6\n']
Note:
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | ```python
s = []
def roll(a):
global s
if all(i <= 0 for i in a):
return s[len(s)-1]+1
else:
for i in range(len(a)):
if a[i] > m:
a[i] = a[i] - m
elif a[i] <= 0:
continue
else:
a[i] = a[i] - m
s.append(i)
return roll(a)
n, m = map(int, input().split(" "))
a = list(map(int, input().split(" ")))
print(roll(a))
``` | 3 | |
888 | C | K-Dominant Character | PROGRAMMING | 1,400 | [
"binary search",
"implementation",
"two pointers"
] | null | null | You are given a string *s* consisting of lowercase Latin letters. Character *c* is called *k*-dominant iff each substring of *s* with length at least *k* contains this character *c*.
You have to find minimum *k* such that there exists at least one *k*-dominant character. | The first line contains string *s* consisting of lowercase Latin letters (1<=≤<=|*s*|<=≤<=100000). | Print one number — the minimum value of *k* such that there exists at least one *k*-dominant character. | [
"abacaba\n",
"zzzzz\n",
"abcde\n"
] | [
"2\n",
"1\n",
"3\n"
] | none | 0 | [
{
"input": "abacaba",
"output": "2"
},
{
"input": "zzzzz",
"output": "1"
},
{
"input": "abcde",
"output": "3"
},
{
"input": "bcaccacaaabaacaabaaabcbbcbcaacacbcbaaaacccacbbcbaabcbacaacbabacacacaccbbccbcbacbbbbccccabcabaaab",
"output": "8"
},
{
"input": "daabcdabbab... | 1,637,171,590 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 2,000 | 307,200 | s = input()
minn = 10 ** 18
for i in s:
tmp = s.split(i)
maxx = max(map(len, tmp))
minn = min(minn, maxx)
print(minn + 1)
# Wed Nov 17 2021 17:53:11 GMT+0000 (Coordinated Universal Time)
| Title: K-Dominant Character
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* consisting of lowercase Latin letters. Character *c* is called *k*-dominant iff each substring of *s* with length at least *k* contains this character *c*.
You have to find minimum *k* such that there exists at least one *k*-dominant character.
Input Specification:
The first line contains string *s* consisting of lowercase Latin letters (1<=≤<=|*s*|<=≤<=100000).
Output Specification:
Print one number — the minimum value of *k* such that there exists at least one *k*-dominant character.
Demo Input:
['abacaba\n', 'zzzzz\n', 'abcde\n']
Demo Output:
['2\n', '1\n', '3\n']
Note:
none | ```python
s = input()
minn = 10 ** 18
for i in s:
tmp = s.split(i)
maxx = max(map(len, tmp))
minn = min(minn, maxx)
print(minn + 1)
# Wed Nov 17 2021 17:53:11 GMT+0000 (Coordinated Universal Time)
``` | 0 | |
921 | 02 | Labyrinth-2 | PROGRAMMING | 3,200 | [] | null | null | See the problem statement here: [http://codeforces.com/contest/921/problem/01](//codeforces.com/contest/921/problem/01). | none | none | [] | [] | none | 0 | [] | 1,517,511,726 | 2,147,483,647 | Python 3 | PARTIAL | TESTS | 0 | 8,000 | 220,569,600 | def bfs(walls, doors, keys, exits, x, y, a, b, n, m):
dist = [[1000000000] * m for i in range(n)]
dist[a][b] = 0
queue = [(a, b)]
used = set()
dx = [1, 0, 0, -1]
dy = [0, 1, -1, 0]
while queue:
(cur_x, cur_y) = queue.pop(0)
if (cur_x, cur_y) not in used:
used.add((cur_x, cur_y))
for i in range(4):
new_x, new_y = cur_x + dx[i], cur_y + dy[i]
if 0 <= new_x < n and 0 <= new_y < m:
if (new_x, new_y) not in used:
if (cur_x, cur_y, new_x, new_y) not in walls:
if (new_x, new_y, cur_x, cur_y) not in walls:
if (cur_x, cur_y, new_x, new_y) not in doors:
if (new_x, new_y, cur_x, cur_y) not in doors:
queue.append((new_x, new_y))
dist[new_x][new_y] = dist[cur_x][cur_y] + 1
return dist
def restore(dist, x, y, n, m):
if dist[x][y] == 0:
return ""
dx = [1, 0, 0, -1]
dy = [0, 1, -1, 0]
prt = ["move-down\n", "move-right\n", "move-left\n", "move-up\n"]
for i in range(4):
new_x, new_y = x + dx[i], y + dy[i]
if 0 <= new_x < n and 0 <= new_y < m:
if dist[x][y] - 1 == dist[new_x][new_y]:
return prt[i] + restore(dist, new_x, new_y, n, m)
(g, w, n, m, x, y, c, d, k, e) = map(int, input().split())
walls = set()
for i in range(c):
(x1, y1, x2, y2) = map(int, input().split())
walls.add((x1, y1, x2, y2))
doors = set()
for i in range(d):
(x1, y1, x2, y2) = map(int, input().split())
doors.add((x1, y1, x2, y2))
keys = set()
for i in range(k):
(x1, y1) = map(int, input().split())
keys.add((x1, y1))
exits = set()
for i in range(e):
(x1, y1) = map(int, input().split())
exits.add((x1, y1))
best_len = 141928123080123
best = ""
for (a, b) in exits:
dist = bfs(walls, doors, keys, exits, x, y, a, b, n, m)
ans = restore(dist, x, y, n, m)
if len(ans) < best_len:
best = ans
best_len = len(ans)
print(best) | Title: Labyrinth-2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
See the problem statement here: [http://codeforces.com/contest/921/problem/01](//codeforces.com/contest/921/problem/01).
Input Specification:
none
Output Specification:
none
Note:
none | ```python
def bfs(walls, doors, keys, exits, x, y, a, b, n, m):
dist = [[1000000000] * m for i in range(n)]
dist[a][b] = 0
queue = [(a, b)]
used = set()
dx = [1, 0, 0, -1]
dy = [0, 1, -1, 0]
while queue:
(cur_x, cur_y) = queue.pop(0)
if (cur_x, cur_y) not in used:
used.add((cur_x, cur_y))
for i in range(4):
new_x, new_y = cur_x + dx[i], cur_y + dy[i]
if 0 <= new_x < n and 0 <= new_y < m:
if (new_x, new_y) not in used:
if (cur_x, cur_y, new_x, new_y) not in walls:
if (new_x, new_y, cur_x, cur_y) not in walls:
if (cur_x, cur_y, new_x, new_y) not in doors:
if (new_x, new_y, cur_x, cur_y) not in doors:
queue.append((new_x, new_y))
dist[new_x][new_y] = dist[cur_x][cur_y] + 1
return dist
def restore(dist, x, y, n, m):
if dist[x][y] == 0:
return ""
dx = [1, 0, 0, -1]
dy = [0, 1, -1, 0]
prt = ["move-down\n", "move-right\n", "move-left\n", "move-up\n"]
for i in range(4):
new_x, new_y = x + dx[i], y + dy[i]
if 0 <= new_x < n and 0 <= new_y < m:
if dist[x][y] - 1 == dist[new_x][new_y]:
return prt[i] + restore(dist, new_x, new_y, n, m)
(g, w, n, m, x, y, c, d, k, e) = map(int, input().split())
walls = set()
for i in range(c):
(x1, y1, x2, y2) = map(int, input().split())
walls.add((x1, y1, x2, y2))
doors = set()
for i in range(d):
(x1, y1, x2, y2) = map(int, input().split())
doors.add((x1, y1, x2, y2))
keys = set()
for i in range(k):
(x1, y1) = map(int, input().split())
keys.add((x1, y1))
exits = set()
for i in range(e):
(x1, y1) = map(int, input().split())
exits.add((x1, y1))
best_len = 141928123080123
best = ""
for (a, b) in exits:
dist = bfs(walls, doors, keys, exits, x, y, a, b, n, m)
ans = restore(dist, x, y, n, m)
if len(ans) < best_len:
best = ans
best_len = len(ans)
print(best)
``` | 2 | |
217 | A | Ice Skating | PROGRAMMING | 1,200 | [
"brute force",
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates. | The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct. | Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one. | [
"2\n2 1\n1 2\n",
"2\n2 1\n4 1\n"
] | [
"1\n",
"0\n"
] | none | 500 | [
{
"input": "2\n2 1\n1 2",
"output": "1"
},
{
"input": "2\n2 1\n4 1",
"output": "0"
},
{
"input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182... | 1,662,095,471 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 41 | 92 | 102,400 | x,y,n=[],[],int(input())
from collections import Counter as cnt
for pratyush in range(n):
a,b=list(map(int,input().split()))
x.append(a)
y.append(b)
c,f,g=cnt(x),0,0
d=cnt(y)
for i in c:
f+=c[i]-1
if c[i]>1:
g+=c[i]
for i in d:
f+=d[i]-1
if d[i]>1:
g+=d[i]
print(n-f-1) | Title: Ice Skating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output Specification:
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
Demo Input:
['2\n2 1\n1 2\n', '2\n2 1\n4 1\n']
Demo Output:
['1\n', '0\n']
Note:
none | ```python
x,y,n=[],[],int(input())
from collections import Counter as cnt
for pratyush in range(n):
a,b=list(map(int,input().split()))
x.append(a)
y.append(b)
c,f,g=cnt(x),0,0
d=cnt(y)
for i in c:
f+=c[i]-1
if c[i]>1:
g+=c[i]
for i in d:
f+=d[i]-1
if d[i]>1:
g+=d[i]
print(n-f-1)
``` | 0 | |
383 | A | Milking cows | PROGRAMMING | 1,600 | [
"data structures",
"greedy"
] | null | null | Iahub helps his grandfather at the farm. Today he must milk the cows. There are *n* cows sitting in a row, numbered from 1 to *n* from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).
Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=200000). The second line contains *n* integers *a*1, *a*2, ..., *a**n*, where *a**i* is 0 if the cow number *i* is facing left, and 1 if it is facing right. | Print a single integer, the minimum amount of lost milk.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"4\n0 0 1 0\n",
"5\n1 0 1 0 1\n"
] | [
"1",
"3"
] | In the first sample Iahub milks the cows in the following order: cow 3, cow 4, cow 2, cow 1. When he milks cow 3, cow 4 loses 1 unit of milk. After that, no more milk is lost. | 500 | [
{
"input": "4\n0 0 1 0",
"output": "1"
},
{
"input": "5\n1 0 1 0 1",
"output": "3"
},
{
"input": "50\n1 1 0 1 1 1 1 1 1 0 0 1 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0 1 1 0 1 0 0 1 0",
"output": "416"
},
{
"input": "100\n1 1 0 0 1 1 1 1 0 1 1 1 1 1 1 1 0 0 ... | 1,653,932,550 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 30 | 0 | n = int(input())
a = [int(x) for x in input().split()]
s = 0
res = 0
for i in a:
if i == 0:
res += s
else:
s += 1
print(s) | Title: Milking cows
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub helps his grandfather at the farm. Today he must milk the cows. There are *n* cows sitting in a row, numbered from 1 to *n* from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).
Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=200000). The second line contains *n* integers *a*1, *a*2, ..., *a**n*, where *a**i* is 0 if the cow number *i* is facing left, and 1 if it is facing right.
Output Specification:
Print a single integer, the minimum amount of lost milk.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['4\n0 0 1 0\n', '5\n1 0 1 0 1\n']
Demo Output:
['1', '3']
Note:
In the first sample Iahub milks the cows in the following order: cow 3, cow 4, cow 2, cow 1. When he milks cow 3, cow 4 loses 1 unit of milk. After that, no more milk is lost. | ```python
n = int(input())
a = [int(x) for x in input().split()]
s = 0
res = 0
for i in a:
if i == 0:
res += s
else:
s += 1
print(s)
``` | 0 | |
967 | A | Mind the Gap | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts $1$ minute.
He was asked to insert one takeoff in the schedule. The takeoff takes $1$ minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least $s$ minutes from both sides.
Find the earliest time when Arkady can insert the takeoff. | The first line of input contains two integers $n$ and $s$ ($1 \le n \le 100$, $1 \le s \le 60$) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next $n$ lines contains two integers $h$ and $m$ ($0 \le h \le 23$, $0 \le m \le 59$) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is $0$ $0$). These times are given in increasing order. | Print two integers $h$ and $m$ — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff. | [
"6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40\n",
"16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59\n",
"3 17\n0 30\n1 0\n12 0\n"
] | [
"6 1\n",
"24 50\n",
"0 0\n"
] | In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than $24$ hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing. | 500 | [
{
"input": "6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40",
"output": "6 1"
},
{
"input": "16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59",
"output": "24 50"
},
{
"input": "3 17\n0 30\n1 0\n12 0",
"output": "0 0"
},
{
... | 1,525,304,036 | 1,736 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 124 | 21,401,600 | n, s = [int(i) for i in input().split()]
T = [[int(i) for i in input().split()] for _ in range(n)]
h, m = T[0]
if h*60 + m >= 1 + s + 1:
print(0, 0)
else:
for i in range(1, n):
(h1, m1), (h2, m2) = T[i-1], T[i]
d = (h2-h1-(m2<m1))*60 + (m2-m1)%60
if d >= 1 + s + 1 + s:
h_ = h1+((m1+1+s)>60)
m_ = (m1+1+s) % 60
print(h_, m_)
break
else:
h, m = T[n-1]
h_ = h+((m+1+s)>60)
m_ = (m+1+s) % 60
print(h_, m_)
| Title: Mind the Gap
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts $1$ minute.
He was asked to insert one takeoff in the schedule. The takeoff takes $1$ minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least $s$ minutes from both sides.
Find the earliest time when Arkady can insert the takeoff.
Input Specification:
The first line of input contains two integers $n$ and $s$ ($1 \le n \le 100$, $1 \le s \le 60$) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next $n$ lines contains two integers $h$ and $m$ ($0 \le h \le 23$, $0 \le m \le 59$) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is $0$ $0$). These times are given in increasing order.
Output Specification:
Print two integers $h$ and $m$ — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff.
Demo Input:
['6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40\n', '16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59\n', '3 17\n0 30\n1 0\n12 0\n']
Demo Output:
['6 1\n', '24 50\n', '0 0\n']
Note:
In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than $24$ hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing. | ```python
n, s = [int(i) for i in input().split()]
T = [[int(i) for i in input().split()] for _ in range(n)]
h, m = T[0]
if h*60 + m >= 1 + s + 1:
print(0, 0)
else:
for i in range(1, n):
(h1, m1), (h2, m2) = T[i-1], T[i]
d = (h2-h1-(m2<m1))*60 + (m2-m1)%60
if d >= 1 + s + 1 + s:
h_ = h1+((m1+1+s)>60)
m_ = (m1+1+s) % 60
print(h_, m_)
break
else:
h, m = T[n-1]
h_ = h+((m+1+s)>60)
m_ = (m+1+s) % 60
print(h_, m_)
``` | 0 | |
158 | A | Next Round | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | "Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round. | The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1). | Output the number of participants who advance to the next round. | [
"8 5\n10 9 8 7 7 7 5 5\n",
"4 2\n0 0 0 0\n"
] | [
"6\n",
"0\n"
] | In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score. | 500 | [
{
"input": "8 5\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "4 2\n0 0 0 0",
"output": "0"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "1 1\n10",
"output": "1"
},
{
"input": "17 14\n16 15... | 1,699,784,108 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 62 | 0 | n, k = map(int, input().split())
scores = list(map(int, input().split()))
count = 0
for score in scores:
if scores[0]==0:
count = 0
elif score >= scores[k-1]:
count += 1
print(count) | Title: Next Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
Output Specification:
Output the number of participants who advance to the next round.
Demo Input:
['8 5\n10 9 8 7 7 7 5 5\n', '4 2\n0 0 0 0\n']
Demo Output:
['6\n', '0\n']
Note:
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score. | ```python
n, k = map(int, input().split())
scores = list(map(int, input().split()))
count = 0
for score in scores:
if scores[0]==0:
count = 0
elif score >= scores[k-1]:
count += 1
print(count)
``` | 0 | |
569 | A | Music | PROGRAMMING | 1,500 | [
"implementation",
"math"
] | null | null | Little Lesha loves listening to music via his smartphone. But the smartphone doesn't have much memory, so Lesha listens to his favorite songs in a well-known social network InTalk.
Unfortunately, internet is not that fast in the city of Ekaterinozavodsk and the song takes a lot of time to download. But Lesha is quite impatient. The song's duration is *T* seconds. Lesha downloads the first *S* seconds of the song and plays it. When the playback reaches the point that has not yet been downloaded, Lesha immediately plays the song from the start (the loaded part of the song stays in his phone, and the download is continued from the same place), and it happens until the song is downloaded completely and Lesha listens to it to the end. For *q* seconds of real time the Internet allows you to download *q*<=-<=1 seconds of the track.
Tell Lesha, for how many times he will start the song, including the very first start. | The single line contains three integers *T*,<=*S*,<=*q* (2<=≤<=*q*<=≤<=104, 1<=≤<=*S*<=<<=*T*<=≤<=105). | Print a single integer — the number of times the song will be restarted. | [
"5 2 2\n",
"5 4 7\n",
"6 2 3\n"
] | [
"2\n",
"1\n",
"1\n"
] | In the first test, the song is played twice faster than it is downloaded, which means that during four first seconds Lesha reaches the moment that has not been downloaded, and starts the song again. After another two seconds, the song is downloaded completely, and thus, Lesha starts the song twice.
In the second test, the song is almost downloaded, and Lesha will start it only once.
In the third sample test the download finishes and Lesha finishes listening at the same moment. Note that song isn't restarted in this case. | 500 | [
{
"input": "5 2 2",
"output": "2"
},
{
"input": "5 4 7",
"output": "1"
},
{
"input": "6 2 3",
"output": "1"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "2 1 3",
"output": "1"
},
{
"input": "2 1 10000",
"output": "1"
},
{
"input": "12... | 1,439,224,827 | 627 | Python 3 | WRONG_ANSWER | PRETESTS | 2 | 30 | 0 | t, s, q = map(int, input().split())
ans = 0
downloaded = s
listened = 0
while downloaded <= t:
listened += q
if listened >= downloaded:
ans += 1
listened -= downloaded
downloaded += q - 1
print(ans)
| Title: Music
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Lesha loves listening to music via his smartphone. But the smartphone doesn't have much memory, so Lesha listens to his favorite songs in a well-known social network InTalk.
Unfortunately, internet is not that fast in the city of Ekaterinozavodsk and the song takes a lot of time to download. But Lesha is quite impatient. The song's duration is *T* seconds. Lesha downloads the first *S* seconds of the song and plays it. When the playback reaches the point that has not yet been downloaded, Lesha immediately plays the song from the start (the loaded part of the song stays in his phone, and the download is continued from the same place), and it happens until the song is downloaded completely and Lesha listens to it to the end. For *q* seconds of real time the Internet allows you to download *q*<=-<=1 seconds of the track.
Tell Lesha, for how many times he will start the song, including the very first start.
Input Specification:
The single line contains three integers *T*,<=*S*,<=*q* (2<=≤<=*q*<=≤<=104, 1<=≤<=*S*<=<<=*T*<=≤<=105).
Output Specification:
Print a single integer — the number of times the song will be restarted.
Demo Input:
['5 2 2\n', '5 4 7\n', '6 2 3\n']
Demo Output:
['2\n', '1\n', '1\n']
Note:
In the first test, the song is played twice faster than it is downloaded, which means that during four first seconds Lesha reaches the moment that has not been downloaded, and starts the song again. After another two seconds, the song is downloaded completely, and thus, Lesha starts the song twice.
In the second test, the song is almost downloaded, and Lesha will start it only once.
In the third sample test the download finishes and Lesha finishes listening at the same moment. Note that song isn't restarted in this case. | ```python
t, s, q = map(int, input().split())
ans = 0
downloaded = s
listened = 0
while downloaded <= t:
listened += q
if listened >= downloaded:
ans += 1
listened -= downloaded
downloaded += q - 1
print(ans)
``` | 0 | |
835 | C | Star sky | PROGRAMMING | 1,600 | [
"dp",
"implementation"
] | null | null | The Cartesian coordinate system is set in the sky. There you can see *n* stars, the *i*-th has coordinates (*x**i*, *y**i*), a maximum brightness *c*, equal for all stars, and an initial brightness *s**i* (0<=≤<=*s**i*<=≤<=*c*).
Over time the stars twinkle. At moment 0 the *i*-th star has brightness *s**i*. Let at moment *t* some star has brightness *x*. Then at moment (*t*<=+<=1) this star will have brightness *x*<=+<=1, if *x*<=+<=1<=≤<=*c*, and 0, otherwise.
You want to look at the sky *q* times. In the *i*-th time you will look at the moment *t**i* and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (*x*1*i*, *y*1*i*) and the upper right — (*x*2*i*, *y*2*i*). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.
A star lies in a rectangle if it lies on its border or lies strictly inside it. | The first line contains three integers *n*, *q*, *c* (1<=≤<=*n*,<=*q*<=≤<=105, 1<=≤<=*c*<=≤<=10) — the number of the stars, the number of the views and the maximum brightness of the stars.
The next *n* lines contain the stars description. The *i*-th from these lines contains three integers *x**i*, *y**i*, *s**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=100, 0<=≤<=*s**i*<=≤<=*c*<=≤<=10) — the coordinates of *i*-th star and its initial brightness.
The next *q* lines contain the views description. The *i*-th from these lines contains five integers *t**i*, *x*1*i*, *y*1*i*, *x*2*i*, *y*2*i* (0<=≤<=*t**i*<=≤<=109, 1<=≤<=*x*1*i*<=<<=*x*2*i*<=≤<=100, 1<=≤<=*y*1*i*<=<<=*y*2*i*<=≤<=100) — the moment of the *i*-th view and the coordinates of the viewed rectangle. | For each view print the total brightness of the viewed stars. | [
"2 3 3\n1 1 1\n3 2 0\n2 1 1 2 2\n0 2 1 4 5\n5 1 1 5 5\n",
"3 4 5\n1 1 2\n2 3 0\n3 3 1\n0 1 1 100 100\n1 2 2 4 4\n2 2 1 4 7\n1 50 50 51 51\n"
] | [
"3\n0\n3\n",
"3\n3\n5\n0\n"
] | Let's consider the first example.
At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.
At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.
At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3. | 1,250 | [
{
"input": "2 3 3\n1 1 1\n3 2 0\n2 1 1 2 2\n0 2 1 4 5\n5 1 1 5 5",
"output": "3\n0\n3"
},
{
"input": "3 4 5\n1 1 2\n2 3 0\n3 3 1\n0 1 1 100 100\n1 2 2 4 4\n2 2 1 4 7\n1 50 50 51 51",
"output": "3\n3\n5\n0"
}
] | 1,700,384,332 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 2,000 | 409,600 | def fun(t, x1, y1, x2, y2, stars, c):
sum=0
for i in stars:
if x1<=i[0]<=x2 and y1<=i[1]<=y2:
if i[2]>0:
if t>(c-i[2]):
alpha = t-(c-i[2]+1)
sum+=(alpha%(c+1))
# print(1, t%(c+1))
else:
sum+=(i[2]+t)
# print(2, i[2]+t)
else:
sum+=(t%(c+1))
# print(3, t, c+1)
return sum
n,q,c = map(lambda x: int(x), input().split())
stars = []
for _ in range(n):
a = list(map(lambda x: int(x), input().split()))
stars.append(a)
for _ in range(q):
t, x1, y1, x2, y2 = map(lambda x: int(x), input().split())
print(fun(t, x1, y1, x2, y2, stars, c)) | Title: Star sky
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Cartesian coordinate system is set in the sky. There you can see *n* stars, the *i*-th has coordinates (*x**i*, *y**i*), a maximum brightness *c*, equal for all stars, and an initial brightness *s**i* (0<=≤<=*s**i*<=≤<=*c*).
Over time the stars twinkle. At moment 0 the *i*-th star has brightness *s**i*. Let at moment *t* some star has brightness *x*. Then at moment (*t*<=+<=1) this star will have brightness *x*<=+<=1, if *x*<=+<=1<=≤<=*c*, and 0, otherwise.
You want to look at the sky *q* times. In the *i*-th time you will look at the moment *t**i* and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (*x*1*i*, *y*1*i*) and the upper right — (*x*2*i*, *y*2*i*). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.
A star lies in a rectangle if it lies on its border or lies strictly inside it.
Input Specification:
The first line contains three integers *n*, *q*, *c* (1<=≤<=*n*,<=*q*<=≤<=105, 1<=≤<=*c*<=≤<=10) — the number of the stars, the number of the views and the maximum brightness of the stars.
The next *n* lines contain the stars description. The *i*-th from these lines contains three integers *x**i*, *y**i*, *s**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=100, 0<=≤<=*s**i*<=≤<=*c*<=≤<=10) — the coordinates of *i*-th star and its initial brightness.
The next *q* lines contain the views description. The *i*-th from these lines contains five integers *t**i*, *x*1*i*, *y*1*i*, *x*2*i*, *y*2*i* (0<=≤<=*t**i*<=≤<=109, 1<=≤<=*x*1*i*<=<<=*x*2*i*<=≤<=100, 1<=≤<=*y*1*i*<=<<=*y*2*i*<=≤<=100) — the moment of the *i*-th view and the coordinates of the viewed rectangle.
Output Specification:
For each view print the total brightness of the viewed stars.
Demo Input:
['2 3 3\n1 1 1\n3 2 0\n2 1 1 2 2\n0 2 1 4 5\n5 1 1 5 5\n', '3 4 5\n1 1 2\n2 3 0\n3 3 1\n0 1 1 100 100\n1 2 2 4 4\n2 2 1 4 7\n1 50 50 51 51\n']
Demo Output:
['3\n0\n3\n', '3\n3\n5\n0\n']
Note:
Let's consider the first example.
At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.
At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.
At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3. | ```python
def fun(t, x1, y1, x2, y2, stars, c):
sum=0
for i in stars:
if x1<=i[0]<=x2 and y1<=i[1]<=y2:
if i[2]>0:
if t>(c-i[2]):
alpha = t-(c-i[2]+1)
sum+=(alpha%(c+1))
# print(1, t%(c+1))
else:
sum+=(i[2]+t)
# print(2, i[2]+t)
else:
sum+=(t%(c+1))
# print(3, t, c+1)
return sum
n,q,c = map(lambda x: int(x), input().split())
stars = []
for _ in range(n):
a = list(map(lambda x: int(x), input().split()))
stars.append(a)
for _ in range(q):
t, x1, y1, x2, y2 = map(lambda x: int(x), input().split())
print(fun(t, x1, y1, x2, y2, stars, c))
``` | 0 | |
248 | A | Cupboards | PROGRAMMING | 800 | [
"implementation"
] | null | null | One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*. | The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces. | In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs. | [
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 ... | 1,658,590,820 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 30 | 0 | c0=0
c1=0
for _ in range(int(input())):
l,r=map(int,input().split())
if l==0:
c0+=1
elif l==1:
c1+=1
print(max(c0,c1)) | Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none | ```python
c0=0
c1=0
for _ in range(int(input())):
l,r=map(int,input().split())
if l==0:
c0+=1
elif l==1:
c1+=1
print(max(c0,c1))
``` | 0 | |
452 | A | Eevee | PROGRAMMING | 1,000 | [
"brute force",
"implementation",
"strings"
] | null | null | You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon.
You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it. | First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string.
Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword). | Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter). | [
"7\nj......\n",
"7\n...feon\n",
"7\n.l.r.o.\n"
] | [
"jolteon\n",
"leafeon\n",
"flareon\n"
] | Here's a set of names in a form you can paste into your solution:
["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]
{"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"} | 500 | [
{
"input": "7\n...feon",
"output": "leafeon"
},
{
"input": "7\n.l.r.o.",
"output": "flareon"
},
{
"input": "6\n.s..o.",
"output": "espeon"
},
{
"input": "7\nglaceon",
"output": "glaceon"
},
{
"input": "8\n.a.o.e.n",
"output": "vaporeon"
},
{
"input": "... | 1,406,989,099 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 31 | 0 | input() #not needed.
names = {
"vaporeon",
"jolteon",
"flareon",
"espeon",
"umbreon",
"leafeon",
"glaceon",
"sylveon"
}
text = input().split('.')
text = {item for item in text if item}
for name in names:
if text < set(name): print(name) | Title: Eevee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon.
You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it.
Input Specification:
First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string.
Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword).
Output Specification:
Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter).
Demo Input:
['7\nj......\n', '7\n...feon\n', '7\n.l.r.o.\n']
Demo Output:
['jolteon\n', 'leafeon\n', 'flareon\n']
Note:
Here's a set of names in a form you can paste into your solution:
["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]
{"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"} | ```python
input() #not needed.
names = {
"vaporeon",
"jolteon",
"flareon",
"espeon",
"umbreon",
"leafeon",
"glaceon",
"sylveon"
}
text = input().split('.')
text = {item for item in text if item}
for name in names:
if text < set(name): print(name)
``` | 0 | |
116 | A | Tram | PROGRAMMING | 800 | [
"implementation"
] | null | null | Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram. | The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0. | Print a single integer denoting the minimum possible capacity of the tram (0 is allowed). | [
"4\n0 3\n2 5\n4 2\n4 0\n"
] | [
"6\n"
] | For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | 500 | [
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1... | 1,688,889,074 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 71 | 218 | 1,945,600 | n = int(input())
summ=0
count=0
for i in range(n):
s = input().split()
summ-=int(s[0])
summ+=int(s[1])
if summ>=count:
count=summ
print(count) | Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | ```python
n = int(input())
summ=0
count=0
for i in range(n):
s = input().split()
summ-=int(s[0])
summ+=int(s[1])
if summ>=count:
count=summ
print(count)
``` | 3 | |
580 | A | Kefa and First Steps | PROGRAMMING | 900 | [
"brute force",
"dp",
"implementation"
] | null | null | Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.
Help Kefa cope with this task! | The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109). | Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*. | [
"6\n2 2 1 3 4 1\n",
"3\n2 2 9\n"
] | [
"3",
"3"
] | In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.
In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one. | 750 | [
{
"input": "6\n2 2 1 3 4 1",
"output": "3"
},
{
"input": "3\n2 2 9",
"output": "3"
},
{
"input": "5\n10 100 111 1 2",
"output": "3"
},
{
"input": "10\n1 2 3 4 1 2 3 4 5 6",
"output": "6"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ... | 1,696,630,974 | 2,147,483,647 | Python 3 | OK | TESTS | 28 | 109 | 13,926,400 | n = int(input())
a = list(map(int, input().split()))
dp = [0] * n
dp[0] = 1
for i in range(1, n):
if a[i] >= a[i - 1]:
dp[i] = dp[ i -1] + 1
else:
dp[i] = 1
print(max(dp)) | Title: Kefa and First Steps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.
Help Kefa cope with this task!
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*.
Demo Input:
['6\n2 2 1 3 4 1\n', '3\n2 2 9\n']
Demo Output:
['3', '3']
Note:
In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.
In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one. | ```python
n = int(input())
a = list(map(int, input().split()))
dp = [0] * n
dp[0] = 1
for i in range(1, n):
if a[i] >= a[i - 1]:
dp[i] = dp[ i -1] + 1
else:
dp[i] = 1
print(max(dp))
``` | 3 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,656,237,355 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 4,505,600 | p=input()
a=0
b=0
c=0
for i in range(p):
arr=map(int,raw_input().split(" "))
a=a+arr[0]
b=b+arr[1]
c=c+arr[2]
if(a==0 and b==0 and c==0):
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
p=input()
a=0
b=0
c=0
for i in range(p):
arr=map(int,raw_input().split(" "))
a=a+arr[0]
b=b+arr[1]
c=c+arr[2]
if(a==0 and b==0 and c==0):
print("YES")
else:
print("NO")
``` | -1 |
794 | B | Cutting Carrot | PROGRAMMING | 1,200 | [
"geometry",
"math"
] | null | null | Igor the analyst has adopted *n* little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into *n* pieces of equal area.
Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to *h*. Igor wants to make *n*<=-<=1 cuts parallel to the base to cut the carrot into *n* pieces. He wants to make sure that all *n* pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area? | The first and only line of input contains two space-separated integers, *n* and *h* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=105). | The output should contain *n*<=-<=1 real numbers *x*1,<=*x*2,<=...,<=*x**n*<=-<=1. The number *x**i* denotes that the *i*-th cut must be made *x**i* units away from the apex of the carrot. In addition, 0<=<<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=-<=1<=<<=*h* must hold.
Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10<=-<=6.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if . | [
"3 2\n",
"2 100000\n"
] | [
"1.154700538379 1.632993161855\n",
"70710.678118654752\n"
] | Definition of isosceles triangle: [https://en.wikipedia.org/wiki/Isosceles_triangle](https://en.wikipedia.org/wiki/Isosceles_triangle). | 1,000 | [
{
"input": "3 2",
"output": "1.154700538379 1.632993161855"
},
{
"input": "2 100000",
"output": "70710.678118654752"
},
{
"input": "1000 100000",
"output": "3162.277660168379 4472.135954999579 5477.225575051661 6324.555320336759 7071.067811865475 7745.966692414834 8366.600265340755 8... | 1,494,671,435 | 3,335 | Python 3 | COMPILATION_ERROR | PRETESTS | 0 | 0 | 0 | #include<bits/stdc++.h>
using namespace std;
int main()
{
//ios::sync_with_stdio(false);
//freopen("out.txt", "w", stdout);
double n, h;
cin >> n >> h;
double s = h / 2;
double ans = 0;
for(int i=1; i<n; ++ i)
{
double res = sqrt(i / n) * h;
if(i == 1)
printf("%.9f", res);
else
printf(" %.9f", res);
}
puts("\n");
return 0;
}
| Title: Cutting Carrot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor the analyst has adopted *n* little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into *n* pieces of equal area.
Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to *h*. Igor wants to make *n*<=-<=1 cuts parallel to the base to cut the carrot into *n* pieces. He wants to make sure that all *n* pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?
Input Specification:
The first and only line of input contains two space-separated integers, *n* and *h* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=105).
Output Specification:
The output should contain *n*<=-<=1 real numbers *x*1,<=*x*2,<=...,<=*x**n*<=-<=1. The number *x**i* denotes that the *i*-th cut must be made *x**i* units away from the apex of the carrot. In addition, 0<=<<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=-<=1<=<<=*h* must hold.
Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10<=-<=6.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
Demo Input:
['3 2\n', '2 100000\n']
Demo Output:
['1.154700538379 1.632993161855\n', '70710.678118654752\n']
Note:
Definition of isosceles triangle: [https://en.wikipedia.org/wiki/Isosceles_triangle](https://en.wikipedia.org/wiki/Isosceles_triangle). | ```python
#include<bits/stdc++.h>
using namespace std;
int main()
{
//ios::sync_with_stdio(false);
//freopen("out.txt", "w", stdout);
double n, h;
cin >> n >> h;
double s = h / 2;
double ans = 0;
for(int i=1; i<n; ++ i)
{
double res = sqrt(i / n) * h;
if(i == 1)
printf("%.9f", res);
else
printf(" %.9f", res);
}
puts("\n");
return 0;
}
``` | -1 | |
766 | A | Mahmoud and Longest Uncommon Subsequence | PROGRAMMING | 1,000 | [
"constructive algorithms",
"strings"
] | null | null | While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest common subsequence. They solved it, and then Ehab challenged Mahmoud with another problem.
Given two strings *a* and *b*, find the length of their longest uncommon subsequence, which is the longest string that is a subsequence of one of them and not a subsequence of the other.
A subsequence of some string is a sequence of characters that appears in the same order in the string, The appearances don't have to be consecutive, for example, strings "ac", "bc", "abc" and "a" are subsequences of string "abc" while strings "abbc" and "acb" are not. The empty string is a subsequence of any string. Any string is a subsequence of itself. | The first line contains string *a*, and the second line — string *b*. Both of these strings are non-empty and consist of lowercase letters of English alphabet. The length of each string is not bigger than 105 characters. | If there's no uncommon subsequence, print "-1". Otherwise print the length of the longest uncommon subsequence of *a* and *b*. | [
"abcd\ndefgh\n",
"a\na\n"
] | [
"5\n",
"-1\n"
] | In the first example: you can choose "defgh" from string *b* as it is the longest subsequence of string *b* that doesn't appear as a subsequence of string *a*. | 500 | [
{
"input": "abcd\ndefgh",
"output": "5"
},
{
"input": "a\na",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaacccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaadddddddddddddddddddddddddddddddddddddddddddd... | 1,683,614,135 | 2,147,483,647 | Python 3 | OK | TESTS | 73 | 46 | 204,800 | i = input()
q = input()
if i == q:
print(-1)
else:
print(max(len(i), len(q)))
| Title: Mahmoud and Longest Uncommon Subsequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest common subsequence. They solved it, and then Ehab challenged Mahmoud with another problem.
Given two strings *a* and *b*, find the length of their longest uncommon subsequence, which is the longest string that is a subsequence of one of them and not a subsequence of the other.
A subsequence of some string is a sequence of characters that appears in the same order in the string, The appearances don't have to be consecutive, for example, strings "ac", "bc", "abc" and "a" are subsequences of string "abc" while strings "abbc" and "acb" are not. The empty string is a subsequence of any string. Any string is a subsequence of itself.
Input Specification:
The first line contains string *a*, and the second line — string *b*. Both of these strings are non-empty and consist of lowercase letters of English alphabet. The length of each string is not bigger than 105 characters.
Output Specification:
If there's no uncommon subsequence, print "-1". Otherwise print the length of the longest uncommon subsequence of *a* and *b*.
Demo Input:
['abcd\ndefgh\n', 'a\na\n']
Demo Output:
['5\n', '-1\n']
Note:
In the first example: you can choose "defgh" from string *b* as it is the longest subsequence of string *b* that doesn't appear as a subsequence of string *a*. | ```python
i = input()
q = input()
if i == q:
print(-1)
else:
print(max(len(i), len(q)))
``` | 3 | |
108 | A | Palindromic Times | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | A. Palindromic Times | 2 | 256 | Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him. | The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits. | Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time. | [
"12:21\n",
"23:59\n"
] | [
"13:31\n",
"00:00\n"
] | none | 500 | [
{
"input": "12:21",
"output": "13:31"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "15:51",
"output": "20:02"
},
{
"input": "10:44",
"output": "11:11"
},
{
"input": "04:02",
"output": "04:40"
},
{
"input": "02:11",
"output": "02:20"
},
... | 1,680,170,780 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 62 | 0 | def st():return str(input())
def ls():return list(map(int, input().split()))
def mx(a):return max(a)
def it():return int(input())
def mn(a):return min(a)
def yes():return print('YES')
def no():return print('NO')
def re(a):return range(a)
def pr(a):return print(a)
def prs(a):return print(str(a))
def mapa():return map(int, input().split())
s = [int(i) for i in input().split(':')]
s = (s[0]*60)+s[1]
m = []
m.append(0)
m.append(70)
m.append(140)
m.append(210)
m.append(280)
m.append(350)
m.append(601)
m.append(671)
m.append(741)
m.append(811)
m.append(881)
m.append(951)
m.append(1202)
m.append(1272)
m.append(1342)
m.append(1412)
v = []
for i in range(len(m)):
if m[i] - s > 0:
m[i] - s
print(str(m[i]//60) + ':' + str(m[i]%60))
break
elif s > 1412:
print('00:00')
break | Title: Palindromic Times
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.
Input Specification:
The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.
Output Specification:
Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.
Demo Input:
['12:21\n', '23:59\n']
Demo Output:
['13:31\n', '00:00\n']
Note:
none | ```python
def st():return str(input())
def ls():return list(map(int, input().split()))
def mx(a):return max(a)
def it():return int(input())
def mn(a):return min(a)
def yes():return print('YES')
def no():return print('NO')
def re(a):return range(a)
def pr(a):return print(a)
def prs(a):return print(str(a))
def mapa():return map(int, input().split())
s = [int(i) for i in input().split(':')]
s = (s[0]*60)+s[1]
m = []
m.append(0)
m.append(70)
m.append(140)
m.append(210)
m.append(280)
m.append(350)
m.append(601)
m.append(671)
m.append(741)
m.append(811)
m.append(881)
m.append(951)
m.append(1202)
m.append(1272)
m.append(1342)
m.append(1412)
v = []
for i in range(len(m)):
if m[i] - s > 0:
m[i] - s
print(str(m[i]//60) + ':' + str(m[i]%60))
break
elif s > 1412:
print('00:00')
break
``` | 0 |
192 | B | Walking in the Rain | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | In Berland the opposition is going to arrange mass walking on the boulevard. The boulevard consists of *n* tiles that are lain in a row and are numbered from 1 to *n* from right to left. The opposition should start walking on the tile number 1 and the finish on the tile number *n*. During the walk it is allowed to move from right to left between adjacent tiles in a row, and jump over a tile. More formally, if you are standing on the tile number *i* (*i*<=<<=*n*<=-<=1), you can reach the tiles number *i*<=+<=1 or the tile number *i*<=+<=2 from it (if you stand on the tile number *n*<=-<=1, you can only reach tile number *n*). We can assume that all the opposition movements occur instantaneously.
In order to thwart an opposition rally, the Berland bloody regime organized the rain. The tiles on the boulevard are of poor quality and they are rapidly destroyed in the rain. We know that the *i*-th tile is destroyed after *a**i* days of rain (on day *a**i* tile isn't destroyed yet, and on day *a**i*<=+<=1 it is already destroyed). Of course, no one is allowed to walk on the destroyed tiles! So the walk of the opposition is considered thwarted, if either the tile number 1 is broken, or the tile number *n* is broken, or it is impossible to reach the tile number *n* from the tile number 1 if we can walk on undestroyed tiles.
The opposition wants to gather more supporters for their walk. Therefore, the more time they have to pack, the better. Help the opposition to calculate how much time they still have and tell us for how many days the walk from the tile number 1 to the tile number *n* will be possible. | The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the boulevard's length in tiles.
The second line contains *n* space-separated integers *a**i* — the number of days after which the *i*-th tile gets destroyed (1<=≤<=*a**i*<=≤<=103). | Print a single number — the sought number of days. | [
"4\n10 3 5 10\n",
"5\n10 2 8 3 5\n"
] | [
"5\n",
"5\n"
] | In the first sample the second tile gets destroyed after day three, and the only path left is 1 → 3 → 4. After day five there is a two-tile gap between the first and the last tile, you can't jump over it.
In the second sample path 1 → 3 → 5 is available up to day five, inclusive. On day six the last tile is destroyed and the walk is thwarted. | 1,000 | [
{
"input": "4\n10 3 5 10",
"output": "5"
},
{
"input": "5\n10 2 8 3 5",
"output": "5"
},
{
"input": "10\n10 3 1 6 7 1 3 3 8 1",
"output": "1"
},
{
"input": "10\n26 72 10 52 2 5 61 2 39 64",
"output": "5"
},
{
"input": "100\n8 2 1 2 8 3 5 8 5 1 9 3 4 1 5 6 4 2 9 10... | 1,691,077,252 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 82 | 124 | 2,867,200 | n=int(input())
l=list(map(int,input().split()))
mi=1000000000
if len(l)==1:
print(l[0])
else:
i=0
j=1
k=2
ans=[]
while k<len(l):
if l[j]>=l[k]:
mi=min(l[i],l[j])
ans.append(mi)
i=j
j=j+1
k=k+1
else:
mi=min(l[i],l[k])
ans.append(mi)
i=k
k=k+2
j=j+2
#print(ans)
print(min(min(ans), l[0], l[n-1]))
| Title: Walking in the Rain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland the opposition is going to arrange mass walking on the boulevard. The boulevard consists of *n* tiles that are lain in a row and are numbered from 1 to *n* from right to left. The opposition should start walking on the tile number 1 and the finish on the tile number *n*. During the walk it is allowed to move from right to left between adjacent tiles in a row, and jump over a tile. More formally, if you are standing on the tile number *i* (*i*<=<<=*n*<=-<=1), you can reach the tiles number *i*<=+<=1 or the tile number *i*<=+<=2 from it (if you stand on the tile number *n*<=-<=1, you can only reach tile number *n*). We can assume that all the opposition movements occur instantaneously.
In order to thwart an opposition rally, the Berland bloody regime organized the rain. The tiles on the boulevard are of poor quality and they are rapidly destroyed in the rain. We know that the *i*-th tile is destroyed after *a**i* days of rain (on day *a**i* tile isn't destroyed yet, and on day *a**i*<=+<=1 it is already destroyed). Of course, no one is allowed to walk on the destroyed tiles! So the walk of the opposition is considered thwarted, if either the tile number 1 is broken, or the tile number *n* is broken, or it is impossible to reach the tile number *n* from the tile number 1 if we can walk on undestroyed tiles.
The opposition wants to gather more supporters for their walk. Therefore, the more time they have to pack, the better. Help the opposition to calculate how much time they still have and tell us for how many days the walk from the tile number 1 to the tile number *n* will be possible.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the boulevard's length in tiles.
The second line contains *n* space-separated integers *a**i* — the number of days after which the *i*-th tile gets destroyed (1<=≤<=*a**i*<=≤<=103).
Output Specification:
Print a single number — the sought number of days.
Demo Input:
['4\n10 3 5 10\n', '5\n10 2 8 3 5\n']
Demo Output:
['5\n', '5\n']
Note:
In the first sample the second tile gets destroyed after day three, and the only path left is 1 → 3 → 4. After day five there is a two-tile gap between the first and the last tile, you can't jump over it.
In the second sample path 1 → 3 → 5 is available up to day five, inclusive. On day six the last tile is destroyed and the walk is thwarted. | ```python
n=int(input())
l=list(map(int,input().split()))
mi=1000000000
if len(l)==1:
print(l[0])
else:
i=0
j=1
k=2
ans=[]
while k<len(l):
if l[j]>=l[k]:
mi=min(l[i],l[j])
ans.append(mi)
i=j
j=j+1
k=k+1
else:
mi=min(l[i],l[k])
ans.append(mi)
i=k
k=k+2
j=j+2
#print(ans)
print(min(min(ans), l[0], l[n-1]))
``` | -1 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,680,062,742 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 92 | 0 | m, n = map(int, input().split())
ans = (m * n) // 2
print(ans)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
m, n = map(int, input().split())
ans = (m * n) // 2
print(ans)
``` | 3.977 |
629 | A | Far Relative’s Birthday Cake | PROGRAMMING | 800 | [
"brute force",
"combinatorics",
"constructive algorithms",
"implementation"
] | null | null | Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column. | In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'. | Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column. | [
"3\n.CC\nC..\nC.C\n",
"4\nCC..\nC..C\n.CC.\n.CC.\n"
] | [
"4\n",
"9\n"
] | If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3) | 500 | [
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "4\nCC..\nC..C\n.CC.\n.CC.",
"output": "9"
},
{
"input": "5\n.CCCC\nCCCCC\n.CCC.\nCC...\n.CC.C",
"output": "46"
},
{
"input": "7\n.CC..CC\nCC.C..C\nC.C..C.\nC...C.C\nCCC.CCC\n.CC...C\n.C.CCC.",
"output": "84"
},... | 1,616,065,395 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 171 | 307,200 | n = int(input())
cke=[]
for i in range(n):
a=[]
s = input()
for j in range(n):
a.append(s[j])
cke.append(a)
c=0
for i in range(n):
for j in range(n):
if cke[i][j]=='C':
for k in range(j+1,n):
if cke[i][k]=='C':
c += 1
for k in range(i+1,n):
if cke[k][j]=='C':
c += 1
print(c) | Title: Far Relative’s Birthday Cake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
Input Specification:
In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
Output Specification:
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
Demo Input:
['3\n.CC\nC..\nC.C\n', '4\nCC..\nC..C\n.CC.\n.CC.\n']
Demo Output:
['4\n', '9\n']
Note:
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3) | ```python
n = int(input())
cke=[]
for i in range(n):
a=[]
s = input()
for j in range(n):
a.append(s[j])
cke.append(a)
c=0
for i in range(n):
for j in range(n):
if cke[i][j]=='C':
for k in range(j+1,n):
if cke[i][k]=='C':
c += 1
for k in range(i+1,n):
if cke[k][j]=='C':
c += 1
print(c)
``` | 3 | |
215 | A | Bicycle Chain | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces. | Print the number of "integer" gears with the maximum ratio among all "integer" gears. | [
"2\n4 5\n3\n12 13 15\n",
"4\n1 2 3 4\n5\n10 11 12 13 14\n"
] | [
"2\n",
"1\n"
] | In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15. | 500 | [
{
"input": "2\n4 5\n3\n12 13 15",
"output": "2"
},
{
"input": "4\n1 2 3 4\n5\n10 11 12 13 14",
"output": "1"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 2\n1\n1",
"output": "1"
},
{
"input": "1\n1\n2\n1 2",
"output": "1"
},
{
"input":... | 1,587,320,796 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 216 | 307,200 | n=int(input())
a=list(map(int,input().split()))
m=int(input())
b=list(map(int,input().split()))
r=1
c=0
mr=1
for i in range(n):
for j in range(m-1,-1,-1):
r=b[j]/a[i]
if r==int(r):
mr=int(r)
break
for i in a:
if mr*i in b:
c+=1
print(c)
| Title: Bicycle Chain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
Output Specification:
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
Demo Input:
['2\n4 5\n3\n12 13 15\n', '4\n1 2 3 4\n5\n10 11 12 13 14\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15. | ```python
n=int(input())
a=list(map(int,input().split()))
m=int(input())
b=list(map(int,input().split()))
r=1
c=0
mr=1
for i in range(n):
for j in range(m-1,-1,-1):
r=b[j]/a[i]
if r==int(r):
mr=int(r)
break
for i in a:
if mr*i in b:
c+=1
print(c)
``` | 0 | |
807 | A | Is it rated? | PROGRAMMING | 900 | [
"implementation",
"sortings"
] | null | null | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings. | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | [
"6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n",
"4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n",
"5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n"
] | [
"rated\n",
"unrated\n",
"maybe\n"
] | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not. | 500 | [
{
"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884",
"output": "rated"
},
{
"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699",
"output": "maybe"
},
{
... | 1,591,503,726 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 10 | 108 | 307,200 | n = int(input())
prev = (-1 , -1)
maybe = False
is_rated = False
for i in range(n):
a, b = list(map(int, input().split()))\
if a != b:
is_rated = True
break
if a == prev[0] and b == prev[1]:
maybe = True
prev = (a, b)
if is_rated:
print('rated')
elif maybe:
print('maybe')
else:
print('unrated')
| Title: Is it rated?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output Specification:
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Demo Input:
['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n']
Demo Output:
['rated\n', 'unrated\n', 'maybe\n']
Note:
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not. | ```python
n = int(input())
prev = (-1 , -1)
maybe = False
is_rated = False
for i in range(n):
a, b = list(map(int, input().split()))\
if a != b:
is_rated = True
break
if a == prev[0] and b == prev[1]:
maybe = True
prev = (a, b)
if is_rated:
print('rated')
elif maybe:
print('maybe')
else:
print('unrated')
``` | 0 | |
44 | E | Anfisa the Monkey | PROGRAMMING | 1,400 | [
"dp"
] | E. Anfisa the Monkey | 2 | 256 | Anfisa the monkey learns to type. She is yet unfamiliar with the "space" key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into *k* lines not shorter than *a* and not longer than *b*, for the text to resemble human speech more. Help Anfisa. | The first line contains three integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=200, 1<=≤<=*a*<=≤<=*b*<=≤<=200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols. | Print *k* lines, each of which contains no less than *a* and no more than *b* symbols — Anfisa's text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print "No solution" (without quotes). | [
"3 2 5\nabrakadabra\n",
"4 1 2\nabrakadabra\n"
] | [
"ab\nrakad\nabra\n",
"No solution\n"
] | none | 0 | [
{
"input": "3 2 5\nabrakadabra",
"output": "abra\nkada\nbra"
},
{
"input": "4 1 2\nabrakadabra",
"output": "No solution"
},
{
"input": "3 1 2\nvgnfpo",
"output": "vg\nnf\npo"
},
{
"input": "5 3 4\nvrrdnhazvexzjfv",
"output": "vrr\ndnh\nazv\nexz\njfv"
},
{
"input":... | 1,604,256,244 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 10 | 372 | 204,800 | import sys, os
from math import sqrt, gcd, ceil, log, floor
from bisect import bisect, bisect_left
from collections import defaultdict, Counter, deque
from heapq import heapify, heappush, heappop
input = sys.stdin.readline
read = lambda: list(map(int, input().strip().split()))
# read_f = lambda file: list(map(int, file.readline().strip().split()))
# from time import time
# sys.setrecursionlimit(10**6)
MOD = 10**9 + 7
def main():
# file1 = open("C:\\Users\\shank\\Desktop\\Comp_Code\\input.txt", "r")
# n = int(file1.readline().strip());
# arr = list(map(int, file1.read().strip().split(" ")))
# file1.close()
k, a, b = read(); s = input().strip()
l = len(s); fl = 0
for i in range(a, b+1):
if l%i == 0 and l//i <= k:
fl = 1
break
elif a <= l%i <= b and l//i < k:
fl = 1
break
if fl:
for j in range(0, l, i):
print(s[j:j+i])
else:
print("No solution")
# file = open("output.txt", "w")
# file.write(ans+"\n")
# file.close()
if __name__ == "__main__":
main()
"""
ind 0 1 2 3
1 = 2 3 3 2
2 = 2 2 3 3
""" | Title: Anfisa the Monkey
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Anfisa the monkey learns to type. She is yet unfamiliar with the "space" key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into *k* lines not shorter than *a* and not longer than *b*, for the text to resemble human speech more. Help Anfisa.
Input Specification:
The first line contains three integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=200, 1<=≤<=*a*<=≤<=*b*<=≤<=200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols.
Output Specification:
Print *k* lines, each of which contains no less than *a* and no more than *b* symbols — Anfisa's text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print "No solution" (without quotes).
Demo Input:
['3 2 5\nabrakadabra\n', '4 1 2\nabrakadabra\n']
Demo Output:
['ab\nrakad\nabra\n', 'No solution\n']
Note:
none | ```python
import sys, os
from math import sqrt, gcd, ceil, log, floor
from bisect import bisect, bisect_left
from collections import defaultdict, Counter, deque
from heapq import heapify, heappush, heappop
input = sys.stdin.readline
read = lambda: list(map(int, input().strip().split()))
# read_f = lambda file: list(map(int, file.readline().strip().split()))
# from time import time
# sys.setrecursionlimit(10**6)
MOD = 10**9 + 7
def main():
# file1 = open("C:\\Users\\shank\\Desktop\\Comp_Code\\input.txt", "r")
# n = int(file1.readline().strip());
# arr = list(map(int, file1.read().strip().split(" ")))
# file1.close()
k, a, b = read(); s = input().strip()
l = len(s); fl = 0
for i in range(a, b+1):
if l%i == 0 and l//i <= k:
fl = 1
break
elif a <= l%i <= b and l//i < k:
fl = 1
break
if fl:
for j in range(0, l, i):
print(s[j:j+i])
else:
print("No solution")
# file = open("output.txt", "w")
# file.write(ans+"\n")
# file.close()
if __name__ == "__main__":
main()
"""
ind 0 1 2 3
1 = 2 3 3 2
2 = 2 2 3 3
"""
``` | 0 |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,665,042,179 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 124 | 409,600 | from re import X
n = int(input())
forces = [0, 0, 0]
for i in range(n):
[x, y, z] = [int(_) for _ in input().split()]
forces[0] += x
forces[1] += y
forces[2] += z
if forces[0] or forces[1] or forces[2]:
print('NO')
else:
print('YES')
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
from re import X
n = int(input())
forces = [0, 0, 0]
for i in range(n):
[x, y, z] = [int(_) for _ in input().split()]
forces[0] += x
forces[1] += y
forces[2] += z
if forces[0] or forces[1] or forces[2]:
print('NO')
else:
print('YES')
``` | 3.968237 |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,688,398,008 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 46 | 0 | str=input()
i,j=0,0
s="hello"
while(i<len(str) and j<5):
if(str[i]==s[j] ):
i=i+1
j=j+1
else:
i=i+1
if(j==5):
print("YES")
else:
print("NO")
i=i+1 # -*- coding: latin-1 -*-j=j+1
| Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
str=input()
i,j=0,0
s="hello"
while(i<len(str) and j<5):
if(str[i]==s[j] ):
i=i+1
j=j+1
else:
i=i+1
if(j==5):
print("YES")
else:
print("NO")
i=i+1 # -*- coding: latin-1 -*-j=j+1
``` | 3.977 |
686 | A | Free Ice Cream | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. | The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. | Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. | [
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] | [
"22 1\n",
"3 2\n"
] | Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | 500 | [
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
... | 1,603,031,369 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 109 | 0 | n, x = list(map(int, input().split(" ")))
b = 0
for i in range(n):
s = input().split(" ")
if s[0] == "+":
x += int(s[1])
else:
if x < int(s[1]):
b+=1
else:
x -= int(s[1])
print(x, b)
| Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | ```python
n, x = list(map(int, input().split(" ")))
b = 0
for i in range(n):
s = input().split(" ")
if s[0] == "+":
x += int(s[1])
else:
if x < int(s[1]):
b+=1
else:
x -= int(s[1])
print(x, b)
``` | 3 | |
316 | A2 | Special Task | PROGRAMMING | 1,400 | [
"math"
] | null | null | Special Agent Smart Beaver works in a secret research department of ABBYY. He's been working there for a long time and is satisfied with his job, as it allows him to eat out in the best restaurants and order the most expensive and exotic wood types there.
The content special agent has got an important task: to get the latest research by British scientists on the English Language. These developments are encoded and stored in a large safe. The Beaver's teeth are strong enough, so the authorities assured that upon arriving at the place the beaver won't have any problems with opening the safe.
And he finishes his aspen sprig and leaves for this important task. Of course, the Beaver arrived at the location without any problems, but alas. He can't open the safe with his strong and big teeth. At this point, the Smart Beaver get a call from the headquarters and learns that opening the safe with the teeth is not necessary, as a reliable source has sent the following information: the safe code consists of digits and has no leading zeroes. There also is a special hint, which can be used to open the safe. The hint is string *s* with the following structure:
- if *s**i* = "?", then the digit that goes *i*-th in the safe code can be anything (between 0 to 9, inclusively); - if *s**i* is a digit (between 0 to 9, inclusively), then it means that there is digit *s**i* on position *i* in code; - if the string contains letters from "A" to "J", then all positions with the same letters must contain the same digits and the positions with distinct letters must contain distinct digits. - The length of the safe code coincides with the length of the hint.
For example, hint "?JGJ9" has such matching safe code variants: "51919", "55959", "12329", "93539" and so on, and has wrong variants such as: "56669", "00111", "03539" and "13666".
After receiving such information, the authorities change the plan and ask the special agents to work quietly and gently and not to try to open the safe by mechanical means, and try to find the password using the given hint.
At a special agent school the Smart Beaver was the fastest in his platoon finding codes for such safes, but now he is not in that shape: the years take their toll ... Help him to determine the number of possible variants of the code to the safe, matching the given hint. After receiving this information, and knowing his own speed of entering codes, the Smart Beaver will be able to determine whether he will have time for tonight's show "Beavers are on the trail" on his favorite TV channel, or he should work for a sleepless night... | The first line contains string *s* — the hint to the safe code. String *s* consists of the following characters: ?, 0-9, A-J. It is guaranteed that the first character of string *s* doesn't equal to character 0.
The input limits for scoring 30 points are (subproblem A1):
- 1<=≤<=|*s*|<=≤<=5.
The input limits for scoring 100 points are (subproblems A1+A2):
- 1<=≤<=|*s*|<=≤<=105.
Here |*s*| means the length of string *s*. | Print the number of codes that match the given hint. | [
"AJ\n",
"1?AA\n"
] | [
"81\n",
"100\n"
] | none | 70 | [
{
"input": "AJ",
"output": "81"
},
{
"input": "1?AA",
"output": "100"
},
{
"input": "?",
"output": "9"
},
{
"input": "7",
"output": "1"
},
{
"input": "A",
"output": "9"
},
{
"input": "BBB?",
"output": "90"
},
{
"input": "BC??",
"output"... | 1,682,420,466 | 2,147,483,647 | PyPy 3-64 | COMPILATION_ERROR | TESTS2 | 0 | 0 | 0 | #include <stdio.h>
#include <string.h>
#include <math.h>
#define MAX_LEN 1000
int count_permutations(char *word) {
int count[128] = {0};
int n = 0, m = 0;
char first_char = word[0];
int i, out;
for (i = 0; i < strlen(word); i++) {
char c = word[i];
if (c >= 'A' && c <= 'J' && count[c] == 0) {
n++;
count[c] = 1;
} else if (c == '?') {
m++;
}
}
if (first_char >= '0' && first_char <= '9') {
out = factorial(10) / factorial(10-n);
printf("%d", out);
for (i = 0; i < m; i++) {
printf("0");
}
printf("\n");
} else if (first_char == '?') {
out = factorial(10) / factorial(10-n) * 9;
printf("%d", out);
for (i = 0; i < m-1; i++) {
printf("0");
}
printf("\n");
} else {
out = 9 * factorial(9) / factorial(10-n);
printf("%d", out);
for (i = 0; i < m; i++) {
printf("0");
}
printf("\n");
}
return 0;
}
int main() {
char word[MAX_LEN];
scanf("%s", word);
count_permutations(word);
return 0;
}
| Title: Special Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Special Agent Smart Beaver works in a secret research department of ABBYY. He's been working there for a long time and is satisfied with his job, as it allows him to eat out in the best restaurants and order the most expensive and exotic wood types there.
The content special agent has got an important task: to get the latest research by British scientists on the English Language. These developments are encoded and stored in a large safe. The Beaver's teeth are strong enough, so the authorities assured that upon arriving at the place the beaver won't have any problems with opening the safe.
And he finishes his aspen sprig and leaves for this important task. Of course, the Beaver arrived at the location without any problems, but alas. He can't open the safe with his strong and big teeth. At this point, the Smart Beaver get a call from the headquarters and learns that opening the safe with the teeth is not necessary, as a reliable source has sent the following information: the safe code consists of digits and has no leading zeroes. There also is a special hint, which can be used to open the safe. The hint is string *s* with the following structure:
- if *s**i* = "?", then the digit that goes *i*-th in the safe code can be anything (between 0 to 9, inclusively); - if *s**i* is a digit (between 0 to 9, inclusively), then it means that there is digit *s**i* on position *i* in code; - if the string contains letters from "A" to "J", then all positions with the same letters must contain the same digits and the positions with distinct letters must contain distinct digits. - The length of the safe code coincides with the length of the hint.
For example, hint "?JGJ9" has such matching safe code variants: "51919", "55959", "12329", "93539" and so on, and has wrong variants such as: "56669", "00111", "03539" and "13666".
After receiving such information, the authorities change the plan and ask the special agents to work quietly and gently and not to try to open the safe by mechanical means, and try to find the password using the given hint.
At a special agent school the Smart Beaver was the fastest in his platoon finding codes for such safes, but now he is not in that shape: the years take their toll ... Help him to determine the number of possible variants of the code to the safe, matching the given hint. After receiving this information, and knowing his own speed of entering codes, the Smart Beaver will be able to determine whether he will have time for tonight's show "Beavers are on the trail" on his favorite TV channel, or he should work for a sleepless night...
Input Specification:
The first line contains string *s* — the hint to the safe code. String *s* consists of the following characters: ?, 0-9, A-J. It is guaranteed that the first character of string *s* doesn't equal to character 0.
The input limits for scoring 30 points are (subproblem A1):
- 1<=≤<=|*s*|<=≤<=5.
The input limits for scoring 100 points are (subproblems A1+A2):
- 1<=≤<=|*s*|<=≤<=105.
Here |*s*| means the length of string *s*.
Output Specification:
Print the number of codes that match the given hint.
Demo Input:
['AJ\n', '1?AA\n']
Demo Output:
['81\n', '100\n']
Note:
none | ```python
#include <stdio.h>
#include <string.h>
#include <math.h>
#define MAX_LEN 1000
int count_permutations(char *word) {
int count[128] = {0};
int n = 0, m = 0;
char first_char = word[0];
int i, out;
for (i = 0; i < strlen(word); i++) {
char c = word[i];
if (c >= 'A' && c <= 'J' && count[c] == 0) {
n++;
count[c] = 1;
} else if (c == '?') {
m++;
}
}
if (first_char >= '0' && first_char <= '9') {
out = factorial(10) / factorial(10-n);
printf("%d", out);
for (i = 0; i < m; i++) {
printf("0");
}
printf("\n");
} else if (first_char == '?') {
out = factorial(10) / factorial(10-n) * 9;
printf("%d", out);
for (i = 0; i < m-1; i++) {
printf("0");
}
printf("\n");
} else {
out = 9 * factorial(9) / factorial(10-n);
printf("%d", out);
for (i = 0; i < m; i++) {
printf("0");
}
printf("\n");
}
return 0;
}
int main() {
char word[MAX_LEN];
scanf("%s", word);
count_permutations(word);
return 0;
}
``` | -1 | |
598 | A | Tricky Sum | PROGRAMMING | 900 | [
"math"
] | null | null | In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum.
For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively.
Calculate the answer for *t* values of *n*. | The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed.
Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109). | Print the requested sum for each of *t* integers *n* given in the input. | [
"2\n4\n1000000000\n"
] | [
"-4\n499999998352516354\n"
] | The answer for the first sample is explained in the statement. | 0 | [
{
"input": "2\n4\n1000000000",
"output": "-4\n499999998352516354"
},
{
"input": "10\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "-1\n-3\n0\n-4\n1\n7\n14\n6\n15\n25"
},
{
"input": "10\n10\n9\n47\n33\n99\n83\n62\n1\n100\n53",
"output": "25\n15\n1002\n435\n4696\n3232\n1827\n-1\n4796\n130... | 1,632,040,801 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 77 | 20,172,800 | import math
for _ in range(int(input())):
n=int(input())
sum=n*(n+1)//2
sum1=0
k=int((math.log(n))//math.log(2))+1
#print(k)
print(sum - 2*2**k + 2)
| Title: Tricky Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum.
For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively.
Calculate the answer for *t* values of *n*.
Input Specification:
The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed.
Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the requested sum for each of *t* integers *n* given in the input.
Demo Input:
['2\n4\n1000000000\n']
Demo Output:
['-4\n499999998352516354\n']
Note:
The answer for the first sample is explained in the statement. | ```python
import math
for _ in range(int(input())):
n=int(input())
sum=n*(n+1)//2
sum1=0
k=int((math.log(n))//math.log(2))+1
#print(k)
print(sum - 2*2**k + 2)
``` | 0 | |
3 | A | Shortest path of the king | PROGRAMMING | 1,000 | [
"greedy",
"shortest paths"
] | A. Shortest path of the king | 1 | 64 | The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this.
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to). | The first line contains the chessboard coordinates of square *s*, the second line — of square *t*.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8. | In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them. | [
"a8\nh1\n"
] | [
"7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n"
] | none | 0 | [
{
"input": "a8\nh1",
"output": "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD"
},
{
"input": "b2\nb4",
"output": "2\nU\nU"
},
{
"input": "a5\na5",
"output": "0"
},
{
"input": "h1\nb2",
"output": "6\nLU\nL\nL\nL\nL\nL"
},
{
"input": "c5\nh2",
"output": "5\nRD\nRD\nRD\nR\nR"
... | 1,698,522,413 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 35 | 124 | 0 | a = input()
b = input()
a = [a[0], int(a[1])]
b = [b[0], int(b[1])]
points = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
b[0] = points.index(b[0]) + 1
a[0] = points.index(a[0]) + 1
print(max(abs(a[0] - b[0]), abs(a[1] - b[1])))
while a[0] != b[0] or a[1] != b[1]:
if a[0] < b[0]:
print('R', end='')
a[0] += 1
elif a[0] > b[0]:
print('L', end='')
a[0] -= 1
if a[1] < b[1]:
print('U', end='')
a[1] += 1
elif a[1] > b[1]:
print('D', end='')
a[1] -= 1
print() | Title: Shortest path of the king
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this.
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).
Input Specification:
The first line contains the chessboard coordinates of square *s*, the second line — of square *t*.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.
Output Specification:
In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.
Demo Input:
['a8\nh1\n']
Demo Output:
['7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n']
Note:
none | ```python
a = input()
b = input()
a = [a[0], int(a[1])]
b = [b[0], int(b[1])]
points = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
b[0] = points.index(b[0]) + 1
a[0] = points.index(a[0]) + 1
print(max(abs(a[0] - b[0]), abs(a[1] - b[1])))
while a[0] != b[0] or a[1] != b[1]:
if a[0] < b[0]:
print('R', end='')
a[0] += 1
elif a[0] > b[0]:
print('L', end='')
a[0] -= 1
if a[1] < b[1]:
print('U', end='')
a[1] += 1
elif a[1] > b[1]:
print('D', end='')
a[1] -= 1
print()
``` | 3.938 |
629 | B | Far Relative’s Problem | PROGRAMMING | 1,100 | [
"brute force"
] | null | null | Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible.
Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party.
Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends.
Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive. | Print the maximum number of people that may come to Famil Door's party. | [
"4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n",
"6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n"
] | [
"2\n",
"4\n"
] | In the first sample, friends 3 and 4 can come on any day in range [117, 128].
In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140. | 1,000 | [
{
"input": "4\nM 151 307\nF 343 352\nF 117 145\nM 24 128",
"output": "2"
},
{
"input": "6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200",
"output": "4"
},
{
"input": "1\nF 68 307",
"output": "0"
},
{
"input": "40\nM 55 363\nF 117 252\nM 157 282\nF 322 345\... | 1,455,988,088 | 1,988 | PyPy 3 | WRONG_ANSWER | PRETESTS | 2 | 233 | 9,216,000 | #!/usr/bin/env python3
# 629B_relative.py - Codeforces.com/problemset/problem/629/B by Sergey 2016
import unittest
import sys
###############################################################################
# Relative Class (Main Program)
###############################################################################
class Relative:
""" Relative representation """
def __init__(self, test_inputs=None):
""" Default constructor """
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
return next(it) if it else sys.stdin.readline().rstrip()
# Reading single elements
[self.n] = map(int, uinput().split())
# Reading multiple number of lines of the same number of elements each
l, s = self.n, 3
inp = (" ".join(uinput() for i in range(l))).split()
self.numm = [[(int(inp[i]) if j != 0 else inp[i] == 'M')
for i in range(j, l*s, s)] for j in range(s)]
self.nums, self.numa, self.numb = self.numm
self.dts = []
for i in range(self.n):
self.dts.append((self.numa[i], 1, self.nums[i]))
self.dts.append((self.numb[i], 0, self.nums[i]))
self.sdts = sorted(self.dts)
self.ms, self.fs = [0], [0]
for i in range(len(self.sdts)):
if self.sdts[i][2]:
if self.sdts[i][1]:
self.ms.append(self.ms[-1] + 1)
else:
self.ms.append(self.ms[-1] - 1)
else:
if self.sdts[i][1]:
self.fs.append(self.fs[-1] + 1)
else:
self.fs.append(self.fs[-1] - 1)
self.cr = []
for i in range(len(self.ms)):
self.cr.append(min(self.ms[i], self.fs[i])*2)
def calculate(self):
""" Main calcualtion function of the class """
result = max(self.cr)
return str(result)
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_single_test(self):
""" Relative class testing """
# Constructor test
test = "4\nM 151 307\nF 343 352\nF 117 145\nM 24 128"
d = Relative(test)
self.assertEqual(d.n, 4)
self.assertEqual(d.numa, [151, 343, 117, 24])
self.assertEqual(d.nums, [1, 0, 0, 1])
# Sample test
self.assertEqual(Relative(test).calculate(), "2")
# Sample test
test = "6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200"
self.assertEqual(Relative(test).calculate(), "4")
# Sample test
test = ""
# self.assertEqual(Relative(test).calculate(), "0")
# My tests
test = ""
# self.assertEqual(Relative(test).calculate(), "0")
# Time limit test
# self.time_limit_test(5000)
def time_limit_test(self, nmax):
""" Timelimit testing """
import random
import timeit
# Random inputs
test = str(nmax) + " " + str(nmax) + "\n"
numnums = [str(i) + " " + str(i+1) for i in range(nmax)]
test += "\n".join(numnums) + "\n"
nums = [random.randint(1, 10000) for i in range(nmax)]
test += " ".join(map(str, nums)) + "\n"
# Run the test
start = timeit.default_timer()
d = Relative(test)
calc = timeit.default_timer()
d.calculate()
stop = timeit.default_timer()
print("\nTimelimit Test: " +
"{0:.3f}s (init {1:.3f}s calc {2:.3f}s)".
format(stop-start, calc-start, stop-calc))
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
sys.stdout.write(Relative().calculate())
| Title: Far Relative’s Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible.
Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party.
Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends.
Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive.
Output Specification:
Print the maximum number of people that may come to Famil Door's party.
Demo Input:
['4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n', '6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample, friends 3 and 4 can come on any day in range [117, 128].
In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140. | ```python
#!/usr/bin/env python3
# 629B_relative.py - Codeforces.com/problemset/problem/629/B by Sergey 2016
import unittest
import sys
###############################################################################
# Relative Class (Main Program)
###############################################################################
class Relative:
""" Relative representation """
def __init__(self, test_inputs=None):
""" Default constructor """
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
return next(it) if it else sys.stdin.readline().rstrip()
# Reading single elements
[self.n] = map(int, uinput().split())
# Reading multiple number of lines of the same number of elements each
l, s = self.n, 3
inp = (" ".join(uinput() for i in range(l))).split()
self.numm = [[(int(inp[i]) if j != 0 else inp[i] == 'M')
for i in range(j, l*s, s)] for j in range(s)]
self.nums, self.numa, self.numb = self.numm
self.dts = []
for i in range(self.n):
self.dts.append((self.numa[i], 1, self.nums[i]))
self.dts.append((self.numb[i], 0, self.nums[i]))
self.sdts = sorted(self.dts)
self.ms, self.fs = [0], [0]
for i in range(len(self.sdts)):
if self.sdts[i][2]:
if self.sdts[i][1]:
self.ms.append(self.ms[-1] + 1)
else:
self.ms.append(self.ms[-1] - 1)
else:
if self.sdts[i][1]:
self.fs.append(self.fs[-1] + 1)
else:
self.fs.append(self.fs[-1] - 1)
self.cr = []
for i in range(len(self.ms)):
self.cr.append(min(self.ms[i], self.fs[i])*2)
def calculate(self):
""" Main calcualtion function of the class """
result = max(self.cr)
return str(result)
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_single_test(self):
""" Relative class testing """
# Constructor test
test = "4\nM 151 307\nF 343 352\nF 117 145\nM 24 128"
d = Relative(test)
self.assertEqual(d.n, 4)
self.assertEqual(d.numa, [151, 343, 117, 24])
self.assertEqual(d.nums, [1, 0, 0, 1])
# Sample test
self.assertEqual(Relative(test).calculate(), "2")
# Sample test
test = "6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200"
self.assertEqual(Relative(test).calculate(), "4")
# Sample test
test = ""
# self.assertEqual(Relative(test).calculate(), "0")
# My tests
test = ""
# self.assertEqual(Relative(test).calculate(), "0")
# Time limit test
# self.time_limit_test(5000)
def time_limit_test(self, nmax):
""" Timelimit testing """
import random
import timeit
# Random inputs
test = str(nmax) + " " + str(nmax) + "\n"
numnums = [str(i) + " " + str(i+1) for i in range(nmax)]
test += "\n".join(numnums) + "\n"
nums = [random.randint(1, 10000) for i in range(nmax)]
test += " ".join(map(str, nums)) + "\n"
# Run the test
start = timeit.default_timer()
d = Relative(test)
calc = timeit.default_timer()
d.calculate()
stop = timeit.default_timer()
print("\nTimelimit Test: " +
"{0:.3f}s (init {1:.3f}s calc {2:.3f}s)".
format(stop-start, calc-start, stop-calc))
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
sys.stdout.write(Relative().calculate())
``` | 0 | |
886 | B | Vlad and Cafes | PROGRAMMING | 1,000 | [] | null | null | Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe. | In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted. | Print one integer — index of the cafe that Vlad hasn't visited for as long as possible. | [
"5\n1 3 2 1 2\n",
"6\n2 1 2 2 4 1\n"
] | [
"3\n",
"2\n"
] | In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes. | 1,000 | [
{
"input": "5\n1 3 2 1 2",
"output": "3"
},
{
"input": "6\n2 1 2 2 4 1",
"output": "2"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n200000",
"output": "200000"
},
{
"input": "2\n2018 2017",
"output": "2018"
},
{
"input": "5\n100 1000 1000 1000... | 1,563,446,395 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 108 | 0 | n = int(input())
was = [0 for i in range(n)]
inp = input().split(" ")
for i in inp:
was = list(map(lambda x: x+1, was))
was[int(i)-1] = 0
print(was.index(max(was)))
| Title: Vlad and Cafes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
Input Specification:
In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
Output Specification:
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.
Demo Input:
['5\n1 3 2 1 2\n', '6\n2 1 2 2 4 1\n']
Demo Output:
['3\n', '2\n']
Note:
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes. | ```python
n = int(input())
was = [0 for i in range(n)]
inp = input().split(" ")
for i in inp:
was = list(map(lambda x: x+1, was))
was[int(i)-1] = 0
print(was.index(max(was)))
``` | -1 | |
385 | A | Bear and Raspberry | PROGRAMMING | 1,000 | [
"brute force",
"greedy",
"implementation"
] | null | null | The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=≤<=*i*<=≤<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry.
Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=≤<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel.
The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan. | The first line contains two space-separated integers, *n* and *c* (2<=≤<=*n*<=≤<=100,<=0<=≤<=*c*<=≤<=100), — the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100), the price of a honey barrel on day *i*. | Print a single integer — the answer to the problem. | [
"5 1\n5 10 7 3 20\n",
"6 2\n100 1 10 40 10 40\n",
"3 0\n1 2 3\n"
] | [
"3\n",
"97\n",
"0\n"
] | In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3.
In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97. | 500 | [
{
"input": "5 1\n5 10 7 3 20",
"output": "3"
},
{
"input": "6 2\n100 1 10 40 10 40",
"output": "97"
},
{
"input": "3 0\n1 2 3",
"output": "0"
},
{
"input": "2 0\n2 1",
"output": "1"
},
{
"input": "10 5\n10 1 11 2 12 3 13 4 14 5",
"output": "4"
},
{
"in... | 1,621,752,327 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 62 | 0 | n, c = map(int, input().split())
price = list(map(int, input().split()))
p = price[0]-price[1]-c
for i in range(1,n-1):
p = max(price[i]-price[i+1]-c, p)
if p<0:
print('0')
else:
print(p) | Title: Bear and Raspberry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=≤<=*i*<=≤<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry.
Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=≤<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel.
The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan.
Input Specification:
The first line contains two space-separated integers, *n* and *c* (2<=≤<=*n*<=≤<=100,<=0<=≤<=*c*<=≤<=100), — the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100), the price of a honey barrel on day *i*.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['5 1\n5 10 7 3 20\n', '6 2\n100 1 10 40 10 40\n', '3 0\n1 2 3\n']
Demo Output:
['3\n', '97\n', '0\n']
Note:
In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3.
In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97. | ```python
n, c = map(int, input().split())
price = list(map(int, input().split()))
p = price[0]-price[1]-c
for i in range(1,n-1):
p = max(price[i]-price[i+1]-c, p)
if p<0:
print('0')
else:
print(p)
``` | 3 | |
869 | A | The Artful Expedient | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game. | The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*. | Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization. | [
"3\n1 2 3\n4 5 6\n",
"5\n2 4 6 8 10\n9 7 5 3 1\n"
] | [
"Karen\n",
"Karen\n"
] | In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again. | 500 | [
{
"input": "3\n1 2 3\n4 5 6",
"output": "Karen"
},
{
"input": "5\n2 4 6 8 10\n9 7 5 3 1",
"output": "Karen"
},
{
"input": "1\n1\n2000000",
"output": "Karen"
},
{
"input": "2\n97153 2000000\n1999998 254",
"output": "Karen"
},
{
"input": "15\n31 30 29 28 27 26 25 24... | 1,549,357,969 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 78 | 0 | n=int(input())
xi=[]
yi=[]
for i in range(n):
xi=xi+list(map(int,input().split()))
for i in range(n):
yi=yi+list(map(int,input().split()))
c=0
x=set(xi)
y=set(yi)
for i in range(n):
for j in range(n):
if(((xi[i]^yi[j])in x)or((xi[i]^yi[j])in y)):
c=c+1
if (c%2==0):
print("Karen")
else:
print("Koyomi") | Title: The Artful Expedient
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game.
Input Specification:
The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*.
Output Specification:
Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization.
Demo Input:
['3\n1 2 3\n4 5 6\n', '5\n2 4 6 8 10\n9 7 5 3 1\n']
Demo Output:
['Karen\n', 'Karen\n']
Note:
In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again. | ```python
n=int(input())
xi=[]
yi=[]
for i in range(n):
xi=xi+list(map(int,input().split()))
for i in range(n):
yi=yi+list(map(int,input().split()))
c=0
x=set(xi)
y=set(yi)
for i in range(n):
for j in range(n):
if(((xi[i]^yi[j])in x)or((xi[i]^yi[j])in y)):
c=c+1
if (c%2==0):
print("Karen")
else:
print("Koyomi")
``` | -1 | |
40 | E | Number Table | PROGRAMMING | 2,500 | [
"combinatorics"
] | E. Number Table | 2 | 216 | As it has been found out recently, all the Berland's current economical state can be described using a simple table *n*<=×<=*m* in size. *n* — the number of days in each Berland month, *m* — the number of months. Thus, a table cell corresponds to a day and a month of the Berland's year. Each cell will contain either 1, or -1, which means the state's gains in a particular month, on a particular day. 1 corresponds to profits, -1 corresponds to losses. It turned out important for successful development to analyze the data on the state of the economy of the previous year, however when the treasurers referred to the archives to retrieve the data, it turned out that the table had been substantially damaged. In some table cells the number values had faded and were impossible to be deciphered. It is known that the number of cells in which the data had been preserved is strictly less than *max*(*n*,<=*m*). However, there is additional information — the product of the numbers in each line and column equaled -1. Your task is to find out how many different tables may conform to the preserved data. As the answer to the task can be quite large, you have to find it modulo *p*. | The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000). The second line contains the integer *k* (0<=≤<=*k*<=<<=*max*(*n*,<=*m*)) — the number of cells in which the data had been preserved. The next *k* lines contain the data on the state of the table in the preserved cells. Each line is of the form "*a* *b* *c*", where *a* (1<=≤<=*a*<=≤<=*n*) — the number of the table row, *b* (1<=≤<=*b*<=≤<=*m*) — the number of the column, *c* — the value containing in the cell (1 or -1). They are numbered starting from 1. It is guaranteed that no two lines with same *a* and *b* values exist. The last line contains an integer *p* (2<=≤<=*p*<=≤<=109<=+<=7). | Print the number of different tables that could conform to the preserved data modulo *p*. | [
"2 2\n0\n100\n",
"2 2\n1\n1 1 -1\n100\n"
] | [
"2\n",
"1\n"
] | none | 2,500 | [
{
"input": "2 2\n0\n100",
"output": "2"
},
{
"input": "2 2\n1\n1 1 -1\n100",
"output": "1"
},
{
"input": "7 6\n4\n3 2 -1\n7 1 1\n2 1 1\n3 4 -1\n179460773",
"output": "0"
},
{
"input": "6 2\n3\n5 1 -1\n4 2 1\n1 2 -1\n958300150",
"output": "4"
},
{
"input": "6 9\n3\... | 1,689,635,992 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | print("_RANDOM_GUESS_1689635992.7509944")# 1689635992.7510135 | Title: Number Table
Time Limit: 2 seconds
Memory Limit: 216 megabytes
Problem Description:
As it has been found out recently, all the Berland's current economical state can be described using a simple table *n*<=×<=*m* in size. *n* — the number of days in each Berland month, *m* — the number of months. Thus, a table cell corresponds to a day and a month of the Berland's year. Each cell will contain either 1, or -1, which means the state's gains in a particular month, on a particular day. 1 corresponds to profits, -1 corresponds to losses. It turned out important for successful development to analyze the data on the state of the economy of the previous year, however when the treasurers referred to the archives to retrieve the data, it turned out that the table had been substantially damaged. In some table cells the number values had faded and were impossible to be deciphered. It is known that the number of cells in which the data had been preserved is strictly less than *max*(*n*,<=*m*). However, there is additional information — the product of the numbers in each line and column equaled -1. Your task is to find out how many different tables may conform to the preserved data. As the answer to the task can be quite large, you have to find it modulo *p*.
Input Specification:
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000). The second line contains the integer *k* (0<=≤<=*k*<=<<=*max*(*n*,<=*m*)) — the number of cells in which the data had been preserved. The next *k* lines contain the data on the state of the table in the preserved cells. Each line is of the form "*a* *b* *c*", where *a* (1<=≤<=*a*<=≤<=*n*) — the number of the table row, *b* (1<=≤<=*b*<=≤<=*m*) — the number of the column, *c* — the value containing in the cell (1 or -1). They are numbered starting from 1. It is guaranteed that no two lines with same *a* and *b* values exist. The last line contains an integer *p* (2<=≤<=*p*<=≤<=109<=+<=7).
Output Specification:
Print the number of different tables that could conform to the preserved data modulo *p*.
Demo Input:
['2 2\n0\n100\n', '2 2\n1\n1 1 -1\n100\n']
Demo Output:
['2\n', '1\n']
Note:
none | ```python
print("_RANDOM_GUESS_1689635992.7509944")# 1689635992.7510135
``` | 0 |
300 | A | Array | PROGRAMMING | 1,100 | [
"brute force",
"constructive algorithms",
"implementation"
] | null | null | Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array. | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements. | In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them. | [
"3\n-1 2 0\n",
"4\n-1 -2 -3 0\n"
] | [
"1 -1\n1 2\n1 0\n",
"1 -1\n2 -3 -2\n1 0\n"
] | none | 500 | [
{
"input": "3\n-1 2 0",
"output": "1 -1\n1 2\n1 0"
},
{
"input": "4\n-1 -2 -3 0",
"output": "1 -1\n2 -3 -2\n1 0"
},
{
"input": "5\n-1 -2 1 2 0",
"output": "1 -1\n2 1 2\n2 0 -2"
},
{
"input": "100\n-64 -51 -75 -98 74 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -1... | 1,611,474,596 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 248 | 102,400 | from math import *
from collections import *
from sys import *
t=stdin.readline
p=stdout.write
def GI(): return map(int, t().strip().split())
def GS(): return map(str, t().strip().split())
def GL(): return list(map(int, t().strip().split()))
def SL(): return list(map(str, t().strip().split()))
n=int(t())
a=sorted(GL())
l1,l2,l3=[],[],[]
while(len(a)>0):
x=a.pop()
if x<0: l1.append(x)
elif x>0: l2.append(x)
else: l3.append(x)
if len(l2)!=0:
print(len(l1),*l1)
print(len(l2),*l2)
print(len(l3),*l3)
else:
print(len(l1)-2,*l1[:-2])
print('2',*l1[-2:])
print(len(l3),*l3) | Title: Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements.
Output Specification:
In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Demo Input:
['3\n-1 2 0\n', '4\n-1 -2 -3 0\n']
Demo Output:
['1 -1\n1 2\n1 0\n', '1 -1\n2 -3 -2\n1 0\n']
Note:
none | ```python
from math import *
from collections import *
from sys import *
t=stdin.readline
p=stdout.write
def GI(): return map(int, t().strip().split())
def GS(): return map(str, t().strip().split())
def GL(): return list(map(int, t().strip().split()))
def SL(): return list(map(str, t().strip().split()))
n=int(t())
a=sorted(GL())
l1,l2,l3=[],[],[]
while(len(a)>0):
x=a.pop()
if x<0: l1.append(x)
elif x>0: l2.append(x)
else: l3.append(x)
if len(l2)!=0:
print(len(l1),*l1)
print(len(l2),*l2)
print(len(l3),*l3)
else:
print(len(l1)-2,*l1[:-2])
print('2',*l1[-2:])
print(len(l3),*l3)
``` | 0 | |
707 | A | Brain's Photos | PROGRAMMING | 800 | [
"implementation"
] | null | null | Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. | Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. | [
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] | [
"#Color",
"#Black&White",
"#Black&White"
] | none | 500 | [
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
... | 1,690,953,693 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 50 | 62 | 1,638,400 | M,N = map(int,input().split())
def COLOR(M):
Flag = False
for _ in range(M):
X = list(input().split())
for i in X:
if i == 'C' or i == 'M' or i == 'Y':
Flag = True
return "#Black&White" if Flag==False else "#Color"
print(COLOR(M)) | Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none | ```python
M,N = map(int,input().split())
def COLOR(M):
Flag = False
for _ in range(M):
X = list(input().split())
for i in X:
if i == 'C' or i == 'M' or i == 'Y':
Flag = True
return "#Black&White" if Flag==False else "#Color"
print(COLOR(M))
``` | 3 | |
467 | A | George and Accommodation | PROGRAMMING | 800 | [
"implementation"
] | null | null | George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity. | Print a single integer — the number of rooms where George and Alex can move in. | [
"3\n1 1\n2 2\n3 3\n",
"3\n1 10\n0 10\n10 10\n"
] | [
"0\n",
"2\n"
] | none | 500 | [
{
"input": "3\n1 1\n2 2\n3 3",
"output": "0"
},
{
"input": "3\n1 10\n0 10\n10 10",
"output": "2"
},
{
"input": "2\n36 67\n61 69",
"output": "2"
},
{
"input": "3\n21 71\n10 88\n43 62",
"output": "3"
},
{
"input": "3\n1 2\n2 3\n3 4",
"output": "0"
},
{
"... | 1,694,781,705 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 46 | 0 | N = int(input())
cnt = 0
for i in range(N):
X,Y = map(int,input().split())
if((Y-X)>=2):
cnt = cnt+1
print(cnt) | Title: George and Accommodation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
Output Specification:
Print a single integer — the number of rooms where George and Alex can move in.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '3\n1 10\n0 10\n10 10\n']
Demo Output:
['0\n', '2\n']
Note:
none | ```python
N = int(input())
cnt = 0
for i in range(N):
X,Y = map(int,input().split())
if((Y-X)>=2):
cnt = cnt+1
print(cnt)
``` | 3 | |
862 | B | Mahmoud and Ehab and the bipartiteness | PROGRAMMING | 1,300 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same . | The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree. | Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions. | [
"3\n1 2\n1 3\n",
"5\n1 2\n2 3\n3 4\n4 5\n"
] | [
"0\n",
"2\n"
] | Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5). | 1,000 | [
{
"input": "3\n1 2\n1 3",
"output": "0"
},
{
"input": "5\n1 2\n2 3\n3 4\n4 5",
"output": "2"
},
{
"input": "10\n3 8\n6 2\n9 7\n10 1\n3 5\n1 3\n6 7\n5 4\n3 6",
"output": "16"
},
{
"input": "10\n7 6\n2 7\n4 1\n8 5\n9 4\n5 3\n8 7\n10 8\n10 4",
"output": "16"
},
{
"in... | 1,595,618,693 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 16 | 1,044 | 34,508,800 | cnt = [0,0]
n = int(input())
v = {i:[] for i in range(n)}
def dfs(node,pnode,color):
global cnt,v
cnt[color] += 1
for i in range(len(v[node])):
if v[node][i] != pnode:
dfs(v[node][i],node,not color)
for _ in range(n-1):
a,b = map(int,input().split())
v[a-1].append(b-1)
v[b-1].append(a-1)
dfs(0,0,0)
print(cnt[0]*cnt[1] - n + 1) | Title: Mahmoud and Ehab and the bipartiteness
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .
Input Specification:
The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree.
Output Specification:
Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.
Demo Input:
['3\n1 2\n1 3\n', '5\n1 2\n2 3\n3 4\n4 5\n']
Demo Output:
['0\n', '2\n']
Note:
Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5). | ```python
cnt = [0,0]
n = int(input())
v = {i:[] for i in range(n)}
def dfs(node,pnode,color):
global cnt,v
cnt[color] += 1
for i in range(len(v[node])):
if v[node][i] != pnode:
dfs(v[node][i],node,not color)
for _ in range(n-1):
a,b = map(int,input().split())
v[a-1].append(b-1)
v[b-1].append(a-1)
dfs(0,0,0)
print(cnt[0]*cnt[1] - n + 1)
``` | -1 | |
116 | A | Tram | PROGRAMMING | 800 | [
"implementation"
] | null | null | Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram. | The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0. | Print a single integer denoting the minimum possible capacity of the tram (0 is allowed). | [
"4\n0 3\n2 5\n4 2\n4 0\n"
] | [
"6\n"
] | For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | 500 | [
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1... | 1,690,969,346 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 122 | 0 | c=0
for i in range (int(input())):
a,b=map(int,input().split())
if b-a>0:
c+=b-a
print(c)
| Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | ```python
c=0
for i in range (int(input())):
a,b=map(int,input().split())
if b-a>0:
c+=b-a
print(c)
``` | 0 | |
16 | B | Burglar and Matches | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | B. Burglar and Matches | 0 | 64 | A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer. | Output the only number — answer to the problem. | [
"7 3\n5 10\n2 5\n3 6\n",
"3 3\n1 3\n2 2\n3 1\n"
] | [
"62\n",
"7\n"
] | none | 0 | [
{
"input": "7 3\n5 10\n2 5\n3 6",
"output": "62"
},
{
"input": "3 3\n1 3\n2 2\n3 1",
"output": "7"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "1 2\n1 9\n1 6",
"output": "9"
},
{
"input": "1 10\n1 1\n1 9\n1 3\n1 9\n1 7\n1 10\n1 4\n1 7\n1 3\n1 1",
"ou... | 1,601,564,069 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 8 | 280 | 2,048,000 | n,m=map(int,input().split())
mb=0
boxes=[]
for i in range(m):
a,b=map(int,input().split())
boxes.append([a,b])
def take_second(elem):
return elem[1]
boxes=sorted(boxes, key=take_second,reverse=True)
i=0
total=0
while mb<n:
curr=boxes[i]
if(curr[0]>0):
curr[0]-=1
mb+=1
total+=curr[1]
else:
i+=1
print(total)
| Title: Burglar and Matches
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
Output Specification:
Output the only number — answer to the problem.
Demo Input:
['7 3\n5 10\n2 5\n3 6\n', '3 3\n1 3\n2 2\n3 1\n']
Demo Output:
['62\n', '7\n']
Note:
none | ```python
n,m=map(int,input().split())
mb=0
boxes=[]
for i in range(m):
a,b=map(int,input().split())
boxes.append([a,b])
def take_second(elem):
return elem[1]
boxes=sorted(boxes, key=take_second,reverse=True)
i=0
total=0
while mb<n:
curr=boxes[i]
if(curr[0]>0):
curr[0]-=1
mb+=1
total+=curr[1]
else:
i+=1
print(total)
``` | -1 |
228 | A | Is your horseshoe on the other hoof? | PROGRAMMING | 800 | [
"implementation"
] | null | null | Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. | The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers. | Print a single integer — the minimum number of horseshoes Valera needs to buy. | [
"1 7 3 3\n",
"7 7 7 7\n"
] | [
"1\n",
"3\n"
] | none | 500 | [
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 58697... | 1,690,597,253 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | number1,number2,number3,number4=input().split()
number1,number2,number3,number4=int(number1),int(number2),int(number3),int(number4)
count=1
list_=[number1,number2,number3,number4]
list_.sort()
for i in range(3):
if(list_[i]!=list_[i+1]):
count+=1
print(4-count) | Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
number1,number2,number3,number4=input().split()
number1,number2,number3,number4=int(number1),int(number2),int(number3),int(number4)
count=1
list_=[number1,number2,number3,number4]
list_.sort()
for i in range(3):
if(list_[i]!=list_[i+1]):
count+=1
print(4-count)
``` | 3 | |
895 | A | Pizza Separation | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360. | Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya. | [
"4\n90 90 90 90\n",
"3\n100 100 160\n",
"1\n360\n",
"4\n170 30 150 10\n"
] | [
"0\n",
"40\n",
"360\n",
"0\n"
] | In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | 500 | [
{
"input": "4\n90 90 90 90",
"output": "0"
},
{
"input": "3\n100 100 160",
"output": "40"
},
{
"input": "1\n360",
"output": "360"
},
{
"input": "4\n170 30 150 10",
"output": "0"
},
{
"input": "5\n10 10 10 10 320",
"output": "280"
},
{
"input": "8\n45 4... | 1,511,757,505 | 2,147,483,647 | Python 3 | OK | TESTS | 93 | 498 | 5,632,000 | n=int(input())
from itertools import permutations
a=list(map(int,input().split()))
b=361
for i in range(n):
for j in range(i,n):
b=min(b,abs(sum(a[::])-sum(a[i:j])*2))
print(b) | Title: Pizza Separation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
Output Specification:
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.
Demo Input:
['4\n90 90 90 90\n', '3\n100 100 160\n', '1\n360\n', '4\n170 30 150 10\n']
Demo Output:
['0\n', '40\n', '360\n', '0\n']
Note:
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | ```python
n=int(input())
from itertools import permutations
a=list(map(int,input().split()))
b=361
for i in range(n):
for j in range(i,n):
b=min(b,abs(sum(a[::])-sum(a[i:j])*2))
print(b)
``` | 3 | |
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"outp... | 1,661,100,006 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 92 | 0 | a = input()
flag=0
x = a[0]
for i in range(len(a)-1):
# x = a[i]
if(x==a[i+1]):
flag+=1
else:
x = a[i+1]
flag+=1
if(flag >= 7):
print('YES')
else:
print('NO') | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
a = input()
flag=0
x = a[0]
for i in range(len(a)-1):
# x = a[i]
if(x==a[i+1]):
flag+=1
else:
x = a[i+1]
flag+=1
if(flag >= 7):
print('YES')
else:
print('NO')
``` | 0 |
233 | A | Perfect Permutation | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*. | A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size. | If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces. | [
"1\n",
"2\n",
"4\n"
] | [
"-1\n",
"2 1 \n",
"2 1 4 3 \n"
] | none | 500 | [
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "4",
"output": "2 1 4 3 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "2 1 4 3 6 5 "
},
{
"input": "7",
... | 1,648,640,432 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | a=int(input())
if a==1:
print(-1)
else:
c=''
for i in range(a):
c+=str(random.randint(1,a))+' '
print(c)
| Title: Perfect Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
Input Specification:
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
Output Specification:
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
Demo Input:
['1\n', '2\n', '4\n']
Demo Output:
['-1\n', '2 1 \n', '2 1 4 3 \n']
Note:
none | ```python
a=int(input())
if a==1:
print(-1)
else:
c=''
for i in range(a):
c+=str(random.randint(1,a))+' '
print(c)
``` | -1 | |
770 | A | New Password | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions. | The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists. | Print any password which satisfies all conditions given by Innokentiy. | [
"4 3\n",
"6 6\n",
"5 2\n"
] | [
"java\n",
"python\n",
"phphp\n"
] | In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | 500 | [
{
"input": "4 3",
"output": "abca"
},
{
"input": "6 6",
"output": "abcdef"
},
{
"input": "5 2",
"output": "ababa"
},
{
"input": "3 2",
"output": "aba"
},
{
"input": "10 2",
"output": "ababababab"
},
{
"input": "26 13",
"output": "abcdefghijklmabcde... | 1,685,119,950 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 61 | 1,126,400 | import string
import random
n , k = map(int , input().split())
def generate(n , k):
p = ''
r = string.ascii_letters[:k]
while(True):
if len(p) < n:
p +=r
else:
break
return p[:n]
print(generate(n,k))
| Title: New Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
Input Specification:
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
Output Specification:
Print any password which satisfies all conditions given by Innokentiy.
Demo Input:
['4 3\n', '6 6\n', '5 2\n']
Demo Output:
['java\n', 'python\n', 'phphp\n']
Note:
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | ```python
import string
import random
n , k = map(int , input().split())
def generate(n , k):
p = ''
r = string.ascii_letters[:k]
while(True):
if len(p) < n:
p +=r
else:
break
return p[:n]
print(generate(n,k))
``` | 3 | |
203 | A | Two Problems | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | A boy Valera registered on site Codeforces as Valera, and wrote his first Codeforces Round #300. He boasted to a friend Arkady about winning as much as *x* points for his first contest. But Arkady did not believe his friend's words and decided to check whether Valera could have shown such a result.
He knows that the contest number 300 was unusual because there were only two problems. The contest lasted for *t* minutes, the minutes are numbered starting from zero. The first problem had the initial cost of *a* points, and every minute its cost reduced by *d**a* points. The second problem had the initial cost of *b* points, and every minute this cost reduced by *d**b* points. Thus, as soon as the zero minute of the contest is over, the first problem will cost *a*<=-<=*d**a* points, and the second problem will cost *b*<=-<=*d**b* points. It is guaranteed that at any moment of the contest each problem has a non-negative cost.
Arkady asks you to find out whether Valera could have got exactly *x* points for this contest. You should assume that Valera could have solved any number of the offered problems. You should also assume that for each problem Valera made no more than one attempt, besides, he could have submitted both problems at the same minute of the contest, starting with minute 0 and ending with minute number *t*<=-<=1. Please note that Valera can't submit a solution exactly *t* minutes after the start of the contest or later. | The single line of the input contains six integers *x*,<=*t*,<=*a*,<=*b*,<=*d**a*,<=*d**b* (0<=≤<=*x*<=≤<=600; 1<=≤<=*t*,<=*a*,<=*b*,<=*d**a*,<=*d**b*<=≤<=300) — Valera's result, the contest's duration, the initial cost of the first problem, the initial cost of the second problem, the number of points that the first and the second problem lose per minute, correspondingly.
It is guaranteed that at each minute of the contest each problem has a non-negative cost, that is, *a*<=-<=*i*·*d**a*<=≥<=0 and *b*<=-<=*i*·*d**b*<=≥<=0 for all 0<=≤<=*i*<=≤<=*t*<=-<=1. | If Valera could have earned exactly *x* points at a contest, print "YES", otherwise print "NO" (without the quotes). | [
"30 5 20 20 3 5\n",
"10 4 100 5 5 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample Valera could have acted like this: he could have submitted the first problem at minute 0 and the second problem — at minute 2. Then the first problem brings him 20 points and the second problem brings him 10 points, that in total gives the required 30 points. | 500 | [
{
"input": "30 5 20 20 3 5",
"output": "YES"
},
{
"input": "10 4 100 5 5 1",
"output": "NO"
},
{
"input": "0 7 30 50 3 4",
"output": "YES"
},
{
"input": "50 10 30 20 1 2",
"output": "YES"
},
{
"input": "40 1 40 5 11 2",
"output": "YES"
},
{
"input": "3... | 1,683,668,676 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 92 | 0 | x, t, a, b, da, db = map(int, input().split())
if x == 0:
print("YES")
else:
ans = False
t = 0
for i in range(da):
cur_b = b
for j in range(db):
if a + cur_b == x:
ans = True
break
t += 1
cur_b -= db
if t == 5:
break
a -= da
t += 1
if t >= 5:
t = 0
if ans:
print("YES")
else:
print("NO") | Title: Two Problems
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A boy Valera registered on site Codeforces as Valera, and wrote his first Codeforces Round #300. He boasted to a friend Arkady about winning as much as *x* points for his first contest. But Arkady did not believe his friend's words and decided to check whether Valera could have shown such a result.
He knows that the contest number 300 was unusual because there were only two problems. The contest lasted for *t* minutes, the minutes are numbered starting from zero. The first problem had the initial cost of *a* points, and every minute its cost reduced by *d**a* points. The second problem had the initial cost of *b* points, and every minute this cost reduced by *d**b* points. Thus, as soon as the zero minute of the contest is over, the first problem will cost *a*<=-<=*d**a* points, and the second problem will cost *b*<=-<=*d**b* points. It is guaranteed that at any moment of the contest each problem has a non-negative cost.
Arkady asks you to find out whether Valera could have got exactly *x* points for this contest. You should assume that Valera could have solved any number of the offered problems. You should also assume that for each problem Valera made no more than one attempt, besides, he could have submitted both problems at the same minute of the contest, starting with minute 0 and ending with minute number *t*<=-<=1. Please note that Valera can't submit a solution exactly *t* minutes after the start of the contest or later.
Input Specification:
The single line of the input contains six integers *x*,<=*t*,<=*a*,<=*b*,<=*d**a*,<=*d**b* (0<=≤<=*x*<=≤<=600; 1<=≤<=*t*,<=*a*,<=*b*,<=*d**a*,<=*d**b*<=≤<=300) — Valera's result, the contest's duration, the initial cost of the first problem, the initial cost of the second problem, the number of points that the first and the second problem lose per minute, correspondingly.
It is guaranteed that at each minute of the contest each problem has a non-negative cost, that is, *a*<=-<=*i*·*d**a*<=≥<=0 and *b*<=-<=*i*·*d**b*<=≥<=0 for all 0<=≤<=*i*<=≤<=*t*<=-<=1.
Output Specification:
If Valera could have earned exactly *x* points at a contest, print "YES", otherwise print "NO" (without the quotes).
Demo Input:
['30 5 20 20 3 5\n', '10 4 100 5 5 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Valera could have acted like this: he could have submitted the first problem at minute 0 and the second problem — at minute 2. Then the first problem brings him 20 points and the second problem brings him 10 points, that in total gives the required 30 points. | ```python
x, t, a, b, da, db = map(int, input().split())
if x == 0:
print("YES")
else:
ans = False
t = 0
for i in range(da):
cur_b = b
for j in range(db):
if a + cur_b == x:
ans = True
break
t += 1
cur_b -= db
if t == 5:
break
a -= da
t += 1
if t >= 5:
t = 0
if ans:
print("YES")
else:
print("NO")
``` | 0 | |
688 | B | Lovely Palindromes | PROGRAMMING | 1,000 | [
"constructive algorithms",
"math"
] | null | null | Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number? | The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000). | Print the *n*-th even-length palindrome number. | [
"1\n",
"10\n"
] | [
"11\n",
"1001\n"
] | The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001. | 1,000 | [
{
"input": "1",
"output": "11"
},
{
"input": "10",
"output": "1001"
},
{
"input": "11",
"output": "1111"
},
{
"input": "12",
"output": "1221"
},
{
"input": "100",
"output": "100001"
},
{
"input": "1321",
"output": "13211231"
},
{
"input": "... | 1,636,193,611 | 2,147,483,647 | PyPy 3 | OK | TESTS | 40 | 109 | 22,835,200 | s = input()
res = ['0'] * (2* len(s))
for i in range(len(s)):
res[i] = res[len(res)-1-i] = s[i]
print(''.join(res))
| Title: Lovely Palindromes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
Output Specification:
Print the *n*-th even-length palindrome number.
Demo Input:
['1\n', '10\n']
Demo Output:
['11\n', '1001\n']
Note:
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001. | ```python
s = input()
res = ['0'] * (2* len(s))
for i in range(len(s)):
res[i] = res[len(res)-1-i] = s[i]
print(''.join(res))
``` | 3 | |
433 | A | Kitahara Haruki's Gift | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends? | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple. | In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes). | [
"3\n100 200 100\n",
"4\n100 100 100 200\n"
] | [
"YES\n",
"NO\n"
] | In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | 500 | [
{
"input": "3\n100 200 100",
"output": "YES"
},
{
"input": "4\n100 100 100 200",
"output": "NO"
},
{
"input": "1\n100",
"output": "NO"
},
{
"input": "1\n200",
"output": "NO"
},
{
"input": "2\n100 100",
"output": "YES"
},
{
"input": "2\n200 200",
"o... | 1,658,899,009 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 77 | 0 | import sys
def _input(): return map(int, input().split())
n = int(input())
lst = list(_input())
if lst.count(100)%2==0: print("YES")
else: print("NO")
| Title: Kitahara Haruki's Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
Output Specification:
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
Demo Input:
['3\n100 200 100\n', '4\n100 100 100 200\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | ```python
import sys
def _input(): return map(int, input().split())
n = int(input())
lst = list(_input())
if lst.count(100)%2==0: print("YES")
else: print("NO")
``` | 0 | |
228 | A | Is your horseshoe on the other hoof? | PROGRAMMING | 800 | [
"implementation"
] | null | null | Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. | The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers. | Print a single integer — the minimum number of horseshoes Valera needs to buy. | [
"1 7 3 3\n",
"7 7 7 7\n"
] | [
"1\n",
"3\n"
] | none | 500 | [
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 58697... | 1,690,865,071 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | def minimum_horseshoes_to_buy(colors):
unique_colors = len(set(colors))
return max(0, 4 - unique_colors)
colors = list(map(int, input().split()))
output_result = minimum_horseshoes_to_buy(colors)
print(output_result)
| Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
def minimum_horseshoes_to_buy(colors):
unique_colors = len(set(colors))
return max(0, 4 - unique_colors)
colors = list(map(int, input().split()))
output_result = minimum_horseshoes_to_buy(colors)
print(output_result)
``` | 3 | |
268 | A | Games | PROGRAMMING | 800 | [
"brute force"
] | null | null | Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question. | The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively. | In a single line print the number of games where the host team is going to play in the guest uniform. | [
"3\n1 2\n2 4\n3 4\n",
"4\n100 42\n42 100\n5 42\n100 5\n",
"2\n1 2\n1 2\n"
] | [
"1\n",
"5\n",
"0\n"
] | In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first). | 500 | [
{
"input": "3\n1 2\n2 4\n3 4",
"output": "1"
},
{
"input": "4\n100 42\n42 100\n5 42\n100 5",
"output": "5"
},
{
"input": "2\n1 2\n1 2",
"output": "0"
},
{
"input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52",
"output": "6"
},
{
"input": "10\n68 42\n1 35\n25 70\n... | 1,682,385,119 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | ls = []
counter = 0
for i in range(int(input())):
ls_temp = [int(x) for x in input().split()]
ls.append(ls_temp)
for home in ls :
for away in ls :
if home[0]==away[1] :
counter += 1
print(counter)
| Title: Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
Output Specification:
In a single line print the number of games where the host team is going to play in the guest uniform.
Demo Input:
['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n']
Demo Output:
['1\n', '5\n', '0\n']
Note:
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first). | ```python
ls = []
counter = 0
for i in range(int(input())):
ls_temp = [int(x) for x in input().split()]
ls.append(ls_temp)
for home in ls :
for away in ls :
if home[0]==away[1] :
counter += 1
print(counter)
``` | 3 | |
912 | B | New Year's Eve | PROGRAMMING | 1,300 | [
"bitmasks",
"constructive algorithms",
"number theory"
] | null | null | Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain. | The sole string contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1018). | Output one number — the largest possible xor-sum. | [
"4 3\n",
"6 6\n"
] | [
"7\n",
"7\n"
] | In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7. | 1,000 | [
{
"input": "4 3",
"output": "7"
},
{
"input": "6 6",
"output": "7"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "1022 10",
"output": "1023"
},
{
"input": "415853337373441 52",
"output": "562949953421311"
},
{
"input": "75 12",
"output": "127"
... | 1,518,550,434 | 2,147,483,647 | PyPy 3 | OK | TESTS | 58 | 77 | 19,968,000 | n, k = [int(f) for f in input().split(' ') if f]
if k == 1:
print(n)
else:
l = 0
while n > 0:
l += 1
n //= 2
print(2 ** l - 1)
| Title: New Year's Eve
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.
Input Specification:
The sole string contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1018).
Output Specification:
Output one number — the largest possible xor-sum.
Demo Input:
['4 3\n', '6 6\n']
Demo Output:
['7\n', '7\n']
Note:
In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7. | ```python
n, k = [int(f) for f in input().split(' ') if f]
if k == 1:
print(n)
else:
l = 0
while n > 0:
l += 1
n //= 2
print(2 ** l - 1)
``` | 3 | |
275 | B | Convex Shape | PROGRAMMING | 1,700 | [
"constructive algorithms",
"implementation"
] | null | null | Consider an *n*<=×<=*m* grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
You're given a painted grid in the input. Tell Lenny if the grid is convex or not. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the size of the grid. Each of the next *n* lines contains *m* characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell. | On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes. | [
"3 4\nWWBW\nBWWW\nWWWB\n",
"3 1\nB\nB\nW\n"
] | [
"NO\n",
"YES\n"
] | none | 1,000 | [
{
"input": "3 4\nWWBW\nBWWW\nWWWB",
"output": "NO"
},
{
"input": "3 1\nB\nB\nW",
"output": "YES"
},
{
"input": "1 1\nB",
"output": "YES"
},
{
"input": "1 2\nBB",
"output": "YES"
},
{
"input": "2 1\nB\nB",
"output": "YES"
},
{
"input": "1 2\nBW",
"o... | 1,658,010,145 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 46 | 280 | 3,993,600 | import bisect
import itertools
import sys
import collections
n, m = map(int, input().split())
grid = [list(input()) for _ in range(n)]
# print('\n'.join(map(str, grid)))
# row check
good = True
for i in range(n):
b = False
w = False
for j in range(m):
if grid[i][j] == 'B' and not b:
b = True
elif grid[i][j] == 'W' and b and not w:
w = True
elif grid[i][j] == 'B' and b and w:
good = False
# print('rowcheck', i, j)
break
if not good:
break
# col check
if good:
for j in range(m):
b = False
w = False
for i in range(n):
if grid[i][j] == 'B' and not b and not w:
b = True
elif grid[i][j] == 'W' and b and not w:
w = True
elif grid[i][j] == 'B' and b and w:
good = False
# print('colcheck', i, j)
break
if not good:
break
# 2 change check
if good:
points = [(i, j) for i in range(n) for j in range(m) if grid[i][j] == 'B']
for (i1, j1), (i2, j2) in itertools.combinations(points, 2):
if grid[i1][j2] == 'W' and grid[i2][j1] == 'W':
# print('2change', i1, j1, i2, j2)
good = False
break
print("YES" if good else "NO") | Title: Convex Shape
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider an *n*<=×<=*m* grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the size of the grid. Each of the next *n* lines contains *m* characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output Specification:
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Demo Input:
['3 4\nWWBW\nBWWW\nWWWB\n', '3 1\nB\nB\nW\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
import bisect
import itertools
import sys
import collections
n, m = map(int, input().split())
grid = [list(input()) for _ in range(n)]
# print('\n'.join(map(str, grid)))
# row check
good = True
for i in range(n):
b = False
w = False
for j in range(m):
if grid[i][j] == 'B' and not b:
b = True
elif grid[i][j] == 'W' and b and not w:
w = True
elif grid[i][j] == 'B' and b and w:
good = False
# print('rowcheck', i, j)
break
if not good:
break
# col check
if good:
for j in range(m):
b = False
w = False
for i in range(n):
if grid[i][j] == 'B' and not b and not w:
b = True
elif grid[i][j] == 'W' and b and not w:
w = True
elif grid[i][j] == 'B' and b and w:
good = False
# print('colcheck', i, j)
break
if not good:
break
# 2 change check
if good:
points = [(i, j) for i in range(n) for j in range(m) if grid[i][j] == 'B']
for (i1, j1), (i2, j2) in itertools.combinations(points, 2):
if grid[i1][j2] == 'W' and grid[i2][j1] == 'W':
# print('2change', i1, j1, i2, j2)
good = False
break
print("YES" if good else "NO")
``` | 3 | |
667 | A | Pouring Rain | PROGRAMMING | 1,100 | [
"geometry",
"math"
] | null | null | A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do.
Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation.
Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom.
You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously.
Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds.
Note one milliliter equals to one cubic centimeter. | The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where:
- *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup. | If it is impossible to make the cup empty, print "NO" (without quotes).
Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104. | [
"1 2 3 100\n",
"1 1 1 1\n"
] | [
"NO\n",
"YES\n3.659792366325\n"
] | In the first example the water fills the cup faster than you can drink from it.
In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds. | 500 | [
{
"input": "1 2 3 100",
"output": "NO"
},
{
"input": "1 1 1 1",
"output": "YES\n3.659792366325"
},
{
"input": "48 7946 7992 72",
"output": "NO"
},
{
"input": "72 6791 8546 46",
"output": "NO"
},
{
"input": "100 5635 9099 23",
"output": "NO"
},
{
"input... | 1,462,210,801 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 62 | 5,120,000 | d,h,v,e = map(int,input().split(' '))
PI = 3.1415926
S = (d/2)*(d/2)*PI
V = h*S
V2 = e*S
if(V2 >= v):
print("NO")
exit
else:
print("YES")
print(abs(V/(V2-v)))
| Title: Pouring Rain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do.
Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation.
Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom.
You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously.
Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds.
Note one milliliter equals to one cubic centimeter.
Input Specification:
The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where:
- *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup.
Output Specification:
If it is impossible to make the cup empty, print "NO" (without quotes).
Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104.
Demo Input:
['1 2 3 100\n', '1 1 1 1\n']
Demo Output:
['NO\n', 'YES\n3.659792366325\n']
Note:
In the first example the water fills the cup faster than you can drink from it.
In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds. | ```python
d,h,v,e = map(int,input().split(' '))
PI = 3.1415926
S = (d/2)*(d/2)*PI
V = h*S
V2 = e*S
if(V2 >= v):
print("NO")
exit
else:
print("YES")
print(abs(V/(V2-v)))
``` | 3 | |
32 | A | Reconnaissance | PROGRAMMING | 800 | [
"brute force"
] | A. Reconnaissance | 2 | 256 | According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1,<=2) and (2,<=1) should be regarded as different. | The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*d*<=≤<=109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109. | Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*. | [
"5 10\n10 20 50 60 65\n",
"5 1\n55 30 29 31 55\n"
] | [
"6\n",
"6\n"
] | none | 500 | [
{
"input": "5 10\n10 20 50 60 65",
"output": "6"
},
{
"input": "5 1\n55 30 29 31 55",
"output": "6"
},
{
"input": "6 10\n4 6 4 1 9 3",
"output": "30"
},
{
"input": "7 100\n19 1694 261 162 1 234 513",
"output": "8"
},
{
"input": "8 42\n37 53 74 187 568 22 5 65",
... | 1,604,070,410 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 156 | 0 | a1 = int(input())
for i in range(a1):
a = input()
x = len(a)
if x <=10:
print(a)
else:
q = len(a[1:x-1])
s1 = a[:1] + str(q) + a[x-1:x]
print(s1)
| Title: Reconnaissance
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1,<=2) and (2,<=1) should be regarded as different.
Input Specification:
The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*d*<=≤<=109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.
Output Specification:
Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*.
Demo Input:
['5 10\n10 20 50 60 65\n', '5 1\n55 30 29 31 55\n']
Demo Output:
['6\n', '6\n']
Note:
none | ```python
a1 = int(input())
for i in range(a1):
a = input()
x = len(a)
if x <=10:
print(a)
else:
q = len(a[1:x-1])
s1 = a[:1] + str(q) + a[x-1:x]
print(s1)
``` | -1 |
166 | A | Rank List | PROGRAMMING | 1,100 | [
"binary search",
"implementation",
"sortings"
] | null | null | Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place. | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces. | In the only line print the sought number of teams that got the *k*-th place in the final results' table. | [
"7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n",
"5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n"
] | [
"3\n",
"4\n"
] | The final results' table for the first sample is:
- 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams. | 500 | [
{
"input": "7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10",
"output": "3"
},
{
"input": "5 4\n3 1\n3 1\n5 3\n3 1\n3 1",
"output": "4"
},
{
"input": "5 1\n2 2\n1 1\n1 1\n1 1\n2 2",
"output": "2"
},
{
"input": "6 3\n2 2\n3 1\n2 2\n4 5\n2 2\n4 5",
"output": "1"
},
{
"i... | 1,669,179,464 | 2,147,483,647 | PyPy 3 | OK | TESTS | 46 | 186 | 307,200 | from collections import defaultdict
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, k = map(int, input().split())
x = []
for _ in range(n):
p, t = map(int, input().split())
x.append(50 * p + 50 - t)
x.sort()
cnt = defaultdict(lambda : 0)
for i in x:
cnt[i] += 1
ans = cnt[x[-k]]
print(ans) | Title: Rank List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces.
Output Specification:
In the only line print the sought number of teams that got the *k*-th place in the final results' table.
Demo Input:
['7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n', '5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n']
Demo Output:
['3\n', '4\n']
Note:
The final results' table for the first sample is:
- 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams. | ```python
from collections import defaultdict
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, k = map(int, input().split())
x = []
for _ in range(n):
p, t = map(int, input().split())
x.append(50 * p + 50 - t)
x.sort()
cnt = defaultdict(lambda : 0)
for i in x:
cnt[i] += 1
ans = cnt[x[-k]]
print(ans)
``` | 3 | |
386 | A | Second-Price Auction | PROGRAMMING | 800 | [
"implementation"
] | null | null | In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different. | The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder. | The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based. | [
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] | [
"2 5\n",
"1 8\n",
"6 9\n"
] | none | 500 | [
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{... | 1,641,466,731 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 42 | 108 | 0 | n = int(input())
arr = list(map(int, input().split()))
arr2 = [i for i in arr]
arr2.sort()
x = arr2[-1]
y = arr2[-2]
print(arr.index(x)+1, end = " ")
print(y) | Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none | ```python
n = int(input())
arr = list(map(int, input().split()))
arr2 = [i for i in arr]
arr2.sort()
x = arr2[-1]
y = arr2[-2]
print(arr.index(x)+1, end = " ")
print(y)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Two bears are playing tic-tac-toe via mail. It's boring for them to play usual tic-tac-toe game, so they are a playing modified version of this game. Here are its rules.
The game is played on the following field.
Players are making moves by turns. At first move a player can put his chip in any cell of any small field. For following moves, there are some restrictions: if during last move the opposite player put his chip to cell with coordinates (*x**l*,<=*y**l*) in some small field, the next move should be done in one of the cells of the small field with coordinates (*x**l*,<=*y**l*). For example, if in the first move a player puts his chip to lower left cell of central field, then the second player on his next move should put his chip into some cell of lower left field (pay attention to the first test case). If there are no free cells in the required field, the player can put his chip to any empty cell on any field.
You are given current state of the game and coordinates of cell in which the last move was done. You should find all cells in which the current player can put his chip.
A hare works as a postman in the forest, he likes to foul bears. Sometimes he changes the game field a bit, so the current state of the game could be unreachable. However, after his changes the cell where the last move was done is not empty. You don't need to find if the state is unreachable or not, just output possible next moves according to the rules. | First 11 lines contains descriptions of table with 9 rows and 9 columns which are divided into 9 small fields by spaces and empty lines. Each small field is described by 9 characters without spaces and empty lines. character "x" (ASCII-code 120) means that the cell is occupied with chip of the first player, character "o" (ASCII-code 111) denotes a field occupied with chip of the second player, character "." (ASCII-code 46) describes empty cell.
The line after the table contains two integers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=9). They describe coordinates of the cell in table where the last move was done. Rows in the table are numbered from up to down and columns are numbered from left to right.
It's guaranteed that cell where the last move was done is filled with "x" or "o". Also, it's guaranteed that there is at least one empty cell. It's not guaranteed that current state of game is reachable. | Output the field in same format with characters "!" (ASCII-code 33) on positions where the current player can put his chip. All other cells should not be modified. | [
"... ... ...\n... ... ...\n... ... ...\n\n... ... ...\n... ... ...\n... x.. ...\n\n... ... ...\n... ... ...\n... ... ...\n6 4\n",
"xoo x.. x..\nooo ... ...\nooo ... ...\n\nx.. x.. x..\n... ... ...\n... ... ...\n\nx.. x.. x..\n... ... ...\n... ... ...\n7 4\n",
"o.. ... ...\n... ... ...\n... ... ...\n\n... xxx ..... | [
"... ... ... \n... ... ... \n... ... ... \n\n... ... ... \n... ... ... \n... x.. ... \n\n!!! ... ... \n!!! ... ... \n!!! ... ... \n\n",
"xoo x!! x!! \nooo !!! !!! \nooo !!! !!! \n\nx!! x!! x!! \n!!! !!! !!! \n!!! !!! !!! \n\nx!! x!! x!! \n!!! !!! !!! \n!!! !!! !!! \n\n",
"o!! !!! !!! \n!!! !!! !!! \n!!! !!! !!!... | In the first test case the first player made a move to lower left cell of central field, so the second player can put a chip only to cells of lower left field.
In the second test case the last move was done to upper left cell of lower central field, however all cells in upper left field are occupied, so the second player can put his chip to any empty cell.
In the third test case the last move was done to central cell of central field, so current player can put his chip to any cell of central field, which is already occupied, so he can move anywhere. Pay attention that this state of the game is unreachable. | 0 | [
{
"input": "... ... ...\n... ... ...\n... ... ...\n\n... ... ...\n... ... ...\n... x.. ...\n\n... ... ...\n... ... ...\n... ... ...\n6 4",
"output": "... ... ... \n... ... ... \n... ... ... \n\n... ... ... \n... ... ... \n... x.. ... \n\n!!! ... ... \n!!! ... ... \n!!! ... ... "
},
{
"input": "xoo x... | 1,514,042,652 | 4,752 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 61 | 5,529,600 | field = [input().split() for _ in range(11)]
del field[7]
del field[3]
last_y,last_x = [int(i)-1 for i in input().split()]
f_x = last_x % 3
f_y = last_y % 3
for i in range(3):
if '.' in field[f_y*3+i][f_x]:
for y in range(3):
field[f_y*3+y][f_x] = field[f_y*3+y][f_x].replace('.','!')
break;
else:
for y in range(9):
for x in range(3):
field[y][x] = field[y][x].replace('.','!')
print(str([str(s)[2:-2].replace("', '", ' ') for s in field])[2:-2].replace("', '",'\n'))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two bears are playing tic-tac-toe via mail. It's boring for them to play usual tic-tac-toe game, so they are a playing modified version of this game. Here are its rules.
The game is played on the following field.
Players are making moves by turns. At first move a player can put his chip in any cell of any small field. For following moves, there are some restrictions: if during last move the opposite player put his chip to cell with coordinates (*x**l*,<=*y**l*) in some small field, the next move should be done in one of the cells of the small field with coordinates (*x**l*,<=*y**l*). For example, if in the first move a player puts his chip to lower left cell of central field, then the second player on his next move should put his chip into some cell of lower left field (pay attention to the first test case). If there are no free cells in the required field, the player can put his chip to any empty cell on any field.
You are given current state of the game and coordinates of cell in which the last move was done. You should find all cells in which the current player can put his chip.
A hare works as a postman in the forest, he likes to foul bears. Sometimes he changes the game field a bit, so the current state of the game could be unreachable. However, after his changes the cell where the last move was done is not empty. You don't need to find if the state is unreachable or not, just output possible next moves according to the rules.
Input Specification:
First 11 lines contains descriptions of table with 9 rows and 9 columns which are divided into 9 small fields by spaces and empty lines. Each small field is described by 9 characters without spaces and empty lines. character "x" (ASCII-code 120) means that the cell is occupied with chip of the first player, character "o" (ASCII-code 111) denotes a field occupied with chip of the second player, character "." (ASCII-code 46) describes empty cell.
The line after the table contains two integers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=9). They describe coordinates of the cell in table where the last move was done. Rows in the table are numbered from up to down and columns are numbered from left to right.
It's guaranteed that cell where the last move was done is filled with "x" or "o". Also, it's guaranteed that there is at least one empty cell. It's not guaranteed that current state of game is reachable.
Output Specification:
Output the field in same format with characters "!" (ASCII-code 33) on positions where the current player can put his chip. All other cells should not be modified.
Demo Input:
['... ... ...\n... ... ...\n... ... ...\n\n... ... ...\n... ... ...\n... x.. ...\n\n... ... ...\n... ... ...\n... ... ...\n6 4\n', 'xoo x.. x..\nooo ... ...\nooo ... ...\n\nx.. x.. x..\n... ... ...\n... ... ...\n\nx.. x.. x..\n... ... ...\n... ... ...\n7 4\n', 'o.. ... ...\n... ... ...\n... ... ...\n\n... xxx ...\n... xox ...\n... ooo ...\n\n... ... ...\n... ... ...\n... ... ...\n5 5\n']
Demo Output:
['... ... ... \n... ... ... \n... ... ... \n\n... ... ... \n... ... ... \n... x.. ... \n\n!!! ... ... \n!!! ... ... \n!!! ... ... \n\n', 'xoo x!! x!! \nooo !!! !!! \nooo !!! !!! \n\nx!! x!! x!! \n!!! !!! !!! \n!!! !!! !!! \n\nx!! x!! x!! \n!!! !!! !!! \n!!! !!! !!! \n\n', 'o!! !!! !!! \n!!! !!! !!! \n!!! !!! !!! \n\n!!! xxx !!! \n!!! xox !!! \n!!! ooo !!! \n\n!!! !!! !!! \n!!! !!! !!! \n!!! !!! !!! \n\n']
Note:
In the first test case the first player made a move to lower left cell of central field, so the second player can put a chip only to cells of lower left field.
In the second test case the last move was done to upper left cell of lower central field, however all cells in upper left field are occupied, so the second player can put his chip to any empty cell.
In the third test case the last move was done to central cell of central field, so current player can put his chip to any cell of central field, which is already occupied, so he can move anywhere. Pay attention that this state of the game is unreachable. | ```python
field = [input().split() for _ in range(11)]
del field[7]
del field[3]
last_y,last_x = [int(i)-1 for i in input().split()]
f_x = last_x % 3
f_y = last_y % 3
for i in range(3):
if '.' in field[f_y*3+i][f_x]:
for y in range(3):
field[f_y*3+y][f_x] = field[f_y*3+y][f_x].replace('.','!')
break;
else:
for y in range(9):
for x in range(3):
field[y][x] = field[y][x].replace('.','!')
print(str([str(s)[2:-2].replace("', '", ' ') for s in field])[2:-2].replace("', '",'\n'))
``` | 0 | |
922 | A | Cloning Toys | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Imp likes his plush toy a lot.
Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies.
Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies. | The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one). | Print "Yes", if the desired configuration is possible, and "No" otherwise.
You can print each letter in arbitrary case (upper or lower). | [
"6 3\n",
"4 2\n",
"1000 1001\n"
] | [
"Yes\n",
"No\n",
"Yes\n"
] | In the first example, Imp has to apply the machine twice to original toys and then twice to copies. | 500 | [
{
"input": "6 3",
"output": "Yes"
},
{
"input": "4 2",
"output": "No"
},
{
"input": "1000 1001",
"output": "Yes"
},
{
"input": "1000000000 999999999",
"output": "Yes"
},
{
"input": "81452244 81452247",
"output": "No"
},
{
"input": "188032448 86524683",... | 1,571,472,484 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 108 | 0 | x,y=input().split()
x=int(x)
y=int(y)
check=abs(x-y)
if(check%2==1):
print("Yes")
else:
print("No") | Title: Cloning Toys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imp likes his plush toy a lot.
Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies.
Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies.
Input Specification:
The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one).
Output Specification:
Print "Yes", if the desired configuration is possible, and "No" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['6 3\n', '4 2\n', '1000 1001\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n']
Note:
In the first example, Imp has to apply the machine twice to original toys and then twice to copies. | ```python
x,y=input().split()
x=int(x)
y=int(y)
check=abs(x-y)
if(check%2==1):
print("Yes")
else:
print("No")
``` | 0 | |
964 | A | Splits | PROGRAMMING | 800 | [
"math"
] | null | null | Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$.
For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$.
The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$.
The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$.
For a given $n$, find out the number of different weights of its splits. | The first line contains one integer $n$ ($1 \leq n \leq 10^9$). | Output one integer — the answer to the problem. | [
"7\n",
"8\n",
"9\n"
] | [
"4\n",
"5\n",
"5\n"
] | In the first sample, there are following possible weights of splits of $7$:
Weight 1: [$\textbf 7$]
Weight 2: [$\textbf 3$, $\textbf 3$, 1]
Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1]
Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$] | 500 | [
{
"input": "7",
"output": "4"
},
{
"input": "8",
"output": "5"
},
{
"input": "9",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "286",
"output": "144"
},
{
"input": "48",
"output": "25"
},
{
"input": "941",
"output": "471... | 1,540,256,287 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 124 | 0 | q = int(input())
a = q//2
print(a+1)
| Title: Splits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$.
For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$.
The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$.
The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$.
For a given $n$, find out the number of different weights of its splits.
Input Specification:
The first line contains one integer $n$ ($1 \leq n \leq 10^9$).
Output Specification:
Output one integer — the answer to the problem.
Demo Input:
['7\n', '8\n', '9\n']
Demo Output:
['4\n', '5\n', '5\n']
Note:
In the first sample, there are following possible weights of splits of $7$:
Weight 1: [$\textbf 7$]
Weight 2: [$\textbf 3$, $\textbf 3$, 1]
Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1]
Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$] | ```python
q = int(input())
a = q//2
print(a+1)
``` | 3 | |
873 | D | Merge Sort | PROGRAMMING | 1,800 | [
"constructive algorithms",
"divide and conquer"
] | null | null | Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants! | The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it. | If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them. | [
"3 3\n",
"4 1\n",
"5 6\n"
] | [
"2 1 3 ",
"1 2 3 4 ",
"-1\n"
] | none | 0 | [
{
"input": "3 3",
"output": "2 1 3 "
},
{
"input": "4 1",
"output": "1 2 3 4 "
},
{
"input": "5 6",
"output": "-1"
},
{
"input": "100 100",
"output": "-1"
},
{
"input": "10000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27... | 1,697,872,053 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 77 | 3,072,000 | from math import log2,ceil
def swap(a,i,j):
x = a[i]
a[i] = a[j]
a[j] = x
def invert(a,q):
swap(a,q-1,q)
def invertrec(a,p,r,k,i):
q = (p+r)//2
if (r-p>1 and i < k):
invert(a,q)
i+=2
j = k - i
if (j == 0): return
if (j == 2): invertrec(a,q,r,k,i)
if (j == 4):
invertrec(a,p,q,k,i)
invertrec(a,q,r,k,i)
def maxc(n):
return 2*n-1
#if n == 1: return 1
#elif(log2(n)-int(log2(n)) == 0):
# return (2*n) -1
#else:
# return 1 + maxc(x:=n//2) + maxc(n-x)
def solvel(k,n):
f = lambda i: 2**(i)-1
l=0
r=n+1
while (r-l>1):
m = (l+r)//2
if(k < f(m)): r=m
else: l=m
return (l,r)
n,k = (int(x) for x in input().split())
if (k%2 == 0 or 2*n-1 < k) : print(-1)
else:
a = list(range(1,n+1))
invertrec(a,0,n,k,1)
print(*a)
| Title: Merge Sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants!
Input Specification:
The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it.
Output Specification:
If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.
Demo Input:
['3 3\n', '4 1\n', '5 6\n']
Demo Output:
['2 1 3 ', '1 2 3 4 ', '-1\n']
Note:
none | ```python
from math import log2,ceil
def swap(a,i,j):
x = a[i]
a[i] = a[j]
a[j] = x
def invert(a,q):
swap(a,q-1,q)
def invertrec(a,p,r,k,i):
q = (p+r)//2
if (r-p>1 and i < k):
invert(a,q)
i+=2
j = k - i
if (j == 0): return
if (j == 2): invertrec(a,q,r,k,i)
if (j == 4):
invertrec(a,p,q,k,i)
invertrec(a,q,r,k,i)
def maxc(n):
return 2*n-1
#if n == 1: return 1
#elif(log2(n)-int(log2(n)) == 0):
# return (2*n) -1
#else:
# return 1 + maxc(x:=n//2) + maxc(n-x)
def solvel(k,n):
f = lambda i: 2**(i)-1
l=0
r=n+1
while (r-l>1):
m = (l+r)//2
if(k < f(m)): r=m
else: l=m
return (l,r)
n,k = (int(x) for x in input().split())
if (k%2 == 0 or 2*n-1 < k) : print(-1)
else:
a = list(range(1,n+1))
invertrec(a,0,n,k,1)
print(*a)
``` | 0 | |
665 | C | Simple Strings | PROGRAMMING | 1,300 | [
"dp",
"greedy",
"strings"
] | null | null | zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task! | The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters. | Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters. | [
"aab\n",
"caaab\n",
"zscoder\n"
] | [
"bab\n",
"cabab\n",
"zscoder\n"
] | none | 0 | [
{
"input": "aab",
"output": "bab"
},
{
"input": "caaab",
"output": "cabab"
},
{
"input": "zscoder",
"output": "zscoder"
},
{
"input": "u",
"output": "u"
},
{
"input": "h",
"output": "h"
},
{
"input": "dtottttotd",
"output": "dtotataotd"
},
{
... | 1,644,110,906 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
#define taskname ""
using namespace std;
using ll=long long;
using ldb=long double;
using ull=unsigned long long;
string a;
int n;
void nhap(){
cin >> a;
n=a.size();
a='?'+a;
}
void solve(){
for(int i=2; i<=n; i++){
if(a[i]==a[i-1]){
if(a[i-1]=='a'){
if(a[i+1]=='b'){
a[i]='c';
}
else{
a[i]='b';
}
}
else{
if(a[i+1]=='a'){
if(a[i-1]=='b'){
a[i]='c';
}
else{
a[i]='b';
}
}
else{
a[i]='a';
}
}
}
}
for(int i=1; i<=n; i++){
cout << a[i];
}
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t=1;
//cin >> t;
while(t--){
nhap();
solve();
}
}
| Title: Simple Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task!
Input Specification:
The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters.
Output Specification:
Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters.
Demo Input:
['aab\n', 'caaab\n', 'zscoder\n']
Demo Output:
['bab\n', 'cabab\n', 'zscoder\n']
Note:
none | ```python
#include <bits/stdc++.h>
#define taskname ""
using namespace std;
using ll=long long;
using ldb=long double;
using ull=unsigned long long;
string a;
int n;
void nhap(){
cin >> a;
n=a.size();
a='?'+a;
}
void solve(){
for(int i=2; i<=n; i++){
if(a[i]==a[i-1]){
if(a[i-1]=='a'){
if(a[i+1]=='b'){
a[i]='c';
}
else{
a[i]='b';
}
}
else{
if(a[i+1]=='a'){
if(a[i-1]=='b'){
a[i]='c';
}
else{
a[i]='b';
}
}
else{
a[i]='a';
}
}
}
}
for(int i=1; i<=n; i++){
cout << a[i];
}
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t=1;
//cin >> t;
while(t--){
nhap();
solve();
}
}
``` | -1 | |
520 | A | Pangram | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters. | Output "YES", if the string is a pangram and "NO" otherwise. | [
"12\ntoosmallword\n",
"35\nTheQuickBrownFoxJumpsOverTheLazyDog\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "12\ntoosmallword",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog",
"output": "YES"
},
{
"input": "1\na",
"output": "NO"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "YES"
},
{
"input": "26\nABCDEFGHIJKLMNOPQRSTUVWXY... | 1,699,895,122 | 2,147,483,647 | Python 3 | OK | TESTS | 59 | 46 | 0 | n = int(input())
string = input().lower()
# Convert the string to lowercase and create a set of unique letters
unique_letters = set(string)
# Check if the count of unique letters is 26
if len(unique_letters) == 26:
print("YES")
else:
print("NO")
| Title: Pangram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
Output Specification:
Output "YES", if the string is a pangram and "NO" otherwise.
Demo Input:
['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
n = int(input())
string = input().lower()
# Convert the string to lowercase and create a set of unique letters
unique_letters = set(string)
# Check if the count of unique letters is 26
if len(unique_letters) == 26:
print("YES")
else:
print("NO")
``` | 3 | |
320 | A | Magic Numbers | PROGRAMMING | 900 | [
"brute force",
"greedy"
] | null | null | A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not. | The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros. | Print "YES" if *n* is a magic number or print "NO" if it's not. | [
"114114\n",
"1111\n",
"441231\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "114114",
"output": "YES"
},
{
"input": "1111",
"output": "YES"
},
{
"input": "441231",
"output": "NO"
},
{
"input": "1",
"output": "YES"
},
{
"input": "14",
"output": "YES"
},
{
"input": "114",
"output": "YES"
},
{
"input": "9",... | 1,635,434,599 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 92 | 28,364,800 | i=int(input())
print("8426"[i%4-1] if i>0 else 1) | Title: Magic Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not.
Input Specification:
The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros.
Output Specification:
Print "YES" if *n* is a magic number or print "NO" if it's not.
Demo Input:
['114114\n', '1111\n', '441231\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none | ```python
i=int(input())
print("8426"[i%4-1] if i>0 else 1)
``` | 0 | |
867 | A | Between the Offices | PROGRAMMING | 800 | [
"implementation"
] | null | null | As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not. | The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence. | Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower). | [
"4\nFSSF\n",
"2\nSF\n",
"10\nFFFFFFFFFF\n",
"10\nSSFFSFFSFF\n"
] | [
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] | In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | 500 | [
{
"input": "4\nFSSF",
"output": "NO"
},
{
"input": "2\nSF",
"output": "YES"
},
{
"input": "10\nFFFFFFFFFF",
"output": "NO"
},
{
"input": "10\nSSFFSFFSFF",
"output": "YES"
},
{
"input": "20\nSFSFFFFSSFFFFSSSSFSS",
"output": "NO"
},
{
"input": "20\nSSFFF... | 1,639,978,022 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 34 | 124 | 0 | # import sys
# sys.stdout = open('DSA/Stacks/output.txt', 'w')
# sys.stdin = open('DSA/Stacks/input.txt', 'r')
n = int(input())
ll = list(input())
se = 0
sf = 0
for i in range(len(ll)-1):
if ll[i]=='S' and ll[i+1]=="F":
sf+=1
elif ll[i]=='F' and ll[i+1]=="S":
se+=1
if sf>se:
print("YES")
else:
print("NO") | Title: Between the Offices
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
Input Specification:
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output Specification:
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['4\nFSSF\n', '2\nSF\n', '10\nFFFFFFFFFF\n', '10\nSSFFSFFSFF\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | ```python
# import sys
# sys.stdout = open('DSA/Stacks/output.txt', 'w')
# sys.stdin = open('DSA/Stacks/input.txt', 'r')
n = int(input())
ll = list(input())
se = 0
sf = 0
for i in range(len(ll)-1):
if ll[i]=='S' and ll[i+1]=="F":
sf+=1
elif ll[i]=='F' and ll[i+1]=="S":
se+=1
if sf>se:
print("YES")
else:
print("NO")
``` | 3 | |
399 | B | Red and Blue Balls | PROGRAMMING | 0 | [] | null | null | User ainta has a stack of *n* red and blue balls. He can apply a certain operation which changes the colors of the balls inside the stack.
- While the top ball inside the stack is red, pop the ball from the top of the stack. - Then replace the blue ball on the top with a red ball. - And finally push some blue balls to the stack until the stack has total of *n* balls inside.
If there are no blue balls inside the stack, ainta can't apply this operation. Given the initial state of the stack, ainta wants to know the maximum number of operations he can repeatedly apply. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=50) — the number of balls inside the stack.
The second line contains a string *s* (|*s*|<==<=*n*) describing the initial state of the stack. The *i*-th character of the string *s* denotes the color of the *i*-th ball (we'll number the balls from top to bottom of the stack). If the character is "R", the color is red. If the character is "B", the color is blue. | Print the maximum number of operations ainta can repeatedly apply.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"3\nRBR\n",
"4\nRBBR\n",
"5\nRBBRR\n"
] | [
"2\n",
"6\n",
"6\n"
] | The first example is depicted below.
The explanation how user ainta applies the first operation. He pops out one red ball, changes the color of the ball in the middle from blue to red, and pushes one blue ball.
The explanation how user ainta applies the second operation. He will not pop out red balls, he simply changes the color of the ball on the top from blue to red.
From now on, ainta can't apply any operation because there are no blue balls inside the stack. ainta applied two operations, so the answer is 2.
The second example is depicted below. The blue arrow denotes a single operation. | 1,000 | [
{
"input": "3\nRBR",
"output": "2"
},
{
"input": "4\nRBBR",
"output": "6"
},
{
"input": "5\nRBBRR",
"output": "6"
},
{
"input": "5\nRBRBR",
"output": "10"
},
{
"input": "10\nRRBRRBBRRR",
"output": "100"
},
{
"input": "10\nBRBRRRRRRR",
"output": "5"... | 1,566,686,277 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 109 | 0 | def do():
n = int(input())
stack = [s for s in input()][::-1]
b = stack.count("B")
res = 0
while b > 0:
while stack and stack[-1] == "R":
stack.pop()
stack[-1] = "R"
b -= 1
if len(stack) < n:
res += (1 << (n - len(stack))) - 1
stack += ["R"] * (n - len(stack))
res += 1
return res
print(do())
| Title: Red and Blue Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
User ainta has a stack of *n* red and blue balls. He can apply a certain operation which changes the colors of the balls inside the stack.
- While the top ball inside the stack is red, pop the ball from the top of the stack. - Then replace the blue ball on the top with a red ball. - And finally push some blue balls to the stack until the stack has total of *n* balls inside.
If there are no blue balls inside the stack, ainta can't apply this operation. Given the initial state of the stack, ainta wants to know the maximum number of operations he can repeatedly apply.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=50) — the number of balls inside the stack.
The second line contains a string *s* (|*s*|<==<=*n*) describing the initial state of the stack. The *i*-th character of the string *s* denotes the color of the *i*-th ball (we'll number the balls from top to bottom of the stack). If the character is "R", the color is red. If the character is "B", the color is blue.
Output Specification:
Print the maximum number of operations ainta can repeatedly apply.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['3\nRBR\n', '4\nRBBR\n', '5\nRBBRR\n']
Demo Output:
['2\n', '6\n', '6\n']
Note:
The first example is depicted below.
The explanation how user ainta applies the first operation. He pops out one red ball, changes the color of the ball in the middle from blue to red, and pushes one blue ball.
The explanation how user ainta applies the second operation. He will not pop out red balls, he simply changes the color of the ball on the top from blue to red.
From now on, ainta can't apply any operation because there are no blue balls inside the stack. ainta applied two operations, so the answer is 2.
The second example is depicted below. The blue arrow denotes a single operation. | ```python
def do():
n = int(input())
stack = [s for s in input()][::-1]
b = stack.count("B")
res = 0
while b > 0:
while stack and stack[-1] == "R":
stack.pop()
stack[-1] = "R"
b -= 1
if len(stack) < n:
res += (1 << (n - len(stack))) - 1
stack += ["R"] * (n - len(stack))
res += 1
return res
print(do())
``` | 3 | |
500 | A | New Year Transportation | PROGRAMMING | 1,000 | [
"dfs and similar",
"graphs",
"implementation"
] | null | null | New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system. | The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World. | If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO". | [
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | 500 | [
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 ... | 1,662,471,387 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 1,740,800 | import math
import sys
# s,n=map(int,input().split())
# t=True
# for m in range(int(input())):
# n=int(input())
# s=input()
n,k=map(int,input().split())
# ar=[a,b]
arr=list(map(int,input().split()))
ans=0
while(ans<k-1):
ans+=arr[ans]
if(ans==k-1):
print("YES")
else:
print("NO") | Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | ```python
import math
import sys
# s,n=map(int,input().split())
# t=True
# for m in range(int(input())):
# n=int(input())
# s=input()
n,k=map(int,input().split())
# ar=[a,b]
arr=list(map(int,input().split()))
ans=0
while(ans<k-1):
ans+=arr[ans]
if(ans==k-1):
print("YES")
else:
print("NO")
``` | 3 | |
935 | A | Fafa and his Company | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly. | The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company. | Print a single integer representing the answer to the problem. | [
"2\n",
"10\n"
] | [
"1\n",
"3\n"
] | In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | 500 | [
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
... | 1,670,776,470 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 33 | 62 | 1,331,200 | # FAFA and his Company
n = int(input())
w = 0
for i in range(1, n+1):
if n-i != 0:
if (n-i) % i == 0:
w = w+1
print(w) | Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | ```python
# FAFA and his Company
n = int(input())
w = 0
for i in range(1, n+1):
if n-i != 0:
if (n-i) % i == 0:
w = w+1
print(w)
``` | 3 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,694,064,546 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | input_string=input()
for char in input_string:
result=char.lower()
print(result) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
input_string=input()
for char in input_string:
result=char.lower()
print(result)
``` | 0 |
678 | A | Johny Likes Numbers | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*. | The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109). | Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*. | [
"5 3\n",
"25 13\n",
"26 13\n"
] | [
"6\n",
"26\n",
"39\n"
] | none | 0 | [
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894"... | 1,485,973,726 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 4,608,000 | print ("Hello World"); | Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none | ```python
print ("Hello World");
``` | 0 | |
909 | C | Python Indentation | PROGRAMMING | 1,800 | [
"dp"
] | null | null | In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program. | The first line contains a single integer *N* (1<=≤<=*N*<=≤<=5000) — the number of commands in the program. *N* lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement. | Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109<=+<=7. | [
"4\ns\nf\nf\ns\n",
"4\nf\ns\nf\ns\n"
] | [
"1\n",
"2\n"
] | In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
or | 1,500 | [
{
"input": "4\ns\nf\nf\ns",
"output": "1"
},
{
"input": "4\nf\ns\nf\ns",
"output": "2"
},
{
"input": "156\nf\ns\nf\ns\nf\ns\ns\ns\ns\nf\ns\ns\nf\nf\ns\nf\nf\nf\nf\ns\ns\ns\nf\ns\ns\nf\nf\nf\nf\nf\nf\ns\ns\ns\ns\nf\ns\nf\ns\nf\ns\nf\nf\nf\nf\ns\ns\nf\nf\ns\ns\ns\ns\nf\ns\nf\ns\nf\ns\nf\ns... | 1,515,051,028 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 124 | 6,553,600 | cmap = dict()
def countStmt(a, l):
if (len(a)==1):
return 1
if (a,l) in cmap:
return cmap[(a,l)]
s = 0
if a[0]=='s':
for i in range(l+1):
s+=countStmt(a[1:], i)
else:
s = countStmt(a[1:], l+1)
cmap[(a,l)] = s
return s
n = int(input())
a = ""
for i in range(n):
s = input()
a += s
print(countStmt(a, 0))
| Title: Python Indentation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
Input Specification:
The first line contains a single integer *N* (1<=≤<=*N*<=≤<=5000) — the number of commands in the program. *N* lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
Output Specification:
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109<=+<=7.
Demo Input:
['4\ns\nf\nf\ns\n', '4\nf\ns\nf\ns\n']
Demo Output:
['1\n', '2\n']
Note:
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
or | ```python
cmap = dict()
def countStmt(a, l):
if (len(a)==1):
return 1
if (a,l) in cmap:
return cmap[(a,l)]
s = 0
if a[0]=='s':
for i in range(l+1):
s+=countStmt(a[1:], i)
else:
s = countStmt(a[1:], l+1)
cmap[(a,l)] = s
return s
n = int(input())
a = ""
for i in range(n):
s = input()
a += s
print(countStmt(a, 0))
``` | 0 | |
509 | A | Maximum in Table | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above. | The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table. | Print a single line containing a positive integer *m* — the maximum value in the table. | [
"1\n",
"5\n"
] | [
"1",
"70"
] | In the second test the rows of the table look as follows: | 0 | [
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "70"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "20"
},
{
"input": "6",
"output": "252"
},
{
"input": "7",
"output": "924"
... | 1,595,171,881 | 2,147,483,647 | Python 3 | OK | TESTS | 10 | 108 | 6,963,200 | t=int(input())
n=[[1 for x in range(t)]for x in range(t)]
for x in range(t-1):
for y in range(t-1-x):
n[x+1][y+x+1]=n[x+1][y+x]+n[x][y+x+1]
n[y+x+1][x+1]=n[x+y][x+1]+n[y+x+1][x]
print(n[-1][-1])
| Title: Maximum in Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
Input Specification:
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Output Specification:
Print a single line containing a positive integer *m* — the maximum value in the table.
Demo Input:
['1\n', '5\n']
Demo Output:
['1', '70']
Note:
In the second test the rows of the table look as follows: | ```python
t=int(input())
n=[[1 for x in range(t)]for x in range(t)]
for x in range(t-1):
for y in range(t-1-x):
n[x+1][y+x+1]=n[x+1][y+x]+n[x][y+x+1]
n[y+x+1][x+1]=n[x+y][x+1]+n[y+x+1][x]
print(n[-1][-1])
``` | 3 | |
496 | B | Secret Combination | PROGRAMMING | 1,500 | [
"brute force",
"constructive algorithms",
"implementation"
] | null | null | You got a box with a combination lock. The lock has a display showing *n* digits. There are two buttons on the box, each button changes digits on the display. You have quickly discovered that the first button adds 1 to all the digits (all digits 9 become digits 0), and the second button shifts all the digits on the display one position to the right (the last digit becomes the first one). For example, if the display is currently showing number 579, then if we push the first button, the display will show 680, and if after that we push the second button, the display will show 068.
You know that the lock will open if the display is showing the smallest possible number that can be obtained by pushing the buttons in some order. The leading zeros are ignored while comparing numbers. Now your task is to find the desired number. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits on the display.
The second line contains *n* digits — the initial state of the display. | Print a single line containing *n* digits — the desired state of the display containing the smallest possible number. | [
"3\n579\n",
"4\n2014\n"
] | [
"024\n",
"0142\n"
] | none | 1,000 | [
{
"input": "3\n579",
"output": "024"
},
{
"input": "4\n2014",
"output": "0142"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "3\n039",
"output": "014"
},
{
"input": "4\n4444",
"output": "0000"
},
{
"input": "5\n46802",
"output": "02468"
},
... | 1,566,438,550 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 14 | 2,000 | 0 | def upshift(s):
for i in range(n):
if s[i] == '9':
s[i] = '0'
else:
s[i] = chr(ord(s[i])+1)
return s
def lshift(s):
k = [s[n-1]]
for i in range (0, n-1):
k.append(s[i])
s = k
return s
def copy(s):
k = []
for i in s:
k.append(i)
return k
n = int(input())
s = list(str(input()))
small = copy(s)
for i in range(n):
s = lshift(s)
for j in range(10):
s = upshift(s)
if small > s:
small = copy(s)
for i in small:
print(i, end='') | Title: Secret Combination
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You got a box with a combination lock. The lock has a display showing *n* digits. There are two buttons on the box, each button changes digits on the display. You have quickly discovered that the first button adds 1 to all the digits (all digits 9 become digits 0), and the second button shifts all the digits on the display one position to the right (the last digit becomes the first one). For example, if the display is currently showing number 579, then if we push the first button, the display will show 680, and if after that we push the second button, the display will show 068.
You know that the lock will open if the display is showing the smallest possible number that can be obtained by pushing the buttons in some order. The leading zeros are ignored while comparing numbers. Now your task is to find the desired number.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits on the display.
The second line contains *n* digits — the initial state of the display.
Output Specification:
Print a single line containing *n* digits — the desired state of the display containing the smallest possible number.
Demo Input:
['3\n579\n', '4\n2014\n']
Demo Output:
['024\n', '0142\n']
Note:
none | ```python
def upshift(s):
for i in range(n):
if s[i] == '9':
s[i] = '0'
else:
s[i] = chr(ord(s[i])+1)
return s
def lshift(s):
k = [s[n-1]]
for i in range (0, n-1):
k.append(s[i])
s = k
return s
def copy(s):
k = []
for i in s:
k.append(i)
return k
n = int(input())
s = list(str(input()))
small = copy(s)
for i in range(n):
s = lshift(s)
for j in range(10):
s = upshift(s)
if small > s:
small = copy(s)
for i in small:
print(i, end='')
``` | 0 | |
989 | A | A Blend of Springtime | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | "What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order. | The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively. | Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower). | [
".BAC.\n",
"AA..CB\n"
] | [
"Yes\n",
"No\n"
] | In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | 500 | [
{
"input": ".BAC.",
"output": "Yes"
},
{
"input": "AA..CB",
"output": "No"
},
{
"input": ".",
"output": "No"
},
{
"input": "ACB.AAAAAA",
"output": "Yes"
},
{
"input": "B.BC.BBBCA",
"output": "Yes"
},
{
"input": "BA..CAB..B",
"output": "Yes"
},
... | 1,597,597,526 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 16 | 124 | 0 | s = input().split(".")
count = 0
for i in s:
if("A" in i and "B" in i and "C" in i):
count += 1
print("Yes")
break
if(count == 0):
print("No") | Title: A Blend of Springtime
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input Specification:
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output Specification:
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['.BAC.\n', 'AA..CB\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | ```python
s = input().split(".")
count = 0
for i in s:
if("A" in i and "B" in i and "C" in i):
count += 1
print("Yes")
break
if(count == 0):
print("No")
``` | 0 | |
515 | A | Drazil and Date | PROGRAMMING | 1,000 | [
"math"
] | null | null | Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0,<=0) and Varda's home is located in point (*a*,<=*b*). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (*x*,<=*y*) he can go to positions (*x*<=+<=1,<=*y*), (*x*<=-<=1,<=*y*), (*x*,<=*y*<=+<=1) or (*x*,<=*y*<=-<=1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (*a*,<=*b*) and continue travelling.
Luckily, Drazil arrived to the position (*a*,<=*b*) successfully. Drazil said to Varda: "It took me exactly *s* steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0,<=0) to (*a*,<=*b*) in exactly *s* steps. Can you find out if it is possible for Varda? | You are given three integers *a*, *b*, and *s* (<=-<=109<=≤<=*a*,<=*b*<=≤<=109, 1<=≤<=*s*<=≤<=2·109) in a single line. | If you think Drazil made a mistake and it is impossible to take exactly *s* steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes". | [
"5 5 11\n",
"10 15 25\n",
"0 5 1\n",
"0 0 2\n"
] | [
"No\n",
"Yes\n",
"No\n",
"Yes\n"
] | In fourth sample case one possible route is: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d30660ddf6eb6c64ffd071055a4e8ddd016cde5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 500 | [
{
"input": "5 5 11",
"output": "No"
},
{
"input": "10 15 25",
"output": "Yes"
},
{
"input": "0 5 1",
"output": "No"
},
{
"input": "0 0 2",
"output": "Yes"
},
{
"input": "999999999 999999999 2000000000",
"output": "Yes"
},
{
"input": "-606037695 9983201... | 1,657,609,016 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 8 | 77 | 0 | a,b,s=map(int,input().split())
if (abs(s-(a+b)))%2==0 and (s>=a and s>=b):
print('YES')
else:
print('NO') | Title: Drazil and Date
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0,<=0) and Varda's home is located in point (*a*,<=*b*). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (*x*,<=*y*) he can go to positions (*x*<=+<=1,<=*y*), (*x*<=-<=1,<=*y*), (*x*,<=*y*<=+<=1) or (*x*,<=*y*<=-<=1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (*a*,<=*b*) and continue travelling.
Luckily, Drazil arrived to the position (*a*,<=*b*) successfully. Drazil said to Varda: "It took me exactly *s* steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0,<=0) to (*a*,<=*b*) in exactly *s* steps. Can you find out if it is possible for Varda?
Input Specification:
You are given three integers *a*, *b*, and *s* (<=-<=109<=≤<=*a*,<=*b*<=≤<=109, 1<=≤<=*s*<=≤<=2·109) in a single line.
Output Specification:
If you think Drazil made a mistake and it is impossible to take exactly *s* steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Demo Input:
['5 5 11\n', '10 15 25\n', '0 5 1\n', '0 0 2\n']
Demo Output:
['No\n', 'Yes\n', 'No\n', 'Yes\n']
Note:
In fourth sample case one possible route is: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d30660ddf6eb6c64ffd071055a4e8ddd016cde5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
a,b,s=map(int,input().split())
if (abs(s-(a+b)))%2==0 and (s>=a and s>=b):
print('YES')
else:
print('NO')
``` | 0 | |
577 | A | Multiplication Table | PROGRAMMING | 1,000 | [
"implementation",
"number theory"
] | null | null | Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*. | The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table. | Print a single number: the number of times *x* occurs in the table. | [
"10 5\n",
"6 12\n",
"5 13\n"
] | [
"2\n",
"4\n",
"0\n"
] | A table for the second sample test is given below. The occurrences of number 12 are marked bold. | 500 | [
{
"input": "10 5",
"output": "2"
},
{
"input": "6 12",
"output": "4"
},
{
"input": "5 13",
"output": "0"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "1 10000000... | 1,673,359,209 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 78 | 0 | n,x=[int(x) for x in input().split()]
res=0
for k in range(1,n+1):
if k*k<=x and x<=k*n and x%k==0:
if x==k*k:
res+=1
else:
res+=2
print(res)
| Title: Multiplication Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
Input Specification:
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
Output Specification:
Print a single number: the number of times *x* occurs in the table.
Demo Input:
['10 5\n', '6 12\n', '5 13\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
A table for the second sample test is given below. The occurrences of number 12 are marked bold. | ```python
n,x=[int(x) for x in input().split()]
res=0
for k in range(1,n+1):
if k*k<=x and x<=k*n and x%k==0:
if x==k*k:
res+=1
else:
res+=2
print(res)
``` | 3 | |
709 | A | Juicer | PROGRAMMING | 900 | [
"implementation"
] | null | null | Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section? | The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer. | Print one integer — the number of times Kolya will have to empty the waste section. | [
"2 7 10\n5 6\n",
"1 5 10\n7\n",
"3 10 10\n5 7 7\n",
"1 1 1\n1\n"
] | [
"1\n",
"0\n",
"1\n",
"0\n"
] | In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all. | 500 | [
{
"input": "2 7 10\n5 6",
"output": "1"
},
{
"input": "1 5 10\n7",
"output": "0"
},
{
"input": "3 10 10\n5 7 7",
"output": "1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 951637 951638\n44069 951637",
"output": "1"
},
{
"input": "50 100 12... | 1,689,849,952 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 93 | 14,438,400 | # -*- coding:utf-8 -*-
from typing import List
def Juicer(A:List[int], b:int, d:int)->int:
# write code here
sum = 0
for weight in A:
if weight <= b:
sum += weight
if sum == 0:
return 0
elif (sum % d) != 0:
ret = sum // d
else:
ret = (sum // d) - 1
return ret
if __name__ == '__main__':
n,b,d = map(int, input().split())
L = list(map(int, input().split()))
print(Juicer(L, b, d))
| Title: Juicer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?
Input Specification:
The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.
Output Specification:
Print one integer — the number of times Kolya will have to empty the waste section.
Demo Input:
['2 7 10\n5 6\n', '1 5 10\n7\n', '3 10 10\n5 7 7\n', '1 1 1\n1\n']
Demo Output:
['1\n', '0\n', '1\n', '0\n']
Note:
In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all. | ```python
# -*- coding:utf-8 -*-
from typing import List
def Juicer(A:List[int], b:int, d:int)->int:
# write code here
sum = 0
for weight in A:
if weight <= b:
sum += weight
if sum == 0:
return 0
elif (sum % d) != 0:
ret = sum // d
else:
ret = (sum // d) - 1
return ret
if __name__ == '__main__':
n,b,d = map(int, input().split())
L = list(map(int, input().split()))
print(Juicer(L, b, d))
``` | 0 | |
353 | B | Two Heaps | PROGRAMMING | 1,900 | [
"combinatorics",
"constructive algorithms",
"greedy",
"implementation",
"math",
"sortings"
] | null | null | Valera has 2·*n* cubes, each cube contains an integer from 10 to 99. He arbitrarily chooses *n* cubes and puts them in the first heap. The remaining cubes form the second heap.
Valera decided to play with cubes. During the game he takes a cube from the first heap and writes down the number it has. Then he takes a cube from the second heap and write out its two digits near two digits he had written (to the right of them). In the end he obtained a single fourdigit integer — the first two digits of it is written on the cube from the first heap, and the second two digits of it is written on the second cube from the second heap.
Valera knows arithmetic very well. So, he can easily count the number of distinct fourdigit numbers he can get in the game. The other question is: how to split cubes into two heaps so that this number (the number of distinct fourdigit integers Valera can get) will be as large as possible? | The first line contains integer *n* (1<=≤<=*n*<=≤<=100). The second line contains 2·*n* space-separated integers *a**i* (10<=≤<=*a**i*<=≤<=99), denoting the numbers on the cubes. | In the first line print a single number — the maximum possible number of distinct four-digit numbers Valera can obtain. In the second line print 2·*n* numbers *b**i* (1<=≤<=*b**i*<=≤<=2). The numbers mean: the *i*-th cube belongs to the *b**i*-th heap in your division.
If there are multiple optimal ways to split the cubes into the heaps, print any of them. | [
"1\n10 99\n",
"2\n13 24 13 45\n"
] | [
"1\n2 1 \n",
"4\n1 2 2 1 \n"
] | In the first test case Valera can put the first cube in the first heap, and second cube — in second heap. In this case he obtain number 1099. If he put the second cube in the first heap, and the first cube in the second heap, then he can obtain number 9910. In both cases the maximum number of distinct integers is equal to one.
In the second test case Valera can obtain numbers 1313, 1345, 2413, 2445. Note, that if he put the first and the third cubes in the first heap, he can obtain only two numbers 1324 and 1345. | 1,500 | [
{
"input": "1\n10 99",
"output": "1\n2 1 "
},
{
"input": "2\n13 24 13 45",
"output": "4\n1 2 2 1 "
},
{
"input": "5\n21 60 18 21 17 39 58 74 62 34",
"output": "25\n1 1 1 2 2 1 2 1 2 2 "
},
{
"input": "10\n26 43 29 92 22 27 95 56 72 55 93 51 91 30 70 77 32 69 87 98",
"outp... | 1,586,335,740 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 5 | 248 | 0 | n = int(input())
a = list(enumerate(input().split()))
x = [0]*(2*n)
a.sort(key=lambda x:x[1])
total = len(set(map(lambda x:x[1],a[::2])))*len(set(map(lambda x:x[1],a[1::2])))
print(total)
c=1
for i,e in a:
x[i]=1+c%2
c+=1
print(*x)
| Title: Two Heaps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera has 2·*n* cubes, each cube contains an integer from 10 to 99. He arbitrarily chooses *n* cubes and puts them in the first heap. The remaining cubes form the second heap.
Valera decided to play with cubes. During the game he takes a cube from the first heap and writes down the number it has. Then he takes a cube from the second heap and write out its two digits near two digits he had written (to the right of them). In the end he obtained a single fourdigit integer — the first two digits of it is written on the cube from the first heap, and the second two digits of it is written on the second cube from the second heap.
Valera knows arithmetic very well. So, he can easily count the number of distinct fourdigit numbers he can get in the game. The other question is: how to split cubes into two heaps so that this number (the number of distinct fourdigit integers Valera can get) will be as large as possible?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100). The second line contains 2·*n* space-separated integers *a**i* (10<=≤<=*a**i*<=≤<=99), denoting the numbers on the cubes.
Output Specification:
In the first line print a single number — the maximum possible number of distinct four-digit numbers Valera can obtain. In the second line print 2·*n* numbers *b**i* (1<=≤<=*b**i*<=≤<=2). The numbers mean: the *i*-th cube belongs to the *b**i*-th heap in your division.
If there are multiple optimal ways to split the cubes into the heaps, print any of them.
Demo Input:
['1\n10 99\n', '2\n13 24 13 45\n']
Demo Output:
['1\n2 1 \n', '4\n1 2 2 1 \n']
Note:
In the first test case Valera can put the first cube in the first heap, and second cube — in second heap. In this case he obtain number 1099. If he put the second cube in the first heap, and the first cube in the second heap, then he can obtain number 9910. In both cases the maximum number of distinct integers is equal to one.
In the second test case Valera can obtain numbers 1313, 1345, 2413, 2445. Note, that if he put the first and the third cubes in the first heap, he can obtain only two numbers 1324 and 1345. | ```python
n = int(input())
a = list(enumerate(input().split()))
x = [0]*(2*n)
a.sort(key=lambda x:x[1])
total = len(set(map(lambda x:x[1],a[::2])))*len(set(map(lambda x:x[1],a[1::2])))
print(total)
c=1
for i,e in a:
x[i]=1+c%2
c+=1
print(*x)
``` | 0 | |
596 | B | Wilbur and Array | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Wilbur the pig is tinkering with arrays again. He has the array *a*1,<=*a*2,<=...,<=*a**n* initially consisting of *n* zeros. At one step, he can choose any index *i* and either add 1 to all elements *a**i*,<=*a**i*<=+<=1,<=... ,<=*a**n* or subtract 1 from all elements *a**i*,<=*a**i*<=+<=1,<=...,<=*a**n*. His goal is to end up with the array *b*1,<=*b*2,<=...,<=*b**n*.
Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the array *a**i*. Initially *a**i*<==<=0 for every position *i*, so this array is not given in the input.
The second line of the input contains *n* integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=109<=≤<=*b**i*<=≤<=109). | Print the minimum number of steps that Wilbur needs to make in order to achieve *a**i*<==<=*b**i* for all *i*. | [
"5\n1 2 3 4 5\n",
"4\n1 2 2 1\n"
] | [
"5",
"3"
] | In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.
In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract 1. | 1,000 | [
{
"input": "5\n1 2 3 4 5",
"output": "5"
},
{
"input": "4\n1 2 2 1",
"output": "3"
},
{
"input": "3\n1 2 4",
"output": "4"
},
{
"input": "6\n1 2 3 6 5 4",
"output": "8"
},
{
"input": "10\n2 1 4 3 6 5 8 7 10 9",
"output": "19"
},
{
"input": "7\n12 6 12 ... | 1,447,606,024 | 724 | Python 3 | OK | TESTS | 76 | 218 | 18,022,400 | n = int(input())
ans = 0
a = [0] + list(map(int, input().split()))
for i in range(n):
ans += abs(a[i + 1] - a[i])
print(ans) | Title: Wilbur and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Wilbur the pig is tinkering with arrays again. He has the array *a*1,<=*a*2,<=...,<=*a**n* initially consisting of *n* zeros. At one step, he can choose any index *i* and either add 1 to all elements *a**i*,<=*a**i*<=+<=1,<=... ,<=*a**n* or subtract 1 from all elements *a**i*,<=*a**i*<=+<=1,<=...,<=*a**n*. His goal is to end up with the array *b*1,<=*b*2,<=...,<=*b**n*.
Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the array *a**i*. Initially *a**i*<==<=0 for every position *i*, so this array is not given in the input.
The second line of the input contains *n* integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=109<=≤<=*b**i*<=≤<=109).
Output Specification:
Print the minimum number of steps that Wilbur needs to make in order to achieve *a**i*<==<=*b**i* for all *i*.
Demo Input:
['5\n1 2 3 4 5\n', '4\n1 2 2 1\n']
Demo Output:
['5', '3']
Note:
In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.
In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract 1. | ```python
n = int(input())
ans = 0
a = [0] + list(map(int, input().split()))
for i in range(n):
ans += abs(a[i + 1] - a[i])
print(ans)
``` | 3 | |
472 | A | Design Tutorial: Learn from Math | PROGRAMMING | 800 | [
"math",
"number theory"
] | null | null | One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers. | The only line contains an integer *n* (12<=≤<=*n*<=≤<=106). | Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them. | [
"12\n",
"15\n",
"23\n",
"1000000\n"
] | [
"4 8\n",
"6 9\n",
"8 15\n",
"500000 500000\n"
] | In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | 500 | [
{
"input": "12",
"output": "4 8"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "23",
"output": "8 15"
},
{
"input": "1000000",
"output": "500000 500000"
},
{
"input": "63874",
"output": "4 63870"
},
{
"input": "14568",
"output": "4 14564"
},
... | 1,691,250,844 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 46 | 0 | def prime(x):
if x <= 1:
return False
for i in range(2, int(x**0.5) + 1):
if x % i == 0:
return False
return True
n = int(input())
x = 4 # Start with the smallest possible composite number
verif = True
while x < n and verif:
x += 1
if not prime(x) and not prime(n - x):
print(x, n - x)
verif = False
| Title: Design Tutorial: Learn from Math
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
Input Specification:
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
Output Specification:
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
Demo Input:
['12\n', '15\n', '23\n', '1000000\n']
Demo Output:
['4 8\n', '6 9\n', '8 15\n', '500000 500000\n']
Note:
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | ```python
def prime(x):
if x <= 1:
return False
for i in range(2, int(x**0.5) + 1):
if x % i == 0:
return False
return True
n = int(input())
x = 4 # Start with the smallest possible composite number
verif = True
while x < n and verif:
x += 1
if not prime(x) and not prime(n - x):
print(x, n - x)
verif = False
``` | 3 | |
155 | A | I_love_\%username\% | PROGRAMMING | 800 | [
"brute force"
] | null | null | Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him. | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000. | Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests. | [
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] | [
"2\n",
"4\n"
] | In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | 500 | [
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
... | 1,696,897,798 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 124 | 204,800 |
n = input()
com = [int(i) for i in input().split()]
# print(com)
rez = 0
k = 1
while k < len(com):
# print(com[:k])
if com[k] > max(com[:k]) or com[k] < min(com[:k]):
rez += 1
k += 1
print(rez) | Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | ```python
n = input()
com = [int(i) for i in input().split()]
# print(com)
rez = 0
k = 1
while k < len(com):
# print(com[:k])
if com[k] > max(com[:k]) or com[k] < min(com[:k]):
rez += 1
k += 1
print(rez)
``` | 3 | |
205 | A | Little Elephant and Rozdil | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities. | Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes). | [
"2\n7 4\n",
"7\n7 4 47 100 4 9 12\n"
] | [
"2\n",
"Still Rozdil\n"
] | In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | 500 | [
{
"input": "2\n7 4",
"output": "2"
},
{
"input": "7\n7 4 47 100 4 9 12",
"output": "Still Rozdil"
},
{
"input": "1\n47",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "Still Rozdil"
},
{
"input": "7\n7 6 5 4 3 2 1",
"output": "7"
},
{
... | 1,627,338,486 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 13,824,000 | c = int(input())
time = list(map(int, input().split()))
match = 0
answer = 0
for x in range(c):
if time[x] == min(time):
match += 1
if match == 2:
answer = 1
break
if answer == 1:
print("Still Rozdil")
elif answer == 0:
for x in range(c):
if time[x] == min(time):
break
print(x+1) | Title: Little Elephant and Rozdil
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities.
Output Specification:
Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes).
Demo Input:
['2\n7 4\n', '7\n7 4 47 100 4 9 12\n']
Demo Output:
['2\n', 'Still Rozdil\n']
Note:
In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | ```python
c = int(input())
time = list(map(int, input().split()))
match = 0
answer = 0
for x in range(c):
if time[x] == min(time):
match += 1
if match == 2:
answer = 1
break
if answer == 1:
print("Still Rozdil")
elif answer == 0:
for x in range(c):
if time[x] == min(time):
break
print(x+1)
``` | 0 | |
408 | B | Garland | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought *n* colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly *m* pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of *m* pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get. | The first line contains a non-empty sequence of *n* (1<=≤<=*n*<=≤<=1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of *m* (1<=≤<=*m*<=≤<=1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make. | Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer. | [
"aaabbac\naabbccac\n",
"a\nz\n"
] | [
"6\n",
"-1"
] | In the first test sample Vasya can make an garland of area 6: he can use both sheets of color *b*, three (but not four) sheets of color *a* and cut a single sheet of color *c* in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color *z*. | 1,000 | [
{
"input": "aaabbac\naabbccac",
"output": "6"
},
{
"input": "a\nz",
"output": "-1"
},
{
"input": "r\nr",
"output": "1"
},
{
"input": "stnsdn\nndnndsn",
"output": "4"
},
{
"input": "yqfqfp\ntttwtqq",
"output": "-1"
},
{
"input": "zzbbrrtrtzr\ntbbtrrrzr"... | 1,659,616,967 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 46 | 0 | x=input()
y=input()
d1={}
d2={}
count=0
for i in x:
d1[i]=x.count(i)
for i in y:
d2[i]=y.count(i)
for i in d2.keys():
if i in d1.keys():
if d1[i]<d2[i]:
count+=d1[i]
else:
count+=d2[i]
else:
count=-1
break
print(count)
| Title: Garland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought *n* colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly *m* pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of *m* pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get.
Input Specification:
The first line contains a non-empty sequence of *n* (1<=≤<=*n*<=≤<=1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of *m* (1<=≤<=*m*<=≤<=1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make.
Output Specification:
Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer.
Demo Input:
['aaabbac\naabbccac\n', 'a\nz\n']
Demo Output:
['6\n', '-1']
Note:
In the first test sample Vasya can make an garland of area 6: he can use both sheets of color *b*, three (but not four) sheets of color *a* and cut a single sheet of color *c* in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color *z*. | ```python
x=input()
y=input()
d1={}
d2={}
count=0
for i in x:
d1[i]=x.count(i)
for i in y:
d2[i]=y.count(i)
for i in d2.keys():
if i in d1.keys():
if d1[i]<d2[i]:
count+=d1[i]
else:
count+=d2[i]
else:
count=-1
break
print(count)
``` | 3 | |
817 | C | Really Big Numbers | PROGRAMMING | 1,600 | [
"binary search",
"brute force",
"dp",
"math"
] | null | null | Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*.
Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations. | The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018). | Print one integer — the quantity of really big numbers that are not greater than *n*. | [
"12 1\n",
"25 20\n",
"10 9\n"
] | [
"3\n",
"0\n",
"1\n"
] | In the first example numbers 10, 11 and 12 are really big.
In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).
In the third example 10 is the only really big number (10 - 1 ≥ 9). | 0 | [
{
"input": "12 1",
"output": "3"
},
{
"input": "25 20",
"output": "0"
},
{
"input": "10 9",
"output": "1"
},
{
"input": "300 1000",
"output": "0"
},
{
"input": "500 1000",
"output": "0"
},
{
"input": "1000 2000",
"output": "0"
},
{
"input":... | 1,569,078,245 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 21 | 156 | 1,228,800 | def digisum(n):
count = 0
while(n>0):
count+=n%10
n = n//10
return count
n,s = list(map(int,input().split()))
l,r = 1,n
flag = 0
while l<r:
m = l+(r-l)//2
digi_sum = digisum(m)
num = m
if num-digi_sum>=s:
flag = 1
cur = m
r = m
else:
l = m+1
if r-l==1:
digi_sum = digisum(l)
num = l
if num-digi_sum>=s:
flag = 1
cur = l
break
digi_sum = digisum(r)
num = r
if num-digi_sum>=s:
flag = 1
cur = r
break
if flag==0:
print(0)
else:
print(n-cur+1) | Title: Really Big Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*.
Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations.
Input Specification:
The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018).
Output Specification:
Print one integer — the quantity of really big numbers that are not greater than *n*.
Demo Input:
['12 1\n', '25 20\n', '10 9\n']
Demo Output:
['3\n', '0\n', '1\n']
Note:
In the first example numbers 10, 11 and 12 are really big.
In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).
In the third example 10 is the only really big number (10 - 1 ≥ 9). | ```python
def digisum(n):
count = 0
while(n>0):
count+=n%10
n = n//10
return count
n,s = list(map(int,input().split()))
l,r = 1,n
flag = 0
while l<r:
m = l+(r-l)//2
digi_sum = digisum(m)
num = m
if num-digi_sum>=s:
flag = 1
cur = m
r = m
else:
l = m+1
if r-l==1:
digi_sum = digisum(l)
num = l
if num-digi_sum>=s:
flag = 1
cur = l
break
digi_sum = digisum(r)
num = r
if num-digi_sum>=s:
flag = 1
cur = r
break
if flag==0:
print(0)
else:
print(n-cur+1)
``` | 0 | |
596 | A | Wilbur and Swimming Pool | PROGRAMMING | 1,100 | [
"geometry",
"implementation"
] | null | null | After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set up coordinate axes, and he wants the sides of the rectangle to be parallel to them. Of course, the area of the rectangle must be positive. Wilbur had all four vertices of the planned pool written on a paper, until his friend came along and erased some of the vertices.
Now Wilbur is wondering, if the remaining *n* vertices of the initial rectangle give enough information to restore the area of the planned swimming pool. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=4) — the number of vertices that were not erased by Wilbur's friend.
Each of the following *n* lines contains two integers *x**i* and *y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000) —the coordinates of the *i*-th vertex that remains. Vertices are given in an arbitrary order.
It's guaranteed that these points are distinct vertices of some rectangle, that has positive area and which sides are parallel to the coordinate axes. | Print the area of the initial rectangle if it could be uniquely determined by the points remaining. Otherwise, print <=-<=1. | [
"2\n0 0\n1 1\n",
"1\n1 1\n"
] | [
"1\n",
"-1\n"
] | In the first sample, two opposite corners of the initial rectangle are given, and that gives enough information to say that the rectangle is actually a unit square.
In the second sample there is only one vertex left and this is definitely not enough to uniquely define the area. | 500 | [
{
"input": "2\n0 0\n1 1",
"output": "1"
},
{
"input": "1\n1 1",
"output": "-1"
},
{
"input": "1\n-188 17",
"output": "-1"
},
{
"input": "1\n71 -740",
"output": "-1"
},
{
"input": "4\n-56 -858\n-56 -174\n778 -858\n778 -174",
"output": "570456"
},
{
"inp... | 1,472,735,494 | 2,147,483,647 | Python 3 | OK | TESTS | 121 | 93 | 307,200 | if __name__ == '__main__':
n = int(input())
lines = list()
for i in range(n):
line = str(input()).split()
line = [int(it) for it in line]
lines.append(line)
if n == 1 or n == 2 and (lines[0][0] == lines[1][0] or lines[0][1] == lines[1][1]):
print(-1)
elif n == 2:
print(abs((lines[1][0] - lines[0][0]) * (lines[1][1] - lines[0][1])))
else:
zones = list()
if lines[1][0] == lines[0][0] or lines[1][1] == lines[0][1]:
zones.append(abs(lines[1][0] - lines[0][0]) + abs(lines[1][1] - lines[0][1]))
if lines[1][0] == lines[2][0] or lines[1][1] == lines[2][1]:
zones.append(abs(lines[1][0] - lines[2][0]) + abs(lines[1][1] - lines[2][1]))
if lines[0][0] == lines[2][0] or lines[0][1] == lines[2][1]:
zones.append(abs(lines[0][0] - lines[2][0]) + abs(lines[0][1] - lines[2][1]))
print(zones[0] * zones[1])
| Title: Wilbur and Swimming Pool
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set up coordinate axes, and he wants the sides of the rectangle to be parallel to them. Of course, the area of the rectangle must be positive. Wilbur had all four vertices of the planned pool written on a paper, until his friend came along and erased some of the vertices.
Now Wilbur is wondering, if the remaining *n* vertices of the initial rectangle give enough information to restore the area of the planned swimming pool.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=4) — the number of vertices that were not erased by Wilbur's friend.
Each of the following *n* lines contains two integers *x**i* and *y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000) —the coordinates of the *i*-th vertex that remains. Vertices are given in an arbitrary order.
It's guaranteed that these points are distinct vertices of some rectangle, that has positive area and which sides are parallel to the coordinate axes.
Output Specification:
Print the area of the initial rectangle if it could be uniquely determined by the points remaining. Otherwise, print <=-<=1.
Demo Input:
['2\n0 0\n1 1\n', '1\n1 1\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first sample, two opposite corners of the initial rectangle are given, and that gives enough information to say that the rectangle is actually a unit square.
In the second sample there is only one vertex left and this is definitely not enough to uniquely define the area. | ```python
if __name__ == '__main__':
n = int(input())
lines = list()
for i in range(n):
line = str(input()).split()
line = [int(it) for it in line]
lines.append(line)
if n == 1 or n == 2 and (lines[0][0] == lines[1][0] or lines[0][1] == lines[1][1]):
print(-1)
elif n == 2:
print(abs((lines[1][0] - lines[0][0]) * (lines[1][1] - lines[0][1])))
else:
zones = list()
if lines[1][0] == lines[0][0] or lines[1][1] == lines[0][1]:
zones.append(abs(lines[1][0] - lines[0][0]) + abs(lines[1][1] - lines[0][1]))
if lines[1][0] == lines[2][0] or lines[1][1] == lines[2][1]:
zones.append(abs(lines[1][0] - lines[2][0]) + abs(lines[1][1] - lines[2][1]))
if lines[0][0] == lines[2][0] or lines[0][1] == lines[2][1]:
zones.append(abs(lines[0][0] - lines[2][0]) + abs(lines[0][1] - lines[2][1]))
print(zones[0] * zones[1])
``` | 3 | |
119 | A | Epic Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game. | The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. | If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). | [
"3 5 9\n",
"1 1 100\n"
] | [
"0",
"1"
] | The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | 500 | [
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"inpu... | 1,672,083,371 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 92 | 0 | def gcd(a, b):
while a!= 0 and b != 0:
if a >b:
a %= b
else:
b %= a
return a + b
nums = list(map(int, input().split()))
a = nums[0]
b = nums[1]
n = nums[2]
turn = 1
while n > 0:
turn = 1 - turn
if turn == 0:
n -= gcd(n, a)
else:
n -= gcd(n, b)
print(turn) | Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | ```python
def gcd(a, b):
while a!= 0 and b != 0:
if a >b:
a %= b
else:
b %= a
return a + b
nums = list(map(int, input().split()))
a = nums[0]
b = nums[1]
n = nums[2]
turn = 1
while n > 0:
turn = 1 - turn
if turn == 0:
n -= gcd(n, a)
else:
n -= gcd(n, b)
print(turn)
``` | 3 | |
709 | A | Juicer | PROGRAMMING | 900 | [
"implementation"
] | null | null | Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section? | The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer. | Print one integer — the number of times Kolya will have to empty the waste section. | [
"2 7 10\n5 6\n",
"1 5 10\n7\n",
"3 10 10\n5 7 7\n",
"1 1 1\n1\n"
] | [
"1\n",
"0\n",
"1\n",
"0\n"
] | In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all. | 500 | [
{
"input": "2 7 10\n5 6",
"output": "1"
},
{
"input": "1 5 10\n7",
"output": "0"
},
{
"input": "3 10 10\n5 7 7",
"output": "1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 951637 951638\n44069 951637",
"output": "1"
},
{
"input": "50 100 12... | 1,613,646,339 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 78 | 7,270,400 | _, b, d = map(int, input().split())
a = list(map(int, input().split()))
total_waste = sum([i for i in a if i < b])
print(total_waste // (d + 1)) | Title: Juicer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?
Input Specification:
The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.
Output Specification:
Print one integer — the number of times Kolya will have to empty the waste section.
Demo Input:
['2 7 10\n5 6\n', '1 5 10\n7\n', '3 10 10\n5 7 7\n', '1 1 1\n1\n']
Demo Output:
['1\n', '0\n', '1\n', '0\n']
Note:
In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all. | ```python
_, b, d = map(int, input().split())
a = list(map(int, input().split()))
total_waste = sum([i for i in a if i < b])
print(total_waste // (d + 1))
``` | 0 | |
729 | B | Spotlights | PROGRAMMING | 1,200 | [
"dp",
"implementation"
] | null | null | Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ. | The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan. | Print one integer — the number of good positions for placing the spotlight. | [
"2 4\n0 1 0 0\n1 0 1 0\n",
"4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n"
] | [
"9\n",
"20\n"
] | In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example. | 1,000 | [
{
"input": "2 4\n0 1 0 0\n1 0 1 0",
"output": "9"
},
{
"input": "4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0",
"output": "20"
},
{
"input": "1 5\n1 1 0 0 0",
"output": "3"
},
{
"input": "2 10\n0 0 0 0 0 0 0 1 0 0\n1 0 0 0 0 0 0 0 0 0",
"output": "20"
},
{
"input": "3 ... | 1,497,345,149 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 307,200 | """
B. Прожекторы
ограничение по времени на тест
1 секунда
ограничение по памяти на тест
256 мегабайт
ввод
стандартный ввод
вывод
стандартный вывод
Театральная сцена представляет собой прямоугольное поле размером n × m. Директор театра выдал вам план сцены, согласно
которому на ней будут располагаться актёры. На плане отмечено в каких клетках будут стоять актёры, а в каких нет.
Прожектор, установленный на сцену, будет светить в одном из четырёх направлений (если смотреть на план сцены сверху) —
влево, вверх, вправо или вниз. Таким образом, под позицией прожектора понимается клетка,
в которую он установлен, а также направление, в котором он светит.
Перед вами стоит задача поставить на сцену прожектор в хорошую позицию.
Позиция называется хорошей, если одновременно выполняются два условия:
в соответствующей ей клетке нет актёра;
в направлении, в котором светит прожектор, находится хотя бы один актёр.
Перед вами стоит задача посчитать количество хороших позиций для установки прожектора.
Две позиции установки прожектора считаются различными, если отличаются клетки расположения прожектора,
или направление, в котором он светит.
Входные данные
В первой строке следует два целых положительных числа n и m (1 ≤ n, m ≤ 1000) —
количество строк и количество столбцов в плане.
В следующих n строках следует по m целых чисел, каждое равно либо 0, либо 1, —
описание плана. Если очередное число равно 1, то в соответствующей клетке находится актёр,
а если 0, то клетка останется пустой. Гарантируется, что в плане есть хотя бы один актёр.
Выходные данные
Выведите единственное целое число — количество хороших позиций для установки прожектора.
Примеры
Входные данные
2 4
0 1 0 0
1 0 1 0
Выходные данные
9
Входные данные
4 4
0 0 0 0
1 0 0 1
0 1 1 0
0 1 0 0
Выходные данные
20
Примечание
В первом примере хорошими позициями для установки прожектора являются:
клетка (1, 1) и направление вправо;
клетка (1, 1) и направление вниз;
клетка (1, 3) и направление влево;
клетка (1, 3) и направление вниз;
клетка (1, 4) и направление влево;
клетка (2, 2) и направление влево;
клетка (2, 2) и направление вверх;
клетка (2, 2) и направление вправо;
клетка (2, 4) и направление влево.
Таким образом, в данном примере всего 9 хороших позиций.
"""
def count_neighbors(row, col):
result = 0
temp_offset = 1
while (row - temp_offset) >= 0:
if input_array[row - temp_offset][col] == 1: # North
result += 1
break
temp_offset += 1
temp_offset = 1
while (col - temp_offset) >= 0:
if input_array[row][col - temp_offset] == 1: # West
result += 1
break
temp_offset += 1
temp_offset = 1
while (row + temp_offset) < len(input_array):
if input_array[row + temp_offset][col] == 1: # South
result += 1
break
temp_offset += 1
temp_offset = 1
while (col + temp_offset) < len(input_array[row]):
if input_array[row][col + temp_offset] == 1: # East
result += 1
break
temp_offset += 1
return result
def count_best_positions():
result = 0
for i in range(len(input_array)):
for j in range(len(input_array[i])):
if input_array[i][j] != 1:
result += count_neighbors(i, j)
return result
input_array = []
with open('./input.txt') as file:
count_rows = int(file.readline().split()[0])
for number_row in range(count_rows):
input_array.append([int(cell) for cell in file.readline().split()])
with open('./output.txt', 'w') as file:
file.write(str(count_best_positions()))
| Title: Spotlights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
Output Specification:
Print one integer — the number of good positions for placing the spotlight.
Demo Input:
['2 4\n0 1 0 0\n1 0 1 0\n', '4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n']
Demo Output:
['9\n', '20\n']
Note:
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example. | ```python
"""
B. Прожекторы
ограничение по времени на тест
1 секунда
ограничение по памяти на тест
256 мегабайт
ввод
стандартный ввод
вывод
стандартный вывод
Театральная сцена представляет собой прямоугольное поле размером n × m. Директор театра выдал вам план сцены, согласно
которому на ней будут располагаться актёры. На плане отмечено в каких клетках будут стоять актёры, а в каких нет.
Прожектор, установленный на сцену, будет светить в одном из четырёх направлений (если смотреть на план сцены сверху) —
влево, вверх, вправо или вниз. Таким образом, под позицией прожектора понимается клетка,
в которую он установлен, а также направление, в котором он светит.
Перед вами стоит задача поставить на сцену прожектор в хорошую позицию.
Позиция называется хорошей, если одновременно выполняются два условия:
в соответствующей ей клетке нет актёра;
в направлении, в котором светит прожектор, находится хотя бы один актёр.
Перед вами стоит задача посчитать количество хороших позиций для установки прожектора.
Две позиции установки прожектора считаются различными, если отличаются клетки расположения прожектора,
или направление, в котором он светит.
Входные данные
В первой строке следует два целых положительных числа n и m (1 ≤ n, m ≤ 1000) —
количество строк и количество столбцов в плане.
В следующих n строках следует по m целых чисел, каждое равно либо 0, либо 1, —
описание плана. Если очередное число равно 1, то в соответствующей клетке находится актёр,
а если 0, то клетка останется пустой. Гарантируется, что в плане есть хотя бы один актёр.
Выходные данные
Выведите единственное целое число — количество хороших позиций для установки прожектора.
Примеры
Входные данные
2 4
0 1 0 0
1 0 1 0
Выходные данные
9
Входные данные
4 4
0 0 0 0
1 0 0 1
0 1 1 0
0 1 0 0
Выходные данные
20
Примечание
В первом примере хорошими позициями для установки прожектора являются:
клетка (1, 1) и направление вправо;
клетка (1, 1) и направление вниз;
клетка (1, 3) и направление влево;
клетка (1, 3) и направление вниз;
клетка (1, 4) и направление влево;
клетка (2, 2) и направление влево;
клетка (2, 2) и направление вверх;
клетка (2, 2) и направление вправо;
клетка (2, 4) и направление влево.
Таким образом, в данном примере всего 9 хороших позиций.
"""
def count_neighbors(row, col):
result = 0
temp_offset = 1
while (row - temp_offset) >= 0:
if input_array[row - temp_offset][col] == 1: # North
result += 1
break
temp_offset += 1
temp_offset = 1
while (col - temp_offset) >= 0:
if input_array[row][col - temp_offset] == 1: # West
result += 1
break
temp_offset += 1
temp_offset = 1
while (row + temp_offset) < len(input_array):
if input_array[row + temp_offset][col] == 1: # South
result += 1
break
temp_offset += 1
temp_offset = 1
while (col + temp_offset) < len(input_array[row]):
if input_array[row][col + temp_offset] == 1: # East
result += 1
break
temp_offset += 1
return result
def count_best_positions():
result = 0
for i in range(len(input_array)):
for j in range(len(input_array[i])):
if input_array[i][j] != 1:
result += count_neighbors(i, j)
return result
input_array = []
with open('./input.txt') as file:
count_rows = int(file.readline().split()[0])
for number_row in range(count_rows):
input_array.append([int(cell) for cell in file.readline().split()])
with open('./output.txt', 'w') as file:
file.write(str(count_best_positions()))
``` | -1 | |
217 | A | Ice Skating | PROGRAMMING | 1,200 | [
"brute force",
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates. | The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct. | Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one. | [
"2\n2 1\n1 2\n",
"2\n2 1\n4 1\n"
] | [
"1\n",
"0\n"
] | none | 500 | [
{
"input": "2\n2 1\n1 2",
"output": "1"
},
{
"input": "2\n2 1\n4 1",
"output": "0"
},
{
"input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182... | 1,669,473,638 | 2,147,483,647 | PyPy 3 | OK | TESTS | 76 | 184 | 1,536,000 | n = int(input())
a = []
for i in range(n):
x, y = map(int, input().split())
a.append([x,y,False])
def df(x):
x[2]= True
for s in a:
if not s[2] and (x[0] == s[0] or x[1] == s[1]):
df(s)
rj = 0
for i in a:
if not i[2]:
rj += 1
df(i)
print(rj-1)
| Title: Ice Skating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output Specification:
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
Demo Input:
['2\n2 1\n1 2\n', '2\n2 1\n4 1\n']
Demo Output:
['1\n', '0\n']
Note:
none | ```python
n = int(input())
a = []
for i in range(n):
x, y = map(int, input().split())
a.append([x,y,False])
def df(x):
x[2]= True
for s in a:
if not s[2] and (x[0] == s[0] or x[1] == s[1]):
df(s)
rj = 0
for i in a:
if not i[2]:
rj += 1
df(i)
print(rj-1)
``` | 3 | |
707 | A | Brain's Photos | PROGRAMMING | 800 | [
"implementation"
] | null | null | Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. | Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. | [
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] | [
"#Color",
"#Black&White",
"#Black&White"
] | none | 500 | [
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
... | 1,690,114,997 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 31 | 0 | nm = input().split(' ')
n = int(nm[0])
m = int(nm[1])
count = 0
for i in range(n):
colors = input().split(' ')
if ('C' in colors) or ('M' in colors) or ('Y' in colors):
count = 1
if count==1:
print('#Color')
else:
print('#Black&White')
| Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none | ```python
nm = input().split(' ')
n = int(nm[0])
m = int(nm[1])
count = 0
for i in range(n):
colors = input().split(' ')
if ('C' in colors) or ('M' in colors) or ('Y' in colors):
count = 1
if count==1:
print('#Color')
else:
print('#Black&White')
``` | 3 | |
118 | A | String Task | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task. | The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive. | Print the resulting string. It is guaranteed that this string is not empty. | [
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] | [
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] | none | 500 | [
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": "... | 1,694,162,483 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 14 | 92 | 0 | s=input()
s=s.lower()
vowels=['a','e','i','o','u']
res=""
for i in s:
if i not in vowels:
res+='.'
res+=i
print(res)
| Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none | ```python
s=input()
s=s.lower()
vowels=['a','e','i','o','u']
res=""
for i in s:
if i not in vowels:
res+='.'
res+=i
print(res)
``` | 0 | |
304 | A | Pythagorean Theorem II | PROGRAMMING | 1,200 | [
"brute force",
"math"
] | null | null | In mathematics, the Pythagorean theorem — is a relation in Euclidean geometry among the three sides of a right-angled triangle. In terms of areas, it states:
In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
The theorem can be written as an equation relating the lengths of the sides *a*, *b* and *c*, often called the Pythagorean equation:
where *c* represents the length of the hypotenuse, and *a* and *b* represent the lengths of the other two sides.
Given *n*, your task is to count how many right-angled triangles with side-lengths *a*, *b* and *c* that satisfied an inequality 1<=≤<=*a*<=≤<=*b*<=≤<=*c*<=≤<=*n*. | The only line contains one integer *n* (1<=≤<=*n*<=≤<=104) as we mentioned above. | Print a single integer — the answer to the problem. | [
"5\n",
"74\n"
] | [
"1\n",
"35\n"
] | none | 500 | [
{
"input": "5",
"output": "1"
},
{
"input": "74",
"output": "35"
},
{
"input": "1000",
"output": "881"
},
{
"input": "586",
"output": "472"
},
{
"input": "2",
"output": "0"
},
{
"input": "362",
"output": "258"
},
{
"input": "778",
"outp... | 1,663,217,980 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 468 | 0 | num = int(input())
res = 0
for n in range(1,num+1):
for m in range(n+1, num+1,2):
if (m%n==0) and n!=1:
continue
c = m**2 + n**2
k = num//c
res+=k
print(res) | Title: Pythagorean Theorem II
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In mathematics, the Pythagorean theorem — is a relation in Euclidean geometry among the three sides of a right-angled triangle. In terms of areas, it states:
In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
The theorem can be written as an equation relating the lengths of the sides *a*, *b* and *c*, often called the Pythagorean equation:
where *c* represents the length of the hypotenuse, and *a* and *b* represent the lengths of the other two sides.
Given *n*, your task is to count how many right-angled triangles with side-lengths *a*, *b* and *c* that satisfied an inequality 1<=≤<=*a*<=≤<=*b*<=≤<=*c*<=≤<=*n*.
Input Specification:
The only line contains one integer *n* (1<=≤<=*n*<=≤<=104) as we mentioned above.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['5\n', '74\n']
Demo Output:
['1\n', '35\n']
Note:
none | ```python
num = int(input())
res = 0
for n in range(1,num+1):
for m in range(n+1, num+1,2):
if (m%n==0) and n!=1:
continue
c = m**2 + n**2
k = num//c
res+=k
print(res)
``` | 0 | |
2 | A | Winner | PROGRAMMING | 1,500 | [
"hashing",
"implementation"
] | A. Winner | 1 | 64 | The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points. | The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive. | Print the name of the winner. | [
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] | [
"andrew\n",
"andrew\n"
] | none | 0 | [
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},... | 1,554,221,111 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | from sys import *
rt = stdin.read().split('\n')
A = {}
B = {}
C = {}
for s in rt[1:]:
if s != '':
t = s.split()
if t[0] in A:
A[t[0]] = A[t[0]]+int(t[1])
else:
A[t[0]] = int(t[1])
as = max(A.values())
for i in A:
if A[i] == as:
C[i] = A[i]
A = {}
for s in rt[1:]:
if s != '':
t = s.split()
if t[0] in C:
if t[0] in A:
A[t[0]] = A[t[0]]+int(t[1])
else:
A[t[0]] = int(t[1])
if A[t[0]] >= as:
print(t[0])
break | Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none | ```python
from sys import *
rt = stdin.read().split('\n')
A = {}
B = {}
C = {}
for s in rt[1:]:
if s != '':
t = s.split()
if t[0] in A:
A[t[0]] = A[t[0]]+int(t[1])
else:
A[t[0]] = int(t[1])
as = max(A.values())
for i in A:
if A[i] == as:
C[i] = A[i]
A = {}
for s in rt[1:]:
if s != '':
t = s.split()
if t[0] in C:
if t[0] in A:
A[t[0]] = A[t[0]]+int(t[1])
else:
A[t[0]] = int(t[1])
if A[t[0]] >= as:
print(t[0])
break
``` | -1 |
946 | A | Partition | PROGRAMMING | 800 | [
"greedy"
] | null | null | You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences.
Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*? | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*. | Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*. | [
"3\n1 -2 0\n",
"6\n16 23 16 15 42 8\n"
] | [
"3\n",
"120\n"
] | In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3.
In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120. | 0 | [
{
"input": "3\n1 -2 0",
"output": "3"
},
{
"input": "6\n16 23 16 15 42 8",
"output": "120"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "100\n-100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -10... | 1,596,454,254 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 93 | 6,656,000 | a=int(input())
c=list(map(int, input().split()))
x=abs(sum(c))
for i in c:
if i<0:
x+=abs(i)
print(x)
| Title: Partition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences.
Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*.
Output Specification:
Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*.
Demo Input:
['3\n1 -2 0\n', '6\n16 23 16 15 42 8\n']
Demo Output:
['3\n', '120\n']
Note:
In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3.
In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120. | ```python
a=int(input())
c=list(map(int, input().split()))
x=abs(sum(c))
for i in c:
if i<0:
x+=abs(i)
print(x)
``` | 0 |
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