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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
279 | B | Books | PROGRAMMING | 1,400 | [
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null | When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read. | The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book. | Print a single integer — the maximum number of books Valera can read. | [
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] | [
"3\n",
"1\n"
] | none | 1,000 | [
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4... | 1,672,980,444 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 5 | 186 | 2,150,400 | from math import gcd, lcm, floor, sqrt, isqrt
import sys
from collections import defaultdict, deque
from functools import reduce
from heapq import heapify, heappop, heappush
from sys import maxsize as INF
from time import perf_counter
from random import randint
# === i/o ===
def reads(): return sys.stdin.readline().strip("\n")
def readi(): return int(sys.stdin.readline().strip("\n"))
def readf(): return float(sys.stdin.readline().strip("\n"))
def readp(): return sys.stdin.readline().strip("\n").split(" ")
def readpi(): return map(int, sys.stdin.readline().strip("\n").split(" "))
def readli(): return list(map(int, sys.stdin.readline().strip("\n").split(" ")))
def readpt(*args):
s = sys.stdin.readline().strip("\n").split(" ")
return (args[i](s[i]) for i in range(len(s)))
def yes(): sys.stdout.write("YES")
def no(): sys.stdout.write("NO")
def out(s): sys.stdout.write(str(s) + "\n")
def outpl(a):
for n in a: sys.stdout.write(str(n) + " ")
sys.stdout.write("\n")
def outpa(*args):
for n in args: sys.stdout.write(str(n) + " ")
sys.stdout.write("\n")
# === CODE HERE ===
def solve(n, t, a):
a.sort()
ans = 0
for i in a:
if t >= i:
t -= i
ans += 1
else:
break
return ans
n, t = readpi()
a = readli()
ans = solve(n, t, a)
out(ans) | Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
from math import gcd, lcm, floor, sqrt, isqrt
import sys
from collections import defaultdict, deque
from functools import reduce
from heapq import heapify, heappop, heappush
from sys import maxsize as INF
from time import perf_counter
from random import randint
# === i/o ===
def reads(): return sys.stdin.readline().strip("\n")
def readi(): return int(sys.stdin.readline().strip("\n"))
def readf(): return float(sys.stdin.readline().strip("\n"))
def readp(): return sys.stdin.readline().strip("\n").split(" ")
def readpi(): return map(int, sys.stdin.readline().strip("\n").split(" "))
def readli(): return list(map(int, sys.stdin.readline().strip("\n").split(" ")))
def readpt(*args):
s = sys.stdin.readline().strip("\n").split(" ")
return (args[i](s[i]) for i in range(len(s)))
def yes(): sys.stdout.write("YES")
def no(): sys.stdout.write("NO")
def out(s): sys.stdout.write(str(s) + "\n")
def outpl(a):
for n in a: sys.stdout.write(str(n) + " ")
sys.stdout.write("\n")
def outpa(*args):
for n in args: sys.stdout.write(str(n) + " ")
sys.stdout.write("\n")
# === CODE HERE ===
def solve(n, t, a):
a.sort()
ans = 0
for i in a:
if t >= i:
t -= i
ans += 1
else:
break
return ans
n, t = readpi()
a = readli()
ans = solve(n, t, a)
out(ans)
``` | 0 | |
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",
... | 1,544,364,532 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 624 | 0 | n,d = map(int, input().split())
h = [int(x) for x in input().split()]
cnt = 0
for i in range(0,n-1):
for j in range(i+1,n):
if abs(h[i]-h[j]) <= d:
cnt += 2
print(cnt) | 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
n,d = map(int, input().split())
h = [int(x) for x in input().split()]
cnt = 0
for i in range(0,n-1):
for j in range(i+1,n):
if abs(h[i]-h[j]) <= d:
cnt += 2
print(cnt)
``` | 3.844 |
893 | C | Rumor | PROGRAMMING | 1,300 | [
"dfs and similar",
"graphs",
"greedy"
] | null | null | Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely. | The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once. | Print one number — the minimum amount of gold Vova has to spend in order to finish the quest. | [
"5 2\n2 5 3 4 8\n1 4\n4 5\n",
"10 0\n1 2 3 4 5 6 7 8 9 10\n",
"10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n"
] | [
"10\n",
"55\n",
"15\n"
] | In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters. | 0 | [
{
"input": "5 2\n2 5 3 4 8\n1 4\n4 5",
"output": "10"
},
{
"input": "10 0\n1 2 3 4 5 6 7 8 9 10",
"output": "55"
},
{
"input": "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10",
"output": "15"
},
{
"input": "1 0\n0",
"output": "0"
},
{
"input": "1 0\n10000000... | 1,656,502,595 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 93 | 2,150,400 | import sys
input = sys.stdin.readline
n, m = map(int, input().split())
w = list(map(int, input().split()))
d = list(range(n+1))
def F(x):
while d[x] != x:
x = d[x]
return x
for i in range(m):
a, b = map(F, map(int, input().split()))
print(a,b)
if w[a] > w[b]:
d[a] = b
else:
d[b] = a
print(sum(w[i] for i in range(1, n+1) if d[i] == i)) | Title: Rumor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely.
Input Specification:
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.
Output Specification:
Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.
Demo Input:
['5 2\n2 5 3 4 8\n1 4\n4 5\n', '10 0\n1 2 3 4 5 6 7 8 9 10\n', '10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n']
Demo Output:
['10\n', '55\n', '15\n']
Note:
In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters. | ```python
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
w = list(map(int, input().split()))
d = list(range(n+1))
def F(x):
while d[x] != x:
x = d[x]
return x
for i in range(m):
a, b = map(F, map(int, input().split()))
print(a,b)
if w[a] > w[b]:
d[a] = b
else:
d[b] = a
print(sum(w[i] for i in range(1, n+1) if d[i] == i))
``` | -1 | |
604 | B | More Cowbell | PROGRAMMING | 1,400 | [
"binary search",
"greedy"
] | null | null | Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*. | The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order. | Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*. | [
"2 1\n2 5\n",
"4 3\n2 3 5 9\n",
"3 2\n3 5 7\n"
] | [
"7\n",
"9\n",
"8\n"
] | In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | 1,000 | [
{
"input": "2 1\n2 5",
"output": "7"
},
{
"input": "4 3\n2 3 5 9",
"output": "9"
},
{
"input": "3 2\n3 5 7",
"output": "8"
},
{
"input": "20 11\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 10\n3 15 31 61 63 63 68 94 98 100",
"outp... | 1,549,467,395 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 171 | 7,680,000 | n,k = map(int,input().split())
a = list(map(int,input().split()))
if k >= n :
print(max(a))
else:
sum = max(a)
a = a[:2*(n-k)]
#print(a)
for i in range(len(a)):
sum = max(a[i]+a[- (i + 1)],sum)
print(sum) | Title: More Cowbell
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*.
Input Specification:
The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order.
Output Specification:
Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*.
Demo Input:
['2 1\n2 5\n', '4 3\n2 3 5 9\n', '3 2\n3 5 7\n']
Demo Output:
['7\n', '9\n', '8\n']
Note:
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | ```python
n,k = map(int,input().split())
a = list(map(int,input().split()))
if k >= n :
print(max(a))
else:
sum = max(a)
a = a[:2*(n-k)]
#print(a)
for i in range(len(a)):
sum = max(a[i]+a[- (i + 1)],sum)
print(sum)
``` | 3 | |
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... | 1,596,597,090 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 109 | 6,963,200 | x,y =[int(x) for x in input("").split()]
ar= list(map(int, input("").split()))
diff=[]
for i in range(len(ar)-1):
di=ar[i]-ar[i+1]
diff.append(di)
d =diff.index(max(diff))
j=ar[d]-ar[d+1]-y
if(j>0):
print(j)
else:
print(0)
| 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
x,y =[int(x) for x in input("").split()]
ar= list(map(int, input("").split()))
diff=[]
for i in range(len(ar)-1):
di=ar[i]-ar[i+1]
diff.append(di)
d =diff.index(max(diff))
j=ar[d]-ar[d+1]-y
if(j>0):
print(j)
else:
print(0)
``` | 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... | 1,622,300,950 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 154 | 512,000 | #!/usr/bin/env python
# Header
import re
# Input
get_INPUT = input()
# Logic for Word
lower,upper = 0,0
for i in get_INPUT:
if i.isupper():
upper = upper+1
elif i.islower():
lower = lower+1
if lower > upper or lower == upper:
print(get_INPUT.lower())
elif upper > lower:
print(get_INPUT.upper()) | 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
#!/usr/bin/env python
# Header
import re
# Input
get_INPUT = input()
# Logic for Word
lower,upper = 0,0
for i in get_INPUT:
if i.isupper():
upper = upper+1
elif i.islower():
lower = lower+1
if lower > upper or lower == upper:
print(get_INPUT.lower())
elif upper > lower:
print(get_INPUT.upper())
``` | 3.960546 |
998 | B | Cutting | PROGRAMMING | 1,200 | [
"dp",
"greedy",
"sortings"
] | null | null | There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.
There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.
Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ two cuts $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$. On each segment the number of even elements should be equal to the number of odd elements.
The cost of the cut between $x$ and $y$ numbers is $|x - y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins. | First line of the input contains an integer $n$ ($2 \le n \le 100$) and an integer $B$ ($1 \le B \le 100$) — the number of elements in the sequence and the number of bitcoins you have.
Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 100$) — elements of the sequence, which contains the equal number of even and odd numbers | Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins. | [
"6 4\n1 2 5 10 15 20\n",
"4 10\n1 3 2 4\n",
"6 100\n1 2 3 4 5 6\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.
In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.
In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1 + 1 = 2$ bitcoins. | 1,000 | [
{
"input": "6 4\n1 2 5 10 15 20",
"output": "1"
},
{
"input": "4 10\n1 3 2 4",
"output": "0"
},
{
"input": "6 100\n1 2 3 4 5 6",
"output": "2"
},
{
"input": "2 100\n13 78",
"output": "0"
},
{
"input": "10 1\n56 56 98 2 11 64 97 41 95 53",
"output": "0"
},
... | 1,547,630,869 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 109 | 0 | n,m=map(int,input().split())
l=list(map(int,input().split()))
x,y,ans,z,p=0,0,[],0,0
for i in range(n-1):
if l[i]%2==1:
x+=1
else:
y+=1
if x==y:
ans.append(abs(l[i]-l[i+1]))
ans.sort()
for i in range(len(ans)):
if p + ans[i]<=m:
z += 1
p += ans[i]
else:
break
print(z) | Title: Cutting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.
There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.
Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ two cuts $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$. On each segment the number of even elements should be equal to the number of odd elements.
The cost of the cut between $x$ and $y$ numbers is $|x - y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins.
Input Specification:
First line of the input contains an integer $n$ ($2 \le n \le 100$) and an integer $B$ ($1 \le B \le 100$) — the number of elements in the sequence and the number of bitcoins you have.
Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 100$) — elements of the sequence, which contains the equal number of even and odd numbers
Output Specification:
Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins.
Demo Input:
['6 4\n1 2 5 10 15 20\n', '4 10\n1 3 2 4\n', '6 100\n1 2 3 4 5 6\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.
In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.
In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1 + 1 = 2$ bitcoins. | ```python
n,m=map(int,input().split())
l=list(map(int,input().split()))
x,y,ans,z,p=0,0,[],0,0
for i in range(n-1):
if l[i]%2==1:
x+=1
else:
y+=1
if x==y:
ans.append(abs(l[i]-l[i+1]))
ans.sort()
for i in range(len(ans)):
if p + ans[i]<=m:
z += 1
p += ans[i]
else:
break
print(z)
``` | 3 | |
255 | A | Greg's Workout | PROGRAMMING | 800 | [
"implementation"
] | null | null | Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training. | The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises. | Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous. | [
"2\n2 8\n",
"3\n5 1 10\n",
"7\n3 3 2 7 9 6 8\n"
] | [
"biceps\n",
"back\n",
"chest\n"
] | In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise. | 500 | [
{
"input": "2\n2 8",
"output": "biceps"
},
{
"input": "3\n5 1 10",
"output": "back"
},
{
"input": "7\n3 3 2 7 9 6 8",
"output": "chest"
},
{
"input": "4\n5 6 6 2",
"output": "chest"
},
{
"input": "5\n8 2 2 6 3",
"output": "chest"
},
{
"input": "6\n8 7 ... | 1,579,437,636 | 2,147,483,647 | Python 3 | OK | TESTS | 61 | 218 | 512,000 | n = int(input())
chest = 0
bicep = 0
back = 0
line = input()
line = line.split()
for i in range(0,n):
line[i] = int(line[i])
if(i % 3 == 0):
chest = chest + line[i]
elif(i % 3 == 1):
bicep = bicep + line[i]
else:
back = back + line[i]
if(chest > bicep and chest > back):
print("chest")
elif(bicep > chest and bicep > back):
print("biceps")
else:
print("back") | Title: Greg's Workout
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
Output Specification:
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous.
Demo Input:
['2\n2 8\n', '3\n5 1 10\n', '7\n3 3 2 7 9 6 8\n']
Demo Output:
['biceps\n', 'back\n', 'chest\n']
Note:
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise. | ```python
n = int(input())
chest = 0
bicep = 0
back = 0
line = input()
line = line.split()
for i in range(0,n):
line[i] = int(line[i])
if(i % 3 == 0):
chest = chest + line[i]
elif(i % 3 == 1):
bicep = bicep + line[i]
else:
back = back + line[i]
if(chest > bicep and chest > back):
print("chest")
elif(bicep > chest and bicep > back):
print("biceps")
else:
print("back")
``` | 3 | |
375 | A | Divisible by Seven | PROGRAMMING | 1,600 | [
"math",
"number theory"
] | null | null | You have number *a*, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number *a* doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes. | The first line contains positive integer *a* in the decimal record. It is guaranteed that the record of number *a* contains digits: 1, 6, 8, 9. Number *a* doesn't contain any leading zeroes. The decimal representation of number *a* contains at least 4 and at most 106 characters. | Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number *a* in the required manner, print 0. | [
"1689\n",
"18906\n"
] | [
"1869\n",
"18690\n"
] | none | 500 | [
{
"input": "1689",
"output": "1869"
},
{
"input": "18906",
"output": "18690"
},
{
"input": "2419323689",
"output": "2432391689"
},
{
"input": "8589157262",
"output": "5857221986"
},
{
"input": "2717172350336955863014903670481525170997949309274087058935108848979319... | 1,671,640,231 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 102,400 | from collections import Counter
n = input()
d = Counter(n)
if d['1']==d['8'] and d['8']==d['6'] and d['9']==d['6']:
print('1869'*d['1']+'0'*d['0'])
else:
print(0) | Title: Divisible by Seven
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have number *a*, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number *a* doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input Specification:
The first line contains positive integer *a* in the decimal record. It is guaranteed that the record of number *a* contains digits: 1, 6, 8, 9. Number *a* doesn't contain any leading zeroes. The decimal representation of number *a* contains at least 4 and at most 106 characters.
Output Specification:
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number *a* in the required manner, print 0.
Demo Input:
['1689\n', '18906\n']
Demo Output:
['1869\n', '18690\n']
Note:
none | ```python
from collections import Counter
n = input()
d = Counter(n)
if d['1']==d['8'] and d['8']==d['6'] and d['9']==d['6']:
print('1869'*d['1']+'0'*d['0'])
else:
print(0)
``` | 0 | |
445 | A | DZY Loves Chessboard | PROGRAMMING | 1,200 | [
"dfs and similar",
"implementation"
] | null | null | DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad. | Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists. | [
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] | [
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] | In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | 500 | [
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n--------... | 1,615,367,839 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 77 | 0 | n,m=map(int,input().split())
matrix=[]
for i in range(n):
matrix.append(list(map(str,input())))
list1=['W','B']
flag=0
for i in range(n):
for j in range(m):
if matrix[i][j]=='.':
matrix[i][j]=list1[flag]
if j!=m-1:
flag^=1
flag^=1
for row in matrix:
print(''.join(row))
| Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | ```python
n,m=map(int,input().split())
matrix=[]
for i in range(n):
matrix.append(list(map(str,input())))
list1=['W','B']
flag=0
for i in range(n):
for j in range(m):
if matrix[i][j]=='.':
matrix[i][j]=list1[flag]
if j!=m-1:
flag^=1
flag^=1
for row in matrix:
print(''.join(row))
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | You are given two integers $a$ and $b$. Moreover, you are given a sequence $s_0, s_1, \dots, s_{n}$. All values in $s$ are integers $1$ or $-1$. It's known that sequence is $k$-periodic and $k$ divides $n+1$. In other words, for each $k \leq i \leq n$ it's satisfied that $s_{i} = s_{i - k}$.
Find out the non-negative remainder of division of $\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}$ by $10^{9} + 9$.
Note that the modulo is unusual! | The first line contains four integers $n, a, b$ and $k$ $(1 \leq n \leq 10^{9}, 1 \leq a, b \leq 10^{9}, 1 \leq k \leq 10^{5})$.
The second line contains a sequence of length $k$ consisting of characters '+' and '-'.
If the $i$-th character (0-indexed) is '+', then $s_{i} = 1$, otherwise $s_{i} = -1$.
Note that only the first $k$ members of the sequence are given, the rest can be obtained using the periodicity property. | Output a single integer — value of given expression modulo $10^{9} + 9$. | [
"2 2 3 3\n+-+\n",
"4 1 5 1\n-\n"
] | [
"7\n",
"999999228\n"
] | In the first example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i})$ = $2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2}$ = 7
In the second example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 \equiv 999999228 \pmod{10^{9} + 9}$. | 0 | [
{
"input": "2 2 3 3\n+-+",
"output": "7"
},
{
"input": "4 1 5 1\n-",
"output": "999999228"
},
{
"input": "1 1 4 2\n-+",
"output": "3"
},
{
"input": "3 1 4 4\n+--+",
"output": "45"
},
{
"input": "5 1 1 6\n++---+",
"output": "0"
},
{
"input": "5 2 2 6\n+... | 1,691,790,651 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 77 | 2,867,200 | def binary_search(a, first, last, v):
while first <= last:
mid = (first + last) // 2
if a[mid] < v:
first = mid + 1
else:
last = mid - 1
return last
def dist(a, b):
return abs(a - b)
def go(pos, elv):
if y0 == y1:
return dist(x0, x1)
res = dist(x0, pos) + dist(pos, x1) + (dist(y0, y1) + elv - 1) // elv
return res
n, m, cl, ce, v = map(int, input().split())
l = sorted(list(map(int, input().split())))
e = sorted(list(map(int, input().split())))
q = int(input())
for _ in range(q):
y0, x0, y1, x1 = map(int, input().split())
lid = binary_search(l, 0, cl - 1, x0)
eid = binary_search(e, 0, ce - 1, x0)
res = float('inf')
if 0 <= lid < cl:
res = min(res, go(l[lid], 1))
if 0 <= lid + 1 < cl:
res = min(res, go(l[lid + 1], 1))
if 0 <= eid < ce:
res = min(res, go(e[eid], v))
if 0 <= eid + 1 < ce:
res = min(res, go(e[eid + 1], v))
print(res)# 1691790651.247151 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two integers $a$ and $b$. Moreover, you are given a sequence $s_0, s_1, \dots, s_{n}$. All values in $s$ are integers $1$ or $-1$. It's known that sequence is $k$-periodic and $k$ divides $n+1$. In other words, for each $k \leq i \leq n$ it's satisfied that $s_{i} = s_{i - k}$.
Find out the non-negative remainder of division of $\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}$ by $10^{9} + 9$.
Note that the modulo is unusual!
Input Specification:
The first line contains four integers $n, a, b$ and $k$ $(1 \leq n \leq 10^{9}, 1 \leq a, b \leq 10^{9}, 1 \leq k \leq 10^{5})$.
The second line contains a sequence of length $k$ consisting of characters '+' and '-'.
If the $i$-th character (0-indexed) is '+', then $s_{i} = 1$, otherwise $s_{i} = -1$.
Note that only the first $k$ members of the sequence are given, the rest can be obtained using the periodicity property.
Output Specification:
Output a single integer — value of given expression modulo $10^{9} + 9$.
Demo Input:
['2 2 3 3\n+-+\n', '4 1 5 1\n-\n']
Demo Output:
['7\n', '999999228\n']
Note:
In the first example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i})$ = $2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2}$ = 7
In the second example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 \equiv 999999228 \pmod{10^{9} + 9}$. | ```python
def binary_search(a, first, last, v):
while first <= last:
mid = (first + last) // 2
if a[mid] < v:
first = mid + 1
else:
last = mid - 1
return last
def dist(a, b):
return abs(a - b)
def go(pos, elv):
if y0 == y1:
return dist(x0, x1)
res = dist(x0, pos) + dist(pos, x1) + (dist(y0, y1) + elv - 1) // elv
return res
n, m, cl, ce, v = map(int, input().split())
l = sorted(list(map(int, input().split())))
e = sorted(list(map(int, input().split())))
q = int(input())
for _ in range(q):
y0, x0, y1, x1 = map(int, input().split())
lid = binary_search(l, 0, cl - 1, x0)
eid = binary_search(e, 0, ce - 1, x0)
res = float('inf')
if 0 <= lid < cl:
res = min(res, go(l[lid], 1))
if 0 <= lid + 1 < cl:
res = min(res, go(l[lid + 1], 1))
if 0 <= eid < ce:
res = min(res, go(e[eid], v))
if 0 <= eid + 1 < ce:
res = min(res, go(e[eid + 1], v))
print(res)# 1691790651.247151
``` | -1 | |
513 | A | Game | PROGRAMMING | 800 | [
"constructive algorithms",
"math"
] | null | null | Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally. | The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission. | Output "First" if the first player wins and "Second" otherwise. | [
"2 2 1 2\n",
"2 1 1 1\n"
] | [
"Second\n",
"First\n"
] | Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely. | 3 | [
{
"input": "2 2 1 2",
"output": "Second"
},
{
"input": "2 1 1 1",
"output": "First"
},
{
"input": "5 7 4 1",
"output": "Second"
},
{
"input": "5 7 1 4",
"output": "Second"
},
{
"input": "5 7 10 10",
"output": "Second"
},
{
"input": "5 7 1 10",
"out... | 1,652,696,391 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | a,b,c,d=input().split()
a=int(a)
b=int(b)
c=int(c)
d=int(d)
if a<=b:
print("Second")
else:
print("First")
#,loh,bhopgfbgfmpobmdfopbhopgftbmpgftrmbnogtopp
| Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
Input Specification:
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
Output Specification:
Output "First" if the first player wins and "Second" otherwise.
Demo Input:
['2 2 1 2\n', '2 1 1 1\n']
Demo Output:
['Second\n', 'First\n']
Note:
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely. | ```python
a,b,c,d=input().split()
a=int(a)
b=int(b)
c=int(c)
d=int(d)
if a<=b:
print("Second")
else:
print("First")
#,loh,bhopgfbgfmpobmdfopbhopgftbmpgftrmbnogtopp
``` | -1 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input":... | 1,688,269,726 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | s=str(input())
l=len(s)
print(s[0],end='')
print(l-2,end='')
print(s[-1]) | Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
s=str(input())
l=len(s)
print(s[0],end='')
print(l-2,end='')
print(s[-1])
``` | 0 |
784 | E | Twisted Circuit | PROGRAMMING | 2,000 | [
"*special",
"brute force",
"implementation"
] | null | null | The input consists of four lines, each line containing a single digit 0 or 1. | Output a single digit, 0 or 1. | [
"0\n1\n1\n0\n"
] | [
"0\n"
] | none | 0 | [
{
"input": "0\n1\n1\n0",
"output": "0"
},
{
"input": "0\n0\n0\n0",
"output": "0"
},
{
"input": "1\n0\n0\n0",
"output": "1"
},
{
"input": "0\n1\n0\n0",
"output": "0"
},
{
"input": "1\n1\n0\n0",
"output": "1"
},
{
"input": "0\n0\n1\n0",
"output": "0"... | 1,491,825,009 | 6,309 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 5,529,600 | a = int(input())
b = int(input())
c = int(input())
d = int(input())
n = a*8+b*4+c*2+d
# 0123456789012345
a = "1111110101111111"
print(a[n]) | Title: Twisted Circuit
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Input Specification:
The input consists of four lines, each line containing a single digit 0 or 1.
Output Specification:
Output a single digit, 0 or 1.
Demo Input:
['0\n1\n1\n0\n']
Demo Output:
['0\n']
Note:
none | ```python
a = int(input())
b = int(input())
c = int(input())
d = int(input())
n = a*8+b*4+c*2+d
# 0123456789012345
a = "1111110101111111"
print(a[n])
``` | 0 | ||
776 | C | Molly's Chemicals | PROGRAMMING | 1,800 | [
"binary search",
"brute force",
"data structures",
"implementation",
"math"
] | null | null | Molly Hooper has *n* different kinds of chemicals arranged in a line. Each of the chemicals has an affection value, The *i*-th of them has affection value *a**i*.
Molly wants Sherlock to fall in love with her. She intends to do this by mixing a contiguous segment of chemicals together to make a love potion with total affection value as a non-negative integer power of *k*. Total affection value of a continuous segment of chemicals is the sum of affection values of each chemical in that segment.
Help her to do so in finding the total number of such segments. | The first line of input contains two integers, *n* and *k*, the number of chemicals and the number, such that the total affection value is a non-negative power of this number *k*. (1<=≤<=*n*<=≤<=105, 1<=≤<=|*k*|<=≤<=10).
Next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — affection values of chemicals. | Output a single integer — the number of valid segments. | [
"4 2\n2 2 2 2\n",
"4 -3\n3 -6 -3 12\n"
] | [
"8\n",
"3\n"
] | Do keep in mind that *k*<sup class="upper-index">0</sup> = 1.
In the first sample, Molly can get following different affection values:
- 2: segments [1, 1], [2, 2], [3, 3], [4, 4]; - 4: segments [1, 2], [2, 3], [3, 4]; - 6: segments [1, 3], [2, 4]; - 8: segments [1, 4].
Out of these, 2, 4 and 8 are powers of *k* = 2. Therefore, the answer is 8.
In the second sample, Molly can choose segments [1, 2], [3, 3], [3, 4]. | 1,500 | [
{
"input": "4 2\n2 2 2 2",
"output": "8"
},
{
"input": "4 -3\n3 -6 -3 12",
"output": "3"
},
{
"input": "14 -9\n-2 -4 62 53 90 41 35 21 85 74 85 57 10 39",
"output": "0"
},
{
"input": "20 9\n90 21 -6 -61 14 -21 -17 -65 -84 -75 -48 56 67 -50 16 65 -79 -61 92 85",
"output": ... | 1,694,868,271 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | a=list(map(int,input().split()))
d={0:1}
p=[]
c=0
for i in range(47):
p.append(pow(k,i))
ps=0
for i in a:
ps+=i
d[ps]=d.get(ps,0)+1
for x in p:
c+=d.get(ps -x,0)
print(c) | Title: Molly's Chemicals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Molly Hooper has *n* different kinds of chemicals arranged in a line. Each of the chemicals has an affection value, The *i*-th of them has affection value *a**i*.
Molly wants Sherlock to fall in love with her. She intends to do this by mixing a contiguous segment of chemicals together to make a love potion with total affection value as a non-negative integer power of *k*. Total affection value of a continuous segment of chemicals is the sum of affection values of each chemical in that segment.
Help her to do so in finding the total number of such segments.
Input Specification:
The first line of input contains two integers, *n* and *k*, the number of chemicals and the number, such that the total affection value is a non-negative power of this number *k*. (1<=≤<=*n*<=≤<=105, 1<=≤<=|*k*|<=≤<=10).
Next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — affection values of chemicals.
Output Specification:
Output a single integer — the number of valid segments.
Demo Input:
['4 2\n2 2 2 2\n', '4 -3\n3 -6 -3 12\n']
Demo Output:
['8\n', '3\n']
Note:
Do keep in mind that *k*<sup class="upper-index">0</sup> = 1.
In the first sample, Molly can get following different affection values:
- 2: segments [1, 1], [2, 2], [3, 3], [4, 4]; - 4: segments [1, 2], [2, 3], [3, 4]; - 6: segments [1, 3], [2, 4]; - 8: segments [1, 4].
Out of these, 2, 4 and 8 are powers of *k* = 2. Therefore, the answer is 8.
In the second sample, Molly can choose segments [1, 2], [3, 3], [3, 4]. | ```python
a=list(map(int,input().split()))
d={0:1}
p=[]
c=0
for i in range(47):
p.append(pow(k,i))
ps=0
for i in a:
ps+=i
d[ps]=d.get(ps,0)+1
for x in p:
c+=d.get(ps -x,0)
print(c)
``` | -1 | |
994 | B | Knights of a Polygonal Table | PROGRAMMING | 1,400 | [
"greedy",
"implementation",
"sortings"
] | null | null | Unlike Knights of a Round Table, Knights of a Polygonal Table deprived of nobility and happy to kill each other. But each knight has some power and a knight can kill another knight if and only if his power is greater than the power of victim. However, even such a knight will torment his conscience, so he can kill no more than $k$ other knights. Also, each knight has some number of coins. After a kill, a knight can pick up all victim's coins.
Now each knight ponders: how many coins he can have if only he kills other knights?
You should answer this question for each knight. | The first line contains two integers $n$ and $k$ $(1 \le n \le 10^5, 0 \le k \le \min(n-1,10))$ — the number of knights and the number $k$ from the statement.
The second line contains $n$ integers $p_1, p_2 ,\ldots,p_n$ $(1 \le p_i \le 10^9)$ — powers of the knights. All $p_i$ are distinct.
The third line contains $n$ integers $c_1, c_2 ,\ldots,c_n$ $(0 \le c_i \le 10^9)$ — the number of coins each knight has. | Print $n$ integers — the maximum number of coins each knight can have it only he kills other knights. | [
"4 2\n4 5 9 7\n1 2 11 33\n",
"5 1\n1 2 3 4 5\n1 2 3 4 5\n",
"1 0\n2\n3\n"
] | [
"1 3 46 36 ",
"1 3 5 7 9 ",
"3 "
] | Consider the first example.
- The first knight is the weakest, so he can't kill anyone. That leaves him with the only coin he initially has. - The second knight can kill the first knight and add his coin to his own two. - The third knight is the strongest, but he can't kill more than $k = 2$ other knights. It is optimal to kill the second and the fourth knights: $2+11+33 = 46$. - The fourth knight should kill the first and the second knights: $33+1+2 = 36$.
In the second example the first knight can't kill anyone, while all the others should kill the one with the index less by one than their own.
In the third example there is only one knight, so he can't kill anyone. | 1,000 | [
{
"input": "4 2\n4 5 9 7\n1 2 11 33",
"output": "1 3 46 36 "
},
{
"input": "5 1\n1 2 3 4 5\n1 2 3 4 5",
"output": "1 3 5 7 9 "
},
{
"input": "1 0\n2\n3",
"output": "3 "
},
{
"input": "7 1\n2 3 4 5 7 8 9\n0 3 7 9 5 8 9",
"output": "0 3 10 16 14 17 18 "
},
{
"input"... | 1,529,171,420 | 4,520 | Python 3 | WRONG_ANSWER | PRETESTS | 4 | 77 | 0 | some_chars=input()
sc=some_chars.split()
knight=int(sc[0])
limit=int(sc[1])
Knight_Force=['']*knight
money=['']*knight
ans=['']*knight
Knight_Force_str=input()
money_str=input()
money=money_str.split()
Knight_Force=Knight_Force_str.split()
looted=[0]*(knight-1)
ind=0
for a in range(knight):
for i in range(knight):
if a!=i and Knight_Force[i]<Knight_Force[a]:
looted[ind]=int(money[i])
ind+=1
ind=0
i=0
looted.sort()
looted.reverse()
conscience=limit
ans[a]=int(money[a])
h=0
while conscience!=0:
if conscience!=0 :
ans[a]=int(ans[a])+looted[h]
h+=1
conscience-=1
looted.clear()
looted=[0]*(knight-1)
j=knight-1
while j>=0:
ans[j]=str(ans[j])
j-=1
ans_str=' '.join(ans)
print(ans_str)
| Title: Knights of a Polygonal Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Unlike Knights of a Round Table, Knights of a Polygonal Table deprived of nobility and happy to kill each other. But each knight has some power and a knight can kill another knight if and only if his power is greater than the power of victim. However, even such a knight will torment his conscience, so he can kill no more than $k$ other knights. Also, each knight has some number of coins. After a kill, a knight can pick up all victim's coins.
Now each knight ponders: how many coins he can have if only he kills other knights?
You should answer this question for each knight.
Input Specification:
The first line contains two integers $n$ and $k$ $(1 \le n \le 10^5, 0 \le k \le \min(n-1,10))$ — the number of knights and the number $k$ from the statement.
The second line contains $n$ integers $p_1, p_2 ,\ldots,p_n$ $(1 \le p_i \le 10^9)$ — powers of the knights. All $p_i$ are distinct.
The third line contains $n$ integers $c_1, c_2 ,\ldots,c_n$ $(0 \le c_i \le 10^9)$ — the number of coins each knight has.
Output Specification:
Print $n$ integers — the maximum number of coins each knight can have it only he kills other knights.
Demo Input:
['4 2\n4 5 9 7\n1 2 11 33\n', '5 1\n1 2 3 4 5\n1 2 3 4 5\n', '1 0\n2\n3\n']
Demo Output:
['1 3 46 36 ', '1 3 5 7 9 ', '3 ']
Note:
Consider the first example.
- The first knight is the weakest, so he can't kill anyone. That leaves him with the only coin he initially has. - The second knight can kill the first knight and add his coin to his own two. - The third knight is the strongest, but he can't kill more than $k = 2$ other knights. It is optimal to kill the second and the fourth knights: $2+11+33 = 46$. - The fourth knight should kill the first and the second knights: $33+1+2 = 36$.
In the second example the first knight can't kill anyone, while all the others should kill the one with the index less by one than their own.
In the third example there is only one knight, so he can't kill anyone. | ```python
some_chars=input()
sc=some_chars.split()
knight=int(sc[0])
limit=int(sc[1])
Knight_Force=['']*knight
money=['']*knight
ans=['']*knight
Knight_Force_str=input()
money_str=input()
money=money_str.split()
Knight_Force=Knight_Force_str.split()
looted=[0]*(knight-1)
ind=0
for a in range(knight):
for i in range(knight):
if a!=i and Knight_Force[i]<Knight_Force[a]:
looted[ind]=int(money[i])
ind+=1
ind=0
i=0
looted.sort()
looted.reverse()
conscience=limit
ans[a]=int(money[a])
h=0
while conscience!=0:
if conscience!=0 :
ans[a]=int(ans[a])+looted[h]
h+=1
conscience-=1
looted.clear()
looted=[0]*(knight-1)
j=knight-1
while j>=0:
ans[j]=str(ans[j])
j-=1
ans_str=' '.join(ans)
print(ans_str)
``` | 0 | |
92 | A | Chips | PROGRAMMING | 800 | [
"implementation",
"math"
] | A. Chips | 2 | 256 | There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. | Print the number of chips the presenter ended up with. | [
"4 11\n",
"17 107\n",
"3 8\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | 500 | [
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "... | 1,618,188,790 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 124 | 0 | n,m = map(int,input().split())
k = m%(n*(n+1)//2)
c = 1
while k-c >= 0:
k = k - c
c+=1
print(k) | Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | ```python
n,m = map(int,input().split())
k = m%(n*(n+1)//2)
c = 1
while k-c >= 0:
k = k - c
c+=1
print(k)
``` | 3.969 |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,658,842,577 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 9 | 62 | 0 | n,m,a=map(int,input().split())
if n*m<=a*a:
print(1)
else:
count=0
count+=m//a
ch=n//a
count+=((n//a)-1 if ch else 0)*m
if m%a!=0: count+=1
if n%a!=0: count+=(m//a)+1
print(count)
| Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n,m,a=map(int,input().split())
if n*m<=a*a:
print(1)
else:
count=0
count+=m//a
ch=n//a
count+=((n//a)-1 if ch else 0)*m
if m%a!=0: count+=1
if n%a!=0: count+=(m//a)+1
print(count)
``` | 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"... | 1,618,393,706 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 62 | 0 | n,k=map(int,input().split())
a=n//k
print((a+1)*k)
| 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
n,k=map(int,input().split())
a=n//k
print((a+1)*k)
``` | 3 | |
81 | A | Plug-in | PROGRAMMING | 1,400 | [
"implementation"
] | A. Plug-in | 1 | 256 | Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result. | The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters. | Print the given string after it is processed. It is guaranteed that the result will contain at least one character. | [
"hhoowaaaareyyoouu\n",
"reallazy\n",
"abacabaabacabaa\n"
] | [
"wre",
"rezy",
"a"
] | none | 500 | [
{
"input": "hhoowaaaareyyoouu",
"output": "wre"
},
{
"input": "reallazy",
"output": "rezy"
},
{
"input": "abacabaabacabaa",
"output": "a"
},
{
"input": "xraccabccbry",
"output": "xy"
},
{
"input": "a",
"output": "a"
},
{
"input": "b",
"output": "b"... | 1,574,981,906 | 2,147,483,647 | PyPy 3 | OK | TESTS | 32 | 434 | 15,974,400 | from collections import deque
s = list(input())
a = deque()
for i in s:
if len(a):
if i == a[-1]:
a.pop()
else:
a.append(i)
else:
a.append(i)
print(''.join(map(str, a)))
| Title: Plug-in
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input Specification:
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output Specification:
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Demo Input:
['hhoowaaaareyyoouu\n', 'reallazy\n', 'abacabaabacabaa\n']
Demo Output:
['wre', 'rezy', 'a']
Note:
none | ```python
from collections import deque
s = list(input())
a = deque()
for i in s:
if len(a):
if i == a[-1]:
a.pop()
else:
a.append(i)
else:
a.append(i)
print(''.join(map(str, a)))
``` | 3.753245 |
913 | A | Modular Exponentiation | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | The following problem is well-known: given integers *n* and *m*, calculate
where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*.
You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108).
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108). | Output a single integer — the value of . | [
"4\n42\n",
"1\n58\n",
"98765432\n23456789\n"
] | [
"10\n",
"0\n",
"23456789\n"
] | In the first example, the remainder of division of 42 by 2<sup class="upper-index">4</sup> = 16 is equal to 10.
In the second example, 58 is divisible by 2<sup class="upper-index">1</sup> = 2 without remainder, and the answer is 0. | 500 | [
{
"input": "4\n42",
"output": "10"
},
{
"input": "1\n58",
"output": "0"
},
{
"input": "98765432\n23456789",
"output": "23456789"
},
{
"input": "8\n88127381",
"output": "149"
},
{
"input": "32\n92831989",
"output": "92831989"
},
{
"input": "92831989\n25... | 1,603,695,554 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 1,000 | 46,284,800 | n = int(input())
m = int(input())
p = (2 ** (n // 2))
p = p * p
if n%2==1:
p*=2
r = m%(p)
print(r) | Title: Modular Exponentiation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The following problem is well-known: given integers *n* and *m*, calculate
where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*.
You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108).
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108).
Output Specification:
Output a single integer — the value of .
Demo Input:
['4\n42\n', '1\n58\n', '98765432\n23456789\n']
Demo Output:
['10\n', '0\n', '23456789\n']
Note:
In the first example, the remainder of division of 42 by 2<sup class="upper-index">4</sup> = 16 is equal to 10.
In the second example, 58 is divisible by 2<sup class="upper-index">1</sup> = 2 without remainder, and the answer is 0. | ```python
n = int(input())
m = int(input())
p = (2 ** (n // 2))
p = p * p
if n%2==1:
p*=2
r = m%(p)
print(r)
``` | 0 | |
455 | A | Boredom | PROGRAMMING | 1,500 | [
"dp"
] | null | null | Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105). | Print a single integer — the maximum number of points that Alex can earn. | [
"2\n1 2\n",
"3\n1 2 3\n",
"9\n1 2 1 3 2 2 2 2 3\n"
] | [
"2\n",
"4\n",
"10\n"
] | Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points. | 500 | [
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 2 3",
"output": "4"
},
{
"input": "9\n1 2 1 3 2 2 2 2 3",
"output": "10"
},
{
"input": "5\n3 3 4 5 4",
"output": "11"
},
{
"input": "5\n5 3 5 3 4",
"output": "16"
},
{
"input": "5\n4 2 3 2 5",
... | 1,685,057,660 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 46 | 3,072,000 |
n = int(input())
counts = [0]*(10**5+2)
for num in map(int, input().split()):
counts[num] += 1
dp = [0]*(10**5+2)
for i in range(2, 10**5+2):
dp[i] = max(dp[i-1], dp[i-2] + counts[i]*i)
print(dp[10**5]) | Title: Boredom
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Print a single integer — the maximum number of points that Alex can earn.
Demo Input:
['2\n1 2\n', '3\n1 2 3\n', '9\n1 2 1 3 2 2 2 2 3\n']
Demo Output:
['2\n', '4\n', '10\n']
Note:
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points. | ```python
n = int(input())
counts = [0]*(10**5+2)
for num in map(int, input().split()):
counts[num] += 1
dp = [0]*(10**5+2)
for i in range(2, 10**5+2):
dp[i] = max(dp[i-1], dp[i-2] + counts[i]*i)
print(dp[10**5])
``` | 0 | |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
... | 1,589,993,187 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 218 | 0 | # http://codeforces.com/contest/9/problem/A
x, y = list(map(int, input().split()))
num, den = 7 - max(x, y), 6
gcd = lambda m, n: m if not n else gcd(n, m % n)
div = gcd(num, den)
print(str(num // div) + '/' + str(den // div)) | Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
# http://codeforces.com/contest/9/problem/A
x, y = list(map(int, input().split()))
num, den = 7 - max(x, y), 6
gcd = lambda m, n: m if not n else gcd(n, m % n)
div = gcd(num, den)
print(str(num // div) + '/' + str(den // div))
``` | 3.891 |
453 | A | Little Pony and Expected Maximum | PROGRAMMING | 1,600 | [
"probabilities"
] | null | null | Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times. | A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105). | Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4. | [
"6 1\n",
"6 3\n",
"2 2\n"
] | [
"3.500000000000\n",
"4.958333333333\n",
"1.750000000000\n"
] | Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value | 500 | [
{
"input": "6 1",
"output": "3.500000000000"
},
{
"input": "6 3",
"output": "4.958333333333"
},
{
"input": "2 2",
"output": "1.750000000000"
},
{
"input": "5 4",
"output": "4.433600000000"
},
{
"input": "5 8",
"output": "4.814773760000"
},
{
"input": "... | 1,609,194,745 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 1,000 | 819,200 | def binpow (a, n):
res = 1
while n:
if (n & 1):
res *= a
a *= a
n >>= 1
return res
m, n = list(map(int, input().split()))
res = 0
for k in range(1, m+1):
res += k * (binpow(k,n)-binpow((k-1),n))
res /= binpow(m,n)
print(res)
| Title: Little Pony and Expected Maximum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
Input Specification:
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
Output Specification:
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
Demo Input:
['6 1\n', '6 3\n', '2 2\n']
Demo Output:
['3.500000000000\n', '4.958333333333\n', '1.750000000000\n']
Note:
Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value | ```python
def binpow (a, n):
res = 1
while n:
if (n & 1):
res *= a
a *= a
n >>= 1
return res
m, n = list(map(int, input().split()))
res = 0
for k in range(1, m+1):
res += k * (binpow(k,n)-binpow((k-1),n))
res /= binpow(m,n)
print(res)
``` | 0 | |
494 | A | Treasure | PROGRAMMING | 1,500 | [
"greedy"
] | null | null | Malek has recently found a treasure map. While he was looking for a treasure he found a locked door. There was a string *s* written on the door consisting of characters '(', ')' and '#'. Below there was a manual on how to open the door. After spending a long time Malek managed to decode the manual and found out that the goal is to replace each '#' with one or more ')' characters so that the final string becomes beautiful.
Below there was also written that a string is called beautiful if for each *i* (1<=≤<=*i*<=≤<=|*s*|) there are no more ')' characters than '(' characters among the first *i* characters of *s* and also the total number of '(' characters is equal to the total number of ')' characters.
Help Malek open the door by telling him for each '#' character how many ')' characters he must replace it with. | The first line of the input contains a string *s* (1<=≤<=|*s*|<=≤<=105). Each character of this string is one of the characters '(', ')' or '#'. It is guaranteed that *s* contains at least one '#' character. | If there is no way of replacing '#' characters which leads to a beautiful string print <=-<=1. Otherwise for each character '#' print a separate line containing a positive integer, the number of ')' characters this character must be replaced with.
If there are several possible answers, you may output any of them. | [
"(((#)((#)\n",
"()((#((#(#()\n",
"#\n",
"(#)\n"
] | [
"1\n2\n",
"2\n2\n1",
"-1\n",
"-1\n"
] | |*s*| denotes the length of the string *s*. | 500 | [
{
"input": "(((#)((#)",
"output": "1\n2"
},
{
"input": "()((#((#(#()",
"output": "1\n1\n3"
},
{
"input": "#",
"output": "-1"
},
{
"input": "(#)",
"output": "-1"
},
{
"input": "(((((#(#(#(#()",
"output": "1\n1\n1\n5"
},
{
"input": "#))))",
"output":... | 1,584,357,149 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 13 | 171 | 5,734,400 | s = input()
ret = [None] * len(s)
bal = [None] * len(s)
b = 0
last = None
for i, c in enumerate(s):
if c == '(':
b += 1
elif c == ')':
b -= 1
else:
ret[i] = 1
b -= 1
last = i
if b < 0:
print(-1)
exit()
bal[i] = b
ret[last] += bal[-1]
print('\n'.join(map(str, filter(lambda x: x, ret))) if bal[last] - bal[-1] >= 0 else -1) | Title: Treasure
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Malek has recently found a treasure map. While he was looking for a treasure he found a locked door. There was a string *s* written on the door consisting of characters '(', ')' and '#'. Below there was a manual on how to open the door. After spending a long time Malek managed to decode the manual and found out that the goal is to replace each '#' with one or more ')' characters so that the final string becomes beautiful.
Below there was also written that a string is called beautiful if for each *i* (1<=≤<=*i*<=≤<=|*s*|) there are no more ')' characters than '(' characters among the first *i* characters of *s* and also the total number of '(' characters is equal to the total number of ')' characters.
Help Malek open the door by telling him for each '#' character how many ')' characters he must replace it with.
Input Specification:
The first line of the input contains a string *s* (1<=≤<=|*s*|<=≤<=105). Each character of this string is one of the characters '(', ')' or '#'. It is guaranteed that *s* contains at least one '#' character.
Output Specification:
If there is no way of replacing '#' characters which leads to a beautiful string print <=-<=1. Otherwise for each character '#' print a separate line containing a positive integer, the number of ')' characters this character must be replaced with.
If there are several possible answers, you may output any of them.
Demo Input:
['(((#)((#)\n', '()((#((#(#()\n', '#\n', '(#)\n']
Demo Output:
['1\n2\n', '2\n2\n1', '-1\n', '-1\n']
Note:
|*s*| denotes the length of the string *s*. | ```python
s = input()
ret = [None] * len(s)
bal = [None] * len(s)
b = 0
last = None
for i, c in enumerate(s):
if c == '(':
b += 1
elif c == ')':
b -= 1
else:
ret[i] = 1
b -= 1
last = i
if b < 0:
print(-1)
exit()
bal[i] = b
ret[last] += bal[-1]
print('\n'.join(map(str, filter(lambda x: x, ret))) if bal[last] - bal[-1] >= 0 else -1)
``` | 0 | |
242 | B | Big Segment | PROGRAMMING | 1,100 | [
"implementation",
"sortings"
] | null | null | A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide. | Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input. | [
"3\n1 1\n2 2\n3 3\n",
"6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n"
] | [
"-1\n",
"3\n"
] | none | 1,000 | [
{
"input": "3\n1 1\n2 2\n3 3",
"output": "-1"
},
{
"input": "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10",
"output": "3"
},
{
"input": "4\n1 5\n2 2\n2 4\n2 5",
"output": "1"
},
{
"input": "5\n3 3\n1 3\n2 2\n2 3\n1 2",
"output": "2"
},
{
"input": "7\n7 7\n8 8\n3 7\n1 6\n1 ... | 1,549,416,616 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 18 | 2,000 | 5,222,400 |
a = []
b = []
n = int(input())
for i in range (n): #[first, last)
x, y = map(int, input().split())
a.append(x)
b.append(y)
answer = False
for i in range (n):
if a[i] == min(a) and b[i] == max(b):
print(i+1)
answer = True
if answer == False:
print(-1)
| Title: Big Segment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide.
Output Specification:
Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n']
Demo Output:
['-1\n', '3\n']
Note:
none | ```python
a = []
b = []
n = int(input())
for i in range (n): #[first, last)
x, y = map(int, input().split())
a.append(x)
b.append(y)
answer = False
for i in range (n):
if a[i] == min(a) and b[i] == max(b):
print(i+1)
answer = True
if answer == False:
print(-1)
``` | 0 | |
175 | A | Robot Bicorn Attack | PROGRAMMING | 1,400 | [
"brute force",
"implementation"
] | null | null | Vasya plays Robot Bicorn Attack.
The game consists of three rounds. For each one a non-negative integer amount of points is given. The result of the game is the sum of obtained points. Vasya has already played three rounds and wrote obtained points one by one (without leading zeros) into the string *s*. Vasya decided to brag about his achievement to the friends. However, he has forgotten how many points he got for each round. The only thing he remembers is the string *s*.
Help Vasya to find out what is the maximum amount of points he could get. Take into account that Vasya played Robot Bicorn Attack for the first time, so he could not get more than 1000000 (106) points for one round. | The only line of input contains non-empty string *s* obtained by Vasya. The string consists of digits only. The string length does not exceed 30 characters. | Print the only number — the maximum amount of points Vasya could get. If Vasya is wrong and the string could not be obtained according to the rules then output number -1. | [
"1234\n",
"9000\n",
"0009\n"
] | [
"37\n",
"90\n",
"-1\n"
] | In the first example the string must be split into numbers 1, 2 and 34.
In the second example the string must be split into numbers 90, 0 and 0.
In the third example the string is incorrect, because after splitting the string into 3 numbers number 00 or 09 will be obtained, but numbers cannot have leading zeroes. | 500 | [
{
"input": "1234",
"output": "37"
},
{
"input": "9000",
"output": "90"
},
{
"input": "0009",
"output": "-1"
},
{
"input": "100000010000001000000",
"output": "3000000"
},
{
"input": "1000000011",
"output": "1000011"
},
{
"input": "9991",
"output": "... | 1,556,048,800 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 109 | 0 | x = str(input())
y = len(x)
s=0
p=0
for i in range(y-1):
if(int(x[s] + x[s+1]) > p):
p = int(x[s] + x[s+1])
s+=1
x = x.replace(str(p), '')
aw=sum(map(int,str(x))) + p
print(aw) | Title: Robot Bicorn Attack
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya plays Robot Bicorn Attack.
The game consists of three rounds. For each one a non-negative integer amount of points is given. The result of the game is the sum of obtained points. Vasya has already played three rounds and wrote obtained points one by one (without leading zeros) into the string *s*. Vasya decided to brag about his achievement to the friends. However, he has forgotten how many points he got for each round. The only thing he remembers is the string *s*.
Help Vasya to find out what is the maximum amount of points he could get. Take into account that Vasya played Robot Bicorn Attack for the first time, so he could not get more than 1000000 (106) points for one round.
Input Specification:
The only line of input contains non-empty string *s* obtained by Vasya. The string consists of digits only. The string length does not exceed 30 characters.
Output Specification:
Print the only number — the maximum amount of points Vasya could get. If Vasya is wrong and the string could not be obtained according to the rules then output number -1.
Demo Input:
['1234\n', '9000\n', '0009\n']
Demo Output:
['37\n', '90\n', '-1\n']
Note:
In the first example the string must be split into numbers 1, 2 and 34.
In the second example the string must be split into numbers 90, 0 and 0.
In the third example the string is incorrect, because after splitting the string into 3 numbers number 00 or 09 will be obtained, but numbers cannot have leading zeroes. | ```python
x = str(input())
y = len(x)
s=0
p=0
for i in range(y-1):
if(int(x[s] + x[s+1]) > p):
p = int(x[s] + x[s+1])
s+=1
x = x.replace(str(p), '')
aw=sum(map(int,str(x))) + p
print(aw)
``` | 0 | |
292 | A | SMSC | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Some large corporation where Polycarpus works has its own short message service center (SMSC). The center's task is to send all sorts of crucial information. Polycarpus decided to check the efficiency of the SMSC.
For that, he asked to give him the statistics of the performance of the SMSC for some period of time. In the end, Polycarpus got a list of *n* tasks that went to the SMSC of the corporation. Each task was described by the time it was received by the SMSC and the number of text messages to send. More formally, the *i*-th task was described by two integers *t**i* and *c**i* — the receiving time (the second) and the number of the text messages, correspondingly.
Polycarpus knows that the SMSC cannot send more than one text message per second. The SMSC uses a queue to organize its work. Consider a time moment *x*, the SMSC work at that moment as follows:
1. If at the time moment *x* the queue is not empty, then SMSC sends one message from the queue (SMSC gets the message from the head of the queue). Otherwise it doesn't send messages at the time moment *x*. 1. If at the time moment *x* SMSC receives a task, then it adds to the queue all the messages from this task (SMSC adds messages to the tail of the queue). Note, that the messages from the task cannot be send at time moment *x*. That's because the decision about sending message or not is made at point 1 before adding these messages to the queue.
Given the information about all *n* tasks, Polycarpus wants to count two values: the time when the last text message was sent and the maximum size of the queue at some time. Help him count these two characteristics he needs to evaluate the efficiency of the SMSC. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=103) — the number of tasks of the SMSC. Next *n* lines contain the tasks' descriptions: the *i*-th line contains two space-separated integers *t**i* and *c**i* (1<=≤<=*t**i*,<=*c**i*<=≤<=106) — the time (the second) when the *i*-th task was received and the number of messages to send, correspondingly.
It is guaranteed that all tasks were received at different moments of time. It is guaranteed that the tasks are sorted in the chronological order, that is, *t**i*<=<<=*t**i*<=+<=1 for all integer *i* (1<=≤<=*i*<=<<=*n*). | In a single line print two space-separated integers — the time when the last text message was sent and the maximum queue size at a certain moment of time. | [
"2\n1 1\n2 1\n",
"1\n1000000 10\n",
"3\n3 3\n4 3\n5 3\n"
] | [
"3 1\n",
"1000010 10\n",
"12 7\n"
] | In the first test sample:
- second 1: the first message has appeared in the queue, the queue's size is 1; - second 2: the first message is sent, the second message has been received, the queue's size is 1; - second 3: the second message is sent, the queue's size is 0,
Thus, the maximum size of the queue is 1, the last message was sent at the second 3. | 1,000 | [
{
"input": "2\n1 1\n2 1",
"output": "3 1"
},
{
"input": "1\n1000000 10",
"output": "1000010 10"
},
{
"input": "3\n3 3\n4 3\n5 3",
"output": "12 7"
},
{
"input": "1\n1 1",
"output": "2 1"
},
{
"input": "2\n1 11\n100 10",
"output": "110 11"
},
{
"input":... | 1,574,444,576 | 2,147,483,647 | Python 3 | OK | TESTS | 38 | 218 | 204,800 | n = int(input())
message = 0
m = 0
l = 0
for _ in range(n):
t, c = map(int, input().split())
message = max(0, message-(t-l))
message += c
m = max(message, m)
l = t
print(l+message, m)
| Title: SMSC
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some large corporation where Polycarpus works has its own short message service center (SMSC). The center's task is to send all sorts of crucial information. Polycarpus decided to check the efficiency of the SMSC.
For that, he asked to give him the statistics of the performance of the SMSC for some period of time. In the end, Polycarpus got a list of *n* tasks that went to the SMSC of the corporation. Each task was described by the time it was received by the SMSC and the number of text messages to send. More formally, the *i*-th task was described by two integers *t**i* and *c**i* — the receiving time (the second) and the number of the text messages, correspondingly.
Polycarpus knows that the SMSC cannot send more than one text message per second. The SMSC uses a queue to organize its work. Consider a time moment *x*, the SMSC work at that moment as follows:
1. If at the time moment *x* the queue is not empty, then SMSC sends one message from the queue (SMSC gets the message from the head of the queue). Otherwise it doesn't send messages at the time moment *x*. 1. If at the time moment *x* SMSC receives a task, then it adds to the queue all the messages from this task (SMSC adds messages to the tail of the queue). Note, that the messages from the task cannot be send at time moment *x*. That's because the decision about sending message or not is made at point 1 before adding these messages to the queue.
Given the information about all *n* tasks, Polycarpus wants to count two values: the time when the last text message was sent and the maximum size of the queue at some time. Help him count these two characteristics he needs to evaluate the efficiency of the SMSC.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=103) — the number of tasks of the SMSC. Next *n* lines contain the tasks' descriptions: the *i*-th line contains two space-separated integers *t**i* and *c**i* (1<=≤<=*t**i*,<=*c**i*<=≤<=106) — the time (the second) when the *i*-th task was received and the number of messages to send, correspondingly.
It is guaranteed that all tasks were received at different moments of time. It is guaranteed that the tasks are sorted in the chronological order, that is, *t**i*<=<<=*t**i*<=+<=1 for all integer *i* (1<=≤<=*i*<=<<=*n*).
Output Specification:
In a single line print two space-separated integers — the time when the last text message was sent and the maximum queue size at a certain moment of time.
Demo Input:
['2\n1 1\n2 1\n', '1\n1000000 10\n', '3\n3 3\n4 3\n5 3\n']
Demo Output:
['3 1\n', '1000010 10\n', '12 7\n']
Note:
In the first test sample:
- second 1: the first message has appeared in the queue, the queue's size is 1; - second 2: the first message is sent, the second message has been received, the queue's size is 1; - second 3: the second message is sent, the queue's size is 0,
Thus, the maximum size of the queue is 1, the last message was sent at the second 3. | ```python
n = int(input())
message = 0
m = 0
l = 0
for _ in range(n):
t, c = map(int, input().split())
message = max(0, message-(t-l))
message += c
m = max(message, m)
l = t
print(l+message, m)
``` | 3 | |
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"... | 1,478,525,383 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 62 | 0 | n, k = [int(x) for x in input().split(' ')]
print(n + k-n%k)
| 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
n, k = [int(x) for x in input().split(' ')]
print(n + k-n%k)
``` | 3 | |
604 | A | Uncowed Forces | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is *x*, and Kevin submitted correctly at minute *m* but made *w* wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer. | The first line of the input contains five space-separated integers *m*1, *m*2, *m*3, *m*4, *m*5, where *m**i* (0<=≤<=*m**i*<=≤<=119) is the time of Kevin's last submission for problem *i*. His last submission is always correct and gets accepted.
The second line contains five space-separated integers *w*1, *w*2, *w*3, *w*4, *w*5, where *w**i* (0<=≤<=*w**i*<=≤<=10) is Kevin's number of wrong submissions on problem *i*.
The last line contains two space-separated integers *h**s* and *h**u* (0<=≤<=*h**s*,<=*h**u*<=≤<=20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively. | Print a single integer, the value of Kevin's final score. | [
"20 40 60 80 100\n0 1 2 3 4\n1 0\n",
"119 119 119 119 119\n0 0 0 0 0\n10 0\n"
] | [
"4900\n",
"4930\n"
] | In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/42158dc2bc78cd21fa679530ae9ef8b9ea298d15.png" style="max-width: 100.0%;max-height: 100.0%;"/> of the points on each problem. So his score from solving problems is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fdf392d8508500b57f8057ac0c4c892ab5f925a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Adding in 10·100 = 1000 points from hacks, his total score becomes 3930 + 1000 = 4930. | 500 | [
{
"input": "20 40 60 80 100\n0 1 2 3 4\n1 0",
"output": "4900"
},
{
"input": "119 119 119 119 119\n0 0 0 0 0\n10 0",
"output": "4930"
},
{
"input": "3 6 13 38 60\n6 10 10 3 8\n9 9",
"output": "5088"
},
{
"input": "21 44 11 68 75\n6 2 4 8 4\n2 8",
"output": "4522"
},
{... | 1,448,984,502 | 402 | Python 3 | OK | TESTS | 57 | 61 | 0 | max_scores = [ 500, 1000, 1500, 2000, 2500 ]
corects = [ int(c) for c in input().split() ]
wrongs = [ int(w) for w in input().split() ]
shacks, unshacks = map(int, input().split())
final_score = shacks*100 - unshacks*50
def score(x, m, w):
return max(0.3*x, (1-(m/250))*x-50*w)
for i in range(5):
final_score += score(max_scores[i], corects[i], wrongs[i])
print(int(final_score)) | Title: Uncowed Forces
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is *x*, and Kevin submitted correctly at minute *m* but made *w* wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer.
Input Specification:
The first line of the input contains five space-separated integers *m*1, *m*2, *m*3, *m*4, *m*5, where *m**i* (0<=≤<=*m**i*<=≤<=119) is the time of Kevin's last submission for problem *i*. His last submission is always correct and gets accepted.
The second line contains five space-separated integers *w*1, *w*2, *w*3, *w*4, *w*5, where *w**i* (0<=≤<=*w**i*<=≤<=10) is Kevin's number of wrong submissions on problem *i*.
The last line contains two space-separated integers *h**s* and *h**u* (0<=≤<=*h**s*,<=*h**u*<=≤<=20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively.
Output Specification:
Print a single integer, the value of Kevin's final score.
Demo Input:
['20 40 60 80 100\n0 1 2 3 4\n1 0\n', '119 119 119 119 119\n0 0 0 0 0\n10 0\n']
Demo Output:
['4900\n', '4930\n']
Note:
In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/42158dc2bc78cd21fa679530ae9ef8b9ea298d15.png" style="max-width: 100.0%;max-height: 100.0%;"/> of the points on each problem. So his score from solving problems is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fdf392d8508500b57f8057ac0c4c892ab5f925a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Adding in 10·100 = 1000 points from hacks, his total score becomes 3930 + 1000 = 4930. | ```python
max_scores = [ 500, 1000, 1500, 2000, 2500 ]
corects = [ int(c) for c in input().split() ]
wrongs = [ int(w) for w in input().split() ]
shacks, unshacks = map(int, input().split())
final_score = shacks*100 - unshacks*50
def score(x, m, w):
return max(0.3*x, (1-(m/250))*x-50*w)
for i in range(5):
final_score += score(max_scores[i], corects[i], wrongs[i])
print(int(final_score))
``` | 3 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input":... | 1,687,993,101 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | x = input()
if not x.isdigit():
if len(x) > 10:
print((x[0]) + str(len(x)-2) + (x[-1]))
else:
print(x) | Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
x = input()
if not x.isdigit():
if len(x) > 10:
print((x[0]) + str(len(x)-2) + (x[-1]))
else:
print(x)
``` | 0 |
834 | A | The Useless Toy | PROGRAMMING | 900 | [
"implementation"
] | null | null | Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this. | There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position. | Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise. | [
"^ >\n1\n",
"< ^\n3\n",
"^ v\n6\n"
] | [
"cw\n",
"ccw\n",
"undefined\n"
] | none | 500 | [
{
"input": "^ >\n1",
"output": "cw"
},
{
"input": "< ^\n3",
"output": "ccw"
},
{
"input": "^ v\n6",
"output": "undefined"
},
{
"input": "^ >\n999999999",
"output": "ccw"
},
{
"input": "> v\n1",
"output": "cw"
},
{
"input": "v <\n1",
"output": "cw"
... | 1,501,750,893 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 93 | 23,142,400 | dict = {'v': 0, '<': 1, '^': 2, '>': 3}
i, j = input().split()
num = int(input())
l = dict[i]
r = dict[j]
s = (l-r if l>r else r-l) % 4
if s == 0 or s == 2:
print("undefined")
elif s == 1:
print("cw")
else:
print("ccw")
| Title: The Useless Toy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this.
Input Specification:
There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position.
Output Specification:
Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise.
Demo Input:
['^ >\n1\n', '< ^\n3\n', '^ v\n6\n']
Demo Output:
['cw\n', 'ccw\n', 'undefined\n']
Note:
none | ```python
dict = {'v': 0, '<': 1, '^': 2, '>': 3}
i, j = input().split()
num = int(input())
l = dict[i]
r = dict[j]
s = (l-r if l>r else r-l) % 4
if s == 0 or s == 2:
print("undefined")
elif s == 1:
print("cw")
else:
print("ccw")
``` | 0 | |
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... | 1,655,134,274 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 0 | 0 | X,Y=map(int,input().split())
res=int(A*B/2)
print(res) | 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
X,Y=map(int,input().split())
res=int(A*B/2)
print(res)
``` | -1 |
0 | none | none | none | 0 | [
"none"
] | null | null | Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built *n* towers in a row. The *i*-th tower is made of *h**i* identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109) — sizes of towers. | Print the number of operations needed to destroy all towers. | [
"6\n2 1 4 6 2 2\n",
"7\n3 3 3 1 3 3 3\n"
] | [
"3\n",
"2\n"
] | The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. | 0 | [
{
"input": "6\n2 1 4 6 2 2",
"output": "3"
},
{
"input": "7\n3 3 3 1 3 3 3",
"output": "2"
},
{
"input": "7\n5128 5672 5805 5452 5882 5567 5032",
"output": "4"
},
{
"input": "10\n1 2 2 3 5 5 5 4 2 1",
"output": "5"
},
{
"input": "14\n20 20 20 20 20 20 3 20 20 20 2... | 1,500,489,881 | 6,281 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
#define mp make_pair
#define pb push_back
#define F first
#define S second
const int MAX_SIZE = 1e5 + 10;
const ll inf = 1e14;
vector<pll > graph[MAX_SIZE];
priority_queue< pll, vector<pll > , greater< pll > > pq;
vector<ll> dist(MAX_SIZE);
int n;
void dijkstra(int s)
{
for(int i = 0; i <= n + 1; i++)
dist[i] = inf;
dist[s] = 0;
pq.push({0, s});
while(!pq.empty())
{
int u = pq.top().S;
pq.pop();
for(auto &e: graph[u])
{
int v = e.F;
ll w = e.S;
if(dist[v] > w + dist[u])
{
dist[v] = w + dist[u];
pq.push({dist[v], v});
}
}
}
}
int main()
{
ll h;
cin >> n;
for(int i = 1; i <= n; i++)
{
cin >> h;
graph[n + 2].pb({i, h});
}
graph[n + 2].pb({0, 0});
graph[n + 2].pb({n + 1, 0});
for(int i = 1; i <= n + 1; i++)
{
graph[i].pb({i - 1, 1});
graph[i - 1].pb({i, 1});
}
dijkstra(n + 2);
ll ans = 0;
for(int i = 1; i <= n; i++)
{
ans = max(ans, dist[i]);
}
cout << ans << endl;
return 0;
} | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built *n* towers in a row. The *i*-th tower is made of *h**i* identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109) — sizes of towers.
Output Specification:
Print the number of operations needed to destroy all towers.
Demo Input:
['6\n2 1 4 6 2 2\n', '7\n3 3 3 1 3 3 3\n']
Demo Output:
['3\n', '2\n']
Note:
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. | ```python
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
#define mp make_pair
#define pb push_back
#define F first
#define S second
const int MAX_SIZE = 1e5 + 10;
const ll inf = 1e14;
vector<pll > graph[MAX_SIZE];
priority_queue< pll, vector<pll > , greater< pll > > pq;
vector<ll> dist(MAX_SIZE);
int n;
void dijkstra(int s)
{
for(int i = 0; i <= n + 1; i++)
dist[i] = inf;
dist[s] = 0;
pq.push({0, s});
while(!pq.empty())
{
int u = pq.top().S;
pq.pop();
for(auto &e: graph[u])
{
int v = e.F;
ll w = e.S;
if(dist[v] > w + dist[u])
{
dist[v] = w + dist[u];
pq.push({dist[v], v});
}
}
}
}
int main()
{
ll h;
cin >> n;
for(int i = 1; i <= n; i++)
{
cin >> h;
graph[n + 2].pb({i, h});
}
graph[n + 2].pb({0, 0});
graph[n + 2].pb({n + 1, 0});
for(int i = 1; i <= n + 1; i++)
{
graph[i].pb({i - 1, 1});
graph[i - 1].pb({i, 1});
}
dijkstra(n + 2);
ll ans = 0;
for(int i = 1; i <= n; i++)
{
ans = max(ans, dist[i]);
}
cout << ans << endl;
return 0;
}
``` | -1 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Stepan has the newest electronic device with a display. Different digits can be shown on it. Each digit is shown on a seven-section indicator like it is shown on the picture below.
So, for example, to show the digit 3 on the display, 5 sections must be highlighted; and for the digit 6, 6 sections must be highlighted.
The battery of the newest device allows to highlight at most *n* sections on the display.
Stepan wants to know the maximum possible integer number which can be shown on the display of his newest device. Your task is to determine this number. Note that this number must not contain leading zeros. Assume that the size of the display is enough to show any integer. | The first line contains the integer *n* (2<=≤<=*n*<=≤<=100<=000) — the maximum number of sections which can be highlighted on the display. | Print the maximum integer which can be shown on the display of Stepan's newest device. | [
"2\n",
"3\n"
] | [
"1\n",
"7\n"
] | none | 0 | [
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "7"
},
{
"input": "4",
"output": "11"
},
{
"input": "5",
"output": "71"
},
{
"input": "6",
"output": "111"
},
{
"input": "85651",
"output": "711111111111111111111111111111111111111111111111... | 1,491,409,983 | 3,483 | Python 3 | OK | TESTS | 21 | 62 | 7,782,400 | n = int(input())
number = [1] * (n // 2)
if n % 2:
number[0] = 7
print(''.join(map(str, number)))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Stepan has the newest electronic device with a display. Different digits can be shown on it. Each digit is shown on a seven-section indicator like it is shown on the picture below.
So, for example, to show the digit 3 on the display, 5 sections must be highlighted; and for the digit 6, 6 sections must be highlighted.
The battery of the newest device allows to highlight at most *n* sections on the display.
Stepan wants to know the maximum possible integer number which can be shown on the display of his newest device. Your task is to determine this number. Note that this number must not contain leading zeros. Assume that the size of the display is enough to show any integer.
Input Specification:
The first line contains the integer *n* (2<=≤<=*n*<=≤<=100<=000) — the maximum number of sections which can be highlighted on the display.
Output Specification:
Print the maximum integer which can be shown on the display of Stepan's newest device.
Demo Input:
['2\n', '3\n']
Demo Output:
['1\n', '7\n']
Note:
none | ```python
n = int(input())
number = [1] * (n // 2)
if n % 2:
number[0] = 7
print(''.join(map(str, number)))
``` | 3 | |
903 | A | Hungry Student Problem | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.
CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken.
Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks.
Help Ivan to answer this question for several values of *x*! | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases.
The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat. | Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO. | [
"2\n6\n5\n"
] | [
"YES\nNO\n"
] | In the first example Ivan can buy two small portions.
In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much. | 0 | [
{
"input": "2\n6\n5",
"output": "YES\nNO"
},
{
"input": "100\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n20\n21\n22\n23\n24\n25\n26\n27\n28\n29\n30\n31\n32\n33\n34\n35\n36\n37\n38\n39\n40\n41\n42\n43\n44\n45\n46\n47\n48\n49\n50\n51\n52\n53\n54\n55\n56\n57\n58\n59\n60\n61\n62\... | 1,585,770,848 | 2,147,483,647 | Python 3 | OK | TESTS | 8 | 93 | 0 | t=int(input())
s=[]
for hjbbh in range(0,t):
x=int(input())
f=int(x//7)
h=0
for i in range(0,f+1):
if (x-i*7)%3==0:
s.append('YES')
h+=1
break
if h==0:
s.append('NO')
for i in s:
print(i)
| Title: Hungry Student Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.
CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken.
Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks.
Help Ivan to answer this question for several values of *x*!
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases.
The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat.
Output Specification:
Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO.
Demo Input:
['2\n6\n5\n']
Demo Output:
['YES\nNO\n']
Note:
In the first example Ivan can buy two small portions.
In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much. | ```python
t=int(input())
s=[]
for hjbbh in range(0,t):
x=int(input())
f=int(x//7)
h=0
for i in range(0,f+1):
if (x-i*7)%3==0:
s.append('YES')
h+=1
break
if h==0:
s.append('NO')
for i in s:
print(i)
``` | 3 | |
245 | A | System Administrator | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarpus is a system administrator. There are two servers under his strict guidance — *a* and *b*. To stay informed about the servers' performance, Polycarpus executes commands "ping a" and "ping b". Each ping command sends exactly ten packets to the server specified in the argument of the command. Executing a program results in two integers *x* and *y* (*x*<=+<=*y*<==<=10; *x*,<=*y*<=≥<=0). These numbers mean that *x* packets successfully reached the corresponding server through the network and *y* packets were lost.
Today Polycarpus has performed overall *n* ping commands during his workday. Now for each server Polycarpus wants to know whether the server is "alive" or not. Polycarpus thinks that the server is "alive", if at least half of the packets that we send to this server reached it successfully along the network.
Help Polycarpus, determine for each server, whether it is "alive" or not by the given commands and their results. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of commands Polycarpus has fulfilled. Each of the following *n* lines contains three integers — the description of the commands. The *i*-th of these lines contains three space-separated integers *t**i*, *x**i*, *y**i* (1<=≤<=*t**i*<=≤<=2; *x**i*,<=*y**i*<=≥<=0; *x**i*<=+<=*y**i*<==<=10). If *t**i*<==<=1, then the *i*-th command is "ping a", otherwise the *i*-th command is "ping b". Numbers *x**i*, *y**i* represent the result of executing this command, that is, *x**i* packets reached the corresponding server successfully and *y**i* packets were lost.
It is guaranteed that the input has at least one "ping a" command and at least one "ping b" command. | In the first line print string "LIVE" (without the quotes) if server *a* is "alive", otherwise print "DEAD" (without the quotes).
In the second line print the state of server *b* in the similar format. | [
"2\n1 5 5\n2 6 4\n",
"3\n1 0 10\n2 0 10\n1 10 0\n"
] | [
"LIVE\nLIVE\n",
"LIVE\nDEAD\n"
] | Consider the first test case. There 10 packets were sent to server *a*, 5 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall there were 10 packets sent to server *b*, 6 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network.
Consider the second test case. There were overall 20 packages sent to server *a*, 10 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall 10 packets were sent to server *b*, 0 of them reached it. Therefore, less than half of all packets sent to this server successfully reached it through the network. | 0 | [
{
"input": "2\n1 5 5\n2 6 4",
"output": "LIVE\nLIVE"
},
{
"input": "3\n1 0 10\n2 0 10\n1 10 0",
"output": "LIVE\nDEAD"
},
{
"input": "10\n1 3 7\n2 4 6\n1 2 8\n2 5 5\n2 10 0\n2 10 0\n1 8 2\n2 2 8\n2 10 0\n1 1 9",
"output": "DEAD\nLIVE"
},
{
"input": "11\n1 8 2\n1 6 4\n1 9 1\n1... | 1,580,924,129 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 280 | 0 | n = int(input())
a,b,c,d=0,0,0,0
for i in range(n):
a1,b1,c1 = map(int,input().split())
if a1 ==1:
a += b1
c += c1
if a1 ==2:
b += b1
d += c1
if a >= c:
print("LIVE")
if b >= d:
print("LIVE")
else:
print("DEAD")
| Title: System Administrator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus is a system administrator. There are two servers under his strict guidance — *a* and *b*. To stay informed about the servers' performance, Polycarpus executes commands "ping a" and "ping b". Each ping command sends exactly ten packets to the server specified in the argument of the command. Executing a program results in two integers *x* and *y* (*x*<=+<=*y*<==<=10; *x*,<=*y*<=≥<=0). These numbers mean that *x* packets successfully reached the corresponding server through the network and *y* packets were lost.
Today Polycarpus has performed overall *n* ping commands during his workday. Now for each server Polycarpus wants to know whether the server is "alive" or not. Polycarpus thinks that the server is "alive", if at least half of the packets that we send to this server reached it successfully along the network.
Help Polycarpus, determine for each server, whether it is "alive" or not by the given commands and their results.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of commands Polycarpus has fulfilled. Each of the following *n* lines contains three integers — the description of the commands. The *i*-th of these lines contains three space-separated integers *t**i*, *x**i*, *y**i* (1<=≤<=*t**i*<=≤<=2; *x**i*,<=*y**i*<=≥<=0; *x**i*<=+<=*y**i*<==<=10). If *t**i*<==<=1, then the *i*-th command is "ping a", otherwise the *i*-th command is "ping b". Numbers *x**i*, *y**i* represent the result of executing this command, that is, *x**i* packets reached the corresponding server successfully and *y**i* packets were lost.
It is guaranteed that the input has at least one "ping a" command and at least one "ping b" command.
Output Specification:
In the first line print string "LIVE" (without the quotes) if server *a* is "alive", otherwise print "DEAD" (without the quotes).
In the second line print the state of server *b* in the similar format.
Demo Input:
['2\n1 5 5\n2 6 4\n', '3\n1 0 10\n2 0 10\n1 10 0\n']
Demo Output:
['LIVE\nLIVE\n', 'LIVE\nDEAD\n']
Note:
Consider the first test case. There 10 packets were sent to server *a*, 5 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall there were 10 packets sent to server *b*, 6 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network.
Consider the second test case. There were overall 20 packages sent to server *a*, 10 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall 10 packets were sent to server *b*, 0 of them reached it. Therefore, less than half of all packets sent to this server successfully reached it through the network. | ```python
n = int(input())
a,b,c,d=0,0,0,0
for i in range(n):
a1,b1,c1 = map(int,input().split())
if a1 ==1:
a += b1
c += c1
if a1 ==2:
b += b1
d += c1
if a >= c:
print("LIVE")
if b >= d:
print("LIVE")
else:
print("DEAD")
``` | 0 | |
811 | A | Vladik and Courtesy | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy. | Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively. | Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise. | [
"1 1\n",
"7 6\n"
] | [
"Valera\n",
"Vladik\n"
] | Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 500 | [
{
"input": "1 1",
"output": "Valera"
},
{
"input": "7 6",
"output": "Vladik"
},
{
"input": "25 38",
"output": "Vladik"
},
{
"input": "8311 2468",
"output": "Valera"
},
{
"input": "250708 857756",
"output": "Vladik"
},
{
"input": "957985574 24997558",
... | 1,677,840,145 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 42 | 62 | 1,331,200 | n, k = map(int, input().split())
d = 1
while True:
if n-d >= 0:
n = n -d
d += 1
else:
s = "Vladik"
break
if k-d >= 0:
k = k -d
d += 1
else:
s = "Valera"
break
print(s) | Title: Vladik and Courtesy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
Input Specification:
Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.
Output Specification:
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
Demo Input:
['1 1\n', '7 6\n']
Demo Output:
['Valera\n', 'Vladik\n']
Note:
Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
n, k = map(int, input().split())
d = 1
while True:
if n-d >= 0:
n = n -d
d += 1
else:
s = "Vladik"
break
if k-d >= 0:
k = k -d
d += 1
else:
s = "Valera"
break
print(s)
``` | 3 | |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,593,221,090 | 2,147,483,647 | PyPy 3 | OK | TESTS | 102 | 171 | 20,172,800 | f = input()
m = input()
z = ""
for i in range(len(f)):
if f[i]!=m[i]:
z+="1"
else:
z+="0"
print(z) | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
f = input()
m = input()
z = ""
for i in range(len(f)):
if f[i]!=m[i]:
z+="1"
else:
z+="0"
print(z)
``` | 3.919675 |
295 | A | Greg and Array | PROGRAMMING | 1,400 | [
"data structures",
"implementation"
] | null | null | Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*.
Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array.
Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg. | The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array.
Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105).
Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*).
The numbers in the lines are separated by single spaces. | On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier. | [
"3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n",
"1 1 1\n1\n1 1 1\n1 1\n",
"4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n"
] | [
"9 18 17\n",
"2\n",
"5 18 31 20\n"
] | none | 500 | [
{
"input": "3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3",
"output": "9 18 17"
},
{
"input": "1 1 1\n1\n1 1 1\n1 1",
"output": "2"
},
{
"input": "4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3",
"output": "5 18 31 20"
},
{
"input": "1 1 1\n0\n1 1 0\n1 1... | 1,659,458,504 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 23 | 1,500 | 12,492,800 | """
செயல் பேசும் ஆழம் இங்கே சொற்கள் பேசுமா?
Focus, Determination and Sheer-Will
The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
"""
import os, sys, math, cmath, time, collections
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import nsmallest, nlargest, heapify, heappop, heappush, heapreplace
from math import ceil, floor, log, log2, sqrt, gcd, factorial, pow, pi
from bisect import bisect_left, bisect_right
from functools import reduce
# SOME GENERAL HELPER
def input_as_array():
return list(map(int, input().split()))
def debug():
if os.path.exists("data.in"):
print("*" * 10)
def debug2(value):
if os.path.exists("data.in"):
print(value)
def debug3():
if os.path.exists("data.in"):
print("-" * 10)
def debug4(msg, value):
if os.path.exists("data.in"):
print(msg, value)
def debug5(msg):
if os.path.exists("data.in"):
print(msg)
start_time = time.time()
def solve(A, n, ops, m, Q, k):
diff = [0] * (n+2)
for i in range(n):
if i == 0:
diff[i] = A[i]
else:
diff[i] = A[i] - A[i - 1]
for a, b in Q:
for i in range(a, b + 1):
L, R, D = ops[i][0], ops[i][1], ops[i][2]
diff[L] += D
diff[R+1] -= D
# Finally updating the initial array
for i in range(n):
if i == 0:
A[i] = diff[i]
else:
A[i] = diff[i] + A[i - 1]
print(*A)
def main():
n, m, k = input_as_array()
A = input_as_array()
i = 0
ops = []
while i < m:
l, r, d = input_as_array()
l, r = l - 1, r - 1
ops.append((l, r, d))
i += 1
Q = []
for _ in range(k):
x, y = input_as_array()
x, y = x - 1, y - 1
Q.append((x, y))
solve(A, n, ops, m, Q, k)
debug3()
if __name__ == "__main__":
if os.path.exists("data.in"):
sys.stdin = open("data.in", "r")
sys.stdout = open("data.out", "w")
testcases = 1
for i in range(testcases):
main()
# If it's local - Print this O/P
if os.path.exists("data.in"):
print(f"Time Elapsed: {time.time() - start_time} seconds")
sys.stdout.close()
| Title: Greg and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*.
Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array.
Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg.
Input Specification:
The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array.
Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105).
Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*).
The numbers in the lines are separated by single spaces.
Output Specification:
On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Demo Input:
['3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n', '1 1 1\n1\n1 1 1\n1 1\n', '4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n']
Demo Output:
['9 18 17\n', '2\n', '5 18 31 20\n']
Note:
none | ```python
"""
செயல் பேசும் ஆழம் இங்கே சொற்கள் பேசுமா?
Focus, Determination and Sheer-Will
The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
"""
import os, sys, math, cmath, time, collections
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import nsmallest, nlargest, heapify, heappop, heappush, heapreplace
from math import ceil, floor, log, log2, sqrt, gcd, factorial, pow, pi
from bisect import bisect_left, bisect_right
from functools import reduce
# SOME GENERAL HELPER
def input_as_array():
return list(map(int, input().split()))
def debug():
if os.path.exists("data.in"):
print("*" * 10)
def debug2(value):
if os.path.exists("data.in"):
print(value)
def debug3():
if os.path.exists("data.in"):
print("-" * 10)
def debug4(msg, value):
if os.path.exists("data.in"):
print(msg, value)
def debug5(msg):
if os.path.exists("data.in"):
print(msg)
start_time = time.time()
def solve(A, n, ops, m, Q, k):
diff = [0] * (n+2)
for i in range(n):
if i == 0:
diff[i] = A[i]
else:
diff[i] = A[i] - A[i - 1]
for a, b in Q:
for i in range(a, b + 1):
L, R, D = ops[i][0], ops[i][1], ops[i][2]
diff[L] += D
diff[R+1] -= D
# Finally updating the initial array
for i in range(n):
if i == 0:
A[i] = diff[i]
else:
A[i] = diff[i] + A[i - 1]
print(*A)
def main():
n, m, k = input_as_array()
A = input_as_array()
i = 0
ops = []
while i < m:
l, r, d = input_as_array()
l, r = l - 1, r - 1
ops.append((l, r, d))
i += 1
Q = []
for _ in range(k):
x, y = input_as_array()
x, y = x - 1, y - 1
Q.append((x, y))
solve(A, n, ops, m, Q, k)
debug3()
if __name__ == "__main__":
if os.path.exists("data.in"):
sys.stdin = open("data.in", "r")
sys.stdout = open("data.out", "w")
testcases = 1
for i in range(testcases):
main()
# If it's local - Print this O/P
if os.path.exists("data.in"):
print(f"Time Elapsed: {time.time() - start_time} seconds")
sys.stdout.close()
``` | 0 | |
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... | 1,664,375,703 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 0 | a, b, n = map( int, input().split() )
def nod(x, y):
while(y):
x, x = y, x % y
return x
last_move = True
while n > 1:
n -= nod(a, n) if last_move else nod(b, n)
last_move = False if last_move else True
print(0) if last_move else print(1)
| 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
a, b, n = map( int, input().split() )
def nod(x, y):
while(y):
x, x = y, x % y
return x
last_move = True
while n > 1:
n -= nod(a, n) if last_move else nod(b, n)
last_move = False if last_move else True
print(0) if last_move else print(1)
``` | 0 | |
45 | A | Codecraft III | PROGRAMMING | 900 | [
"implementation"
] | A. Codecraft III | 2 | 256 | Today Vasya visited a widely known site and learned that the continuation of his favourite game Codecraft II will appear after exactly *k* months. He looked at the calendar and learned that at the moment is the month number *s*. Vasya immediately got interested in what month Codecraft III will appear. Help him understand that.
All the twelve months in Vasya's calendar are named using their usual English names: January, February, March, April, May, June, July, August, September, October, November, December. | The first input line contains the name of the current month. It is guaranteed that it is a proper English name of one of twelve months. The first letter is uppercase, the rest are lowercase. The second line contains integer *k* (0<=≤<=*k*<=≤<=100) — the number of months left till the appearance of Codecraft III. | Print starting from an uppercase letter the name of the month in which the continuation of Codeforces II will appear. The printed name must be contained in the list January, February, March, April, May, June, July, August, September, October, November, December. | [
"November\n3\n",
"May\n24\n"
] | [
"February\n",
"May\n"
] | none | 0 | [
{
"input": "November\n3",
"output": "February"
},
{
"input": "May\n24",
"output": "May"
},
{
"input": "April\n0",
"output": "April"
},
{
"input": "September\n0",
"output": "September"
},
{
"input": "August\n0",
"output": "August"
},
{
"input": "June\n1... | 1,670,696,300 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 92 | 0 | s = 'January, February, March, April, May, June, July, August, September, October, November, December'
s = s.split(', ')
a =input()
n = int(input())
p = s.index(a)
p = (p + n) % 12
print (s[p]) | Title: Codecraft III
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Today Vasya visited a widely known site and learned that the continuation of his favourite game Codecraft II will appear after exactly *k* months. He looked at the calendar and learned that at the moment is the month number *s*. Vasya immediately got interested in what month Codecraft III will appear. Help him understand that.
All the twelve months in Vasya's calendar are named using their usual English names: January, February, March, April, May, June, July, August, September, October, November, December.
Input Specification:
The first input line contains the name of the current month. It is guaranteed that it is a proper English name of one of twelve months. The first letter is uppercase, the rest are lowercase. The second line contains integer *k* (0<=≤<=*k*<=≤<=100) — the number of months left till the appearance of Codecraft III.
Output Specification:
Print starting from an uppercase letter the name of the month in which the continuation of Codeforces II will appear. The printed name must be contained in the list January, February, March, April, May, June, July, August, September, October, November, December.
Demo Input:
['November\n3\n', 'May\n24\n']
Demo Output:
['February\n', 'May\n']
Note:
none | ```python
s = 'January, February, March, April, May, June, July, August, September, October, November, December'
s = s.split(', ')
a =input()
n = int(input())
p = s.index(a)
p = (p + n) % 12
print (s[p])
``` | 3.977 |
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"
... | 1,698,315,256 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 92 | 0 | n=int(input())
s=0
a=list(map(int,input().split()))
i=0
mini= a[0]
maxi=a[0]
while i<n:
if a[i]<mini:
mini=a[i]
s+=1
elif a[i]>maxi:
maxi=a[i]
s+=1
i+=1
print(s)
| 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=int(input())
s=0
a=list(map(int,input().split()))
i=0
mini= a[0]
maxi=a[0]
while i<n:
if a[i]<mini:
mini=a[i]
s+=1
elif a[i]>maxi:
maxi=a[i]
s+=1
i+=1
print(s)
``` | 3 | |
313 | B | Ilya and Queries | PROGRAMMING | 1,100 | [
"dp",
"implementation"
] | null | null | Ilya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.
You've got string *s*<==<=*s*1*s*2... *s**n* (*n* is the length of the string), consisting only of characters "." and "#" and *m* queries. Each query is described by a pair of integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*). The answer to the query *l**i*,<=*r**i* is the number of such integers *i* (*l**i*<=≤<=*i*<=<<=*r**i*), that *s**i*<==<=*s**i*<=+<=1.
Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem. | The first line contains string *s* of length *n* (2<=≤<=*n*<=≤<=105). It is guaranteed that the given string only consists of characters "." and "#".
The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains the description of the corresponding query. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*). | Print *m* integers — the answers to the queries in the order in which they are given in the input. | [
"......\n4\n3 4\n2 3\n1 6\n2 6\n",
"#..###\n5\n1 3\n5 6\n1 5\n3 6\n3 4\n"
] | [
"1\n1\n5\n4\n",
"1\n1\n2\n2\n0\n"
] | none | 1,000 | [
{
"input": "......\n4\n3 4\n2 3\n1 6\n2 6",
"output": "1\n1\n5\n4"
},
{
"input": "#..###\n5\n1 3\n5 6\n1 5\n3 6\n3 4",
"output": "1\n1\n2\n2\n0"
},
{
"input": ".#...#..\n6\n1 5\n2 3\n6 7\n2 4\n2 5\n1 3",
"output": "2\n0\n0\n1\n2\n0"
},
{
"input": "#.#.#..\n5\n3 4\n4 5\n5 7\n5... | 1,699,761,894 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 810 | 16,588,800 | s=input()
same={i:0 for i in range(len(s))}
answers=[]
for i in range(len(s)-1):
if s[i] == s[i+1]:
same[i+1]=same[i]+1
else:
same[i+1]=same[i]
for _ in range(int(input())):
r,l=map(int,input().split())
answers.append(same[l-1]-same[r-1])
for a in answers:
print(a) | Title: Ilya and Queries
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.
You've got string *s*<==<=*s*1*s*2... *s**n* (*n* is the length of the string), consisting only of characters "." and "#" and *m* queries. Each query is described by a pair of integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*). The answer to the query *l**i*,<=*r**i* is the number of such integers *i* (*l**i*<=≤<=*i*<=<<=*r**i*), that *s**i*<==<=*s**i*<=+<=1.
Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem.
Input Specification:
The first line contains string *s* of length *n* (2<=≤<=*n*<=≤<=105). It is guaranteed that the given string only consists of characters "." and "#".
The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains the description of the corresponding query. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*).
Output Specification:
Print *m* integers — the answers to the queries in the order in which they are given in the input.
Demo Input:
['......\n4\n3 4\n2 3\n1 6\n2 6\n', '#..###\n5\n1 3\n5 6\n1 5\n3 6\n3 4\n']
Demo Output:
['1\n1\n5\n4\n', '1\n1\n2\n2\n0\n']
Note:
none | ```python
s=input()
same={i:0 for i in range(len(s))}
answers=[]
for i in range(len(s)-1):
if s[i] == s[i+1]:
same[i+1]=same[i]+1
else:
same[i+1]=same[i]
for _ in range(int(input())):
r,l=map(int,input().split())
answers.append(same[l-1]-same[r-1])
for a in answers:
print(a)
``` | 3 | |
389 | A | Fox and Number Game | PROGRAMMING | 1,000 | [
"greedy",
"math"
] | null | null | Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum. | The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100). | Output a single integer — the required minimal sum. | [
"2\n1 2\n",
"3\n2 4 6\n",
"2\n12 18\n",
"5\n45 12 27 30 18\n"
] | [
"2\n",
"6\n",
"12\n",
"15\n"
] | In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>. | 500 | [
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 4 6",
"output": "6"
},
{
"input": "2\n12 18",
"output": "12"
},
{
"input": "5\n45 12 27 30 18",
"output": "15"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "2\n100 100",
"output": "200"
... | 1,680,033,864 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 34 | 62 | 0 | from math import gcd
n = int(input())
lst = list(map(int, input().split()))
lst.sort()
r = lst[0]
for x in lst[1:]:
r = gcd(r, x)
print(n * r) | Title: Fox and Number Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
Output Specification:
Output a single integer — the required minimal sum.
Demo Input:
['2\n1 2\n', '3\n2 4 6\n', '2\n12 18\n', '5\n45 12 27 30 18\n']
Demo Output:
['2\n', '6\n', '12\n', '15\n']
Note:
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>. | ```python
from math import gcd
n = int(input())
lst = list(map(int, input().split()))
lst.sort()
r = lst[0]
for x in lst[1:]:
r = gcd(r, x)
print(n * r)
``` | 3 | |
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"
},
{
... | 1,487,820,189 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 1,903 | 7,372,800 | s=list(input())
dah='abcdefghijklmnopqrstuvwxyz'
for i in range(len(s)-1):
if s[i]==s[i+1]:
if i+2<len(s):
for ch in dah:
if ch!=s[i+2] and ch!=s[i]:
s[i+1]=ch
else:
for ch in dah:
if ch!=s[i]:
s[i+1]=ch
for i in range(len(s)):
print(s[i],end='')
| 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
s=list(input())
dah='abcdefghijklmnopqrstuvwxyz'
for i in range(len(s)-1):
if s[i]==s[i+1]:
if i+2<len(s):
for ch in dah:
if ch!=s[i+2] and ch!=s[i]:
s[i+1]=ch
else:
for ch in dah:
if ch!=s[i]:
s[i+1]=ch
for i in range(len(s)):
print(s[i],end='')
``` | 3 | |
1,008 | A | Romaji | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese. | The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters. | Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower). | [
"sumimasen\n",
"ninja\n",
"codeforces\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | 500 | [
{
"input": "sumimasen",
"output": "YES"
},
{
"input": "ninja",
"output": "YES"
},
{
"input": "codeforces",
"output": "NO"
},
{
"input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen",
"output": "YES"
},
{
"in... | 1,636,957,367 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 4 | 31 | 0 | s = input()
s = s.lower()
ans = 0
if len(s) >= 1 and len(s) <=100:
if len(s)==1 and s not in 'aeioun':
ans = 1
else:
for i in range(len(s)-1):
if s[i] not in 'aeioun':
if s[i+1] not in 'aeiou':
ans = 1
if i == len(s)-2:
if s[-1] not in 'aeion':
ans = 1
if ans != 1:
print('YES')
elif ans == 1:
print('NO') | Title: Romaji
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
Input Specification:
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Output Specification:
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['sumimasen\n', 'ninja\n', 'codeforces\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | ```python
s = input()
s = s.lower()
ans = 0
if len(s) >= 1 and len(s) <=100:
if len(s)==1 and s not in 'aeioun':
ans = 1
else:
for i in range(len(s)-1):
if s[i] not in 'aeioun':
if s[i+1] not in 'aeiou':
ans = 1
if i == len(s)-2:
if s[-1] not in 'aeion':
ans = 1
if ans != 1:
print('YES')
elif ans == 1:
print('NO')
``` | -1 | |
964 | B | Messages | PROGRAMMING | 1,300 | [
"math"
] | null | null | There are *n* incoming messages for Vasya. The *i*-th message is going to be received after *t**i* minutes. Each message has a cost, which equals to *A* initially. After being received, the cost of a message decreases by *B* each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives *C*·*k*, where *k* is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after *T* minutes.
Determine the maximum amount of money Vasya's bank account can hold after *T* minutes. | The first line contains five integers *n*, *A*, *B*, *C* and *T* (1<=≤<=*n*,<=*A*,<=*B*,<=*C*,<=*T*<=≤<=1000).
The second string contains *n* integers *t**i* (1<=≤<=*t**i*<=≤<=*T*). | Output one integer — the answer to the problem. | [
"4 5 5 3 5\n1 5 5 4\n",
"5 3 1 1 3\n2 2 2 1 1\n",
"5 5 3 4 5\n1 2 3 4 5\n"
] | [
"20\n",
"15\n",
"35\n"
] | In the first sample the messages must be read immediately after receiving, Vasya receives *A* points for each message, *n*·*A* = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | 1,000 | [
{
"input": "4 5 5 3 5\n1 5 5 4",
"output": "20"
},
{
"input": "5 3 1 1 3\n2 2 2 1 1",
"output": "15"
},
{
"input": "5 5 3 4 5\n1 2 3 4 5",
"output": "35"
},
{
"input": "1 6 4 3 9\n2",
"output": "6"
},
{
"input": "10 9 7 5 3\n3 3 3 3 2 3 2 2 3 3",
"output": "90... | 1,556,887,397 | 2,147,483,647 | Python 3 | OK | TESTS | 60 | 109 | 0 | n, A, B, C, T = map(int, input().split())
t = sum(map(int, input().split()))
print(max(n * A - (T * n - t) * B + (n * T - t) * C, n * A))
| Title: Messages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* incoming messages for Vasya. The *i*-th message is going to be received after *t**i* minutes. Each message has a cost, which equals to *A* initially. After being received, the cost of a message decreases by *B* each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives *C*·*k*, where *k* is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after *T* minutes.
Determine the maximum amount of money Vasya's bank account can hold after *T* minutes.
Input Specification:
The first line contains five integers *n*, *A*, *B*, *C* and *T* (1<=≤<=*n*,<=*A*,<=*B*,<=*C*,<=*T*<=≤<=1000).
The second string contains *n* integers *t**i* (1<=≤<=*t**i*<=≤<=*T*).
Output Specification:
Output one integer — the answer to the problem.
Demo Input:
['4 5 5 3 5\n1 5 5 4\n', '5 3 1 1 3\n2 2 2 1 1\n', '5 5 3 4 5\n1 2 3 4 5\n']
Demo Output:
['20\n', '15\n', '35\n']
Note:
In the first sample the messages must be read immediately after receiving, Vasya receives *A* points for each message, *n*·*A* = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | ```python
n, A, B, C, T = map(int, input().split())
t = sum(map(int, input().split()))
print(max(n * A - (T * n - t) * B + (n * T - t) * C, n * A))
``` | 3 | |
859 | B | Lazy Security Guard | PROGRAMMING | 1,000 | [
"brute force",
"geometry",
"math"
] | null | null | Your security guard friend recently got a new job at a new security company. The company requires him to patrol an area of the city encompassing exactly *N* city blocks, but they let him choose which blocks. That is, your friend must walk the perimeter of a region whose area is exactly *N* blocks. Your friend is quite lazy and would like your help to find the shortest possible route that meets the requirements. The city is laid out in a square grid pattern, and is large enough that for the sake of the problem it can be considered infinite. | Input will consist of a single integer *N* (1<=≤<=*N*<=≤<=106), the number of city blocks that must be enclosed by the route. | Print the minimum perimeter that can be achieved. | [
"4\n",
"11\n",
"22\n"
] | [
"8\n",
"14\n",
"20\n"
] | Here are some possible shapes for the examples:
<img class="tex-graphics" src="https://espresso.codeforces.com/e11bef2cf82b55dd583cfc97d12b5aee5e483a65.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 750 | [
{
"input": "4",
"output": "8"
},
{
"input": "11",
"output": "14"
},
{
"input": "22",
"output": "20"
},
{
"input": "3",
"output": "8"
},
{
"input": "1024",
"output": "128"
},
{
"input": "101",
"output": "42"
},
{
"input": "30",
"output":... | 1,586,158,037 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 109 | 0 | #----Kuzlyaev-Nikita-Codeforces-----
#------------06.04.2020-------------
alph="abcdefghijklmnopqrstuvwxyz"
#-----------------------------------
n=int(input())
r=int(n**0.5)
h=int(n/r)+int(n%r!=0)
print(2*(h+r)) | Title: Lazy Security Guard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your security guard friend recently got a new job at a new security company. The company requires him to patrol an area of the city encompassing exactly *N* city blocks, but they let him choose which blocks. That is, your friend must walk the perimeter of a region whose area is exactly *N* blocks. Your friend is quite lazy and would like your help to find the shortest possible route that meets the requirements. The city is laid out in a square grid pattern, and is large enough that for the sake of the problem it can be considered infinite.
Input Specification:
Input will consist of a single integer *N* (1<=≤<=*N*<=≤<=106), the number of city blocks that must be enclosed by the route.
Output Specification:
Print the minimum perimeter that can be achieved.
Demo Input:
['4\n', '11\n', '22\n']
Demo Output:
['8\n', '14\n', '20\n']
Note:
Here are some possible shapes for the examples:
<img class="tex-graphics" src="https://espresso.codeforces.com/e11bef2cf82b55dd583cfc97d12b5aee5e483a65.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
#----Kuzlyaev-Nikita-Codeforces-----
#------------06.04.2020-------------
alph="abcdefghijklmnopqrstuvwxyz"
#-----------------------------------
n=int(input())
r=int(n**0.5)
h=int(n/r)+int(n%r!=0)
print(2*(h+r))
``` | 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... | 1,602,211,003 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 248 | 0 | # -*- coding: utf-8 -*-
"""
Created on Fri Oct 9 09:21:45 2020
@author: 86198
"""
def add(x,y):
return x+y
n=int(input())
a=0
m=[0,0,0]
for i in (0,n):
l=[int(i)for i in input().split()]
m=map(add,m,l)
for i in list(m):
if i !=0:
a=a+1
else:
a=0
if a!=0:
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
# -*- coding: utf-8 -*-
"""
Created on Fri Oct 9 09:21:45 2020
@author: 86198
"""
def add(x,y):
return x+y
n=int(input())
a=0
m=[0,0,0]
for i in (0,n):
l=[int(i)for i in input().split()]
m=map(add,m,l)
for i in list(m):
if i !=0:
a=a+1
else:
a=0
if a!=0:
print('NO')
else:
print('YES')
``` | 0 |
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... | 1,650,568,872 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | s = str(input())
upper = sum(map(str.isupper, s))
low = sum(map(str.islower, s))
if upper > low:
t = s.upper()
else:
t = s.lower()
print(t) | 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
s = str(input())
upper = sum(map(str.isupper, s))
low = sum(map(str.islower, s))
if upper > low:
t = s.upper()
else:
t = s.lower()
print(t)
``` | 3.977 |
18 | B | Platforms | PROGRAMMING | 1,700 | [
"brute force",
"math"
] | B. Platforms | 2 | 64 | In one one-dimensional world there are *n* platforms. Platform with index *k* (platforms are numbered from 1) is a segment with coordinates [(*k*<=-<=1)*m*,<=(*k*<=-<=1)*m*<=+<=*l*], and *l*<=<<=*m*. Grasshopper Bob starts to jump along the platforms from point 0, with each jump he moves exactly *d* units right. Find out the coordinate of the point, where Bob will fall down. The grasshopper falls down, if he finds himself not on the platform, but if he finds himself on the edge of the platform, he doesn't fall down. | The first input line contains 4 integer numbers *n*, *d*, *m*, *l* (1<=≤<=*n*,<=*d*,<=*m*,<=*l*<=≤<=106,<=*l*<=<<=*m*) — respectively: amount of platforms, length of the grasshopper Bob's jump, and numbers *m* and *l* needed to find coordinates of the *k*-th platform: [(*k*<=-<=1)*m*,<=(*k*<=-<=1)*m*<=+<=*l*]. | Output the coordinates of the point, where the grosshopper will fall down. Don't forget that if Bob finds himself on the platform edge, he doesn't fall down. | [
"2 2 5 3\n",
"5 4 11 8\n"
] | [
"4\n",
"20\n"
] | none | 0 | [
{
"input": "2 2 5 3",
"output": "4"
},
{
"input": "5 4 11 8",
"output": "20"
},
{
"input": "228385 744978 699604 157872",
"output": "2979912"
},
{
"input": "773663 427904 329049 243542",
"output": "1283712"
},
{
"input": "835293 627183 442142 361649",
"output"... | 1,619,245,598 | 4,298 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 342 | 6,041,600 | import sys
# from itertools import product
import logging
logging.root.setLevel(level=logging.INFO)
n,d,m,l = map(int,sys.stdin.readline().strip().split())
loc = 0
right = n*m+l
while loc % m <= l and loc <= right:
loc += d
print(loc) | Title: Platforms
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
In one one-dimensional world there are *n* platforms. Platform with index *k* (platforms are numbered from 1) is a segment with coordinates [(*k*<=-<=1)*m*,<=(*k*<=-<=1)*m*<=+<=*l*], and *l*<=<<=*m*. Grasshopper Bob starts to jump along the platforms from point 0, with each jump he moves exactly *d* units right. Find out the coordinate of the point, where Bob will fall down. The grasshopper falls down, if he finds himself not on the platform, but if he finds himself on the edge of the platform, he doesn't fall down.
Input Specification:
The first input line contains 4 integer numbers *n*, *d*, *m*, *l* (1<=≤<=*n*,<=*d*,<=*m*,<=*l*<=≤<=106,<=*l*<=<<=*m*) — respectively: amount of platforms, length of the grasshopper Bob's jump, and numbers *m* and *l* needed to find coordinates of the *k*-th platform: [(*k*<=-<=1)*m*,<=(*k*<=-<=1)*m*<=+<=*l*].
Output Specification:
Output the coordinates of the point, where the grosshopper will fall down. Don't forget that if Bob finds himself on the platform edge, he doesn't fall down.
Demo Input:
['2 2 5 3\n', '5 4 11 8\n']
Demo Output:
['4\n', '20\n']
Note:
none | ```python
import sys
# from itertools import product
import logging
logging.root.setLevel(level=logging.INFO)
n,d,m,l = map(int,sys.stdin.readline().strip().split())
loc = 0
right = n*m+l
while loc % m <= l and loc <= right:
loc += d
print(loc)
``` | 0 |
17 | A | Noldbach problem | PROGRAMMING | 1,000 | [
"brute force",
"math",
"number theory"
] | A. Noldbach problem | 2 | 64 | Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong. | The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000). | Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO. | [
"27 2\n",
"45 7\n"
] | [
"YES",
"NO"
] | In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form. | 0 | [
{
"input": "27 2",
"output": "YES"
},
{
"input": "45 7",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
},
{
"input": "15 1",
"output": "YES"
},
{
"input": "17 1",
"output": "YES"
},
{
"input": "34 5",
"output": "NO"
},
{
"input": "37 ... | 1,646,433,007 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 31 | 154 | 2,048,000 | from math import sqrt
arr = [int(item) for item in input().split()]
parr = []
def isPrime(n):
if (n <= 0):
return
else:
for i in range(2, int(sqrt(n)) +1):
if (n % i == 0):
return False
else:
return True
for i in range(2, arr[0]+1): #making a list with all prime nums
if(isPrime(i)):
parr.append(i)
#creating permutations of all possibilites of combinations
def combinations():
sum = 0
if(arr[0] == 2):
return sum
for i in range(1, len(parr)):
summ = parr[i] + parr[i-1] + 1
if(isPrime(summ) and summ in parr):
sum += 1
return sum
var = combinations()
if(var == arr[1]):
print("YES")
else:
print("NO") | Title: Noldbach problem
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong.
Input Specification:
The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).
Output Specification:
Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.
Demo Input:
['27 2\n', '45 7\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form. | ```python
from math import sqrt
arr = [int(item) for item in input().split()]
parr = []
def isPrime(n):
if (n <= 0):
return
else:
for i in range(2, int(sqrt(n)) +1):
if (n % i == 0):
return False
else:
return True
for i in range(2, arr[0]+1): #making a list with all prime nums
if(isPrime(i)):
parr.append(i)
#creating permutations of all possibilites of combinations
def combinations():
sum = 0
if(arr[0] == 2):
return sum
for i in range(1, len(parr)):
summ = parr[i] + parr[i-1] + 1
if(isPrime(summ) and summ in parr):
sum += 1
return sum
var = combinations()
if(var == arr[1]):
print("YES")
else:
print("NO")
``` | 0 |
0 | none | none | none | 0 | [
"none"
] | null | null | Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot? | First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players. | Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise. | [
"4\n75 150 75 50\n",
"3\n100 150 250\n"
] | [
"Yes\n",
"No\n"
] | In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | 0 | [
{
"input": "4\n75 150 75 50",
"output": "Yes"
},
{
"input": "3\n100 150 250",
"output": "No"
},
{
"input": "7\n34 34 68 34 34 68 34",
"output": "Yes"
},
{
"input": "10\n72 96 12 18 81 20 6 2 54 1",
"output": "No"
},
{
"input": "20\n958692492 954966768 77387000 724... | 1,490,126,358 | 2,147,483,647 | Python 3 | OK | TESTS | 70 | 920 | 13,414,400 | n=int(input())
A=[]
A = [int(x) for x in input().split()]
for i in range(n):
while A[i]%2==0:
A[i]=A[i]//2
while A[i]%3==0:
A[i]=A[i]//3
#print (A)
for i in A:
if i!=A[0]:
print("No")
exit()
print("Yes")
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input Specification:
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
Output Specification:
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Demo Input:
['4\n75 150 75 50\n', '3\n100 150 250\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | ```python
n=int(input())
A=[]
A = [int(x) for x in input().split()]
for i in range(n):
while A[i]%2==0:
A[i]=A[i]//2
while A[i]%3==0:
A[i]=A[i]//3
#print (A)
for i in A:
if i!=A[0]:
print("No")
exit()
print("Yes")
``` | 3 | |
540 | C | Ice Cave | PROGRAMMING | 2,000 | [
"dfs and similar"
] | null | null | You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice.
The level of the cave where you are is a rectangular square grid of *n* rows and *m* columns. Each cell consists either from intact or from cracked ice. From each cell you can move to cells that are side-adjacent with yours (due to some limitations of the game engine you cannot make jumps on the same place, i.e. jump from a cell to itself). If you move to the cell with cracked ice, then your character falls down through it and if you move to the cell with intact ice, then the ice on this cell becomes cracked.
Let's number the rows with integers from 1 to *n* from top to bottom and the columns with integers from 1 to *m* from left to right. Let's denote a cell on the intersection of the *r*-th row and the *c*-th column as (*r*,<=*c*).
You are staying in the cell (*r*1,<=*c*1) and this cell is cracked because you've just fallen here from a higher level. You need to fall down through the cell (*r*2,<=*c*2) since the exit to the next level is there. Can you do this? | The first line contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=500) — the number of rows and columns in the cave description.
Each of the next *n* lines describes the initial state of the level of the cave, each line consists of *m* characters "." (that is, intact ice) and "X" (cracked ice).
The next line contains two integers, *r*1 and *c*1 (1<=≤<=*r*1<=≤<=*n*,<=1<=≤<=*c*1<=≤<=*m*) — your initial coordinates. It is guaranteed that the description of the cave contains character 'X' in cell (*r*1,<=*c*1), that is, the ice on the starting cell is initially cracked.
The next line contains two integers *r*2 and *c*2 (1<=≤<=*r*2<=≤<=*n*,<=1<=≤<=*c*2<=≤<=*m*) — the coordinates of the cell through which you need to fall. The final cell may coincide with the starting one. | If you can reach the destination, print 'YES', otherwise print 'NO'. | [
"4 6\nX...XX\n...XX.\n.X..X.\n......\n1 6\n2 2\n",
"5 4\n.X..\n...X\nX.X.\n....\n.XX.\n5 3\n1 1\n",
"4 7\n..X.XX.\n.XX..X.\nX...X..\nX......\n2 2\n1 6\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first sample test one possible path is:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c61f56de718beea14935ccdc85ae2c4ad45c1454.png" style="max-width: 100.0%;max-height: 100.0%;"/>
After the first visit of cell (2, 2) the ice on it cracks and when you step there for the second time, your character falls through the ice as intended. | 1,500 | [
{
"input": "4 6\nX...XX\n...XX.\n.X..X.\n......\n1 6\n2 2",
"output": "YES"
},
{
"input": "5 4\n.X..\n...X\nX.X.\n....\n.XX.\n5 3\n1 1",
"output": "NO"
},
{
"input": "4 7\n..X.XX.\n.XX..X.\nX...X..\nX......\n2 2\n1 6",
"output": "YES"
},
{
"input": "5 3\n.XX\n...\n.X.\n.X.\n.... | 1,597,966,977 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 124 | 0 | def is_valid_tile(r, c, n, m):
return r >= 0 and c >= 0 and r < n and c < m
def gen_adjacents(r, c, n, m):
adjs = []
possibles = [(r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)]
for possible in possibles:
if is_valid_tile(*possible, n, m):
adjs.append(possible)
return adjs
def solve(table, r1, c1, r2, c2, n, m):
if r1 == r2 and c1 == c2:
adjacents = gen_adjacents(r1, c1, n, m)
for adjacent in adjacents:
if table[adjacent[0]][adjacent[1]] == "X":
return False
return True
if table[r2][c2] == ".":
adjacents = gen_adjacents(r2, c2, n, m)
solid = 0
for adj in adjacents:
if table[adj[0]][adj[1]] == ".":
solid += 1
if solid < 2:
return False
forward_visited = set([(r1, c1)]); backward_visited = set([(r2, c2)])
forward_nodes = [(r1, c1)]; backward_nodes = [(r2, c2)]
while forward_nodes and backward_nodes:
for i in range(len(forward_nodes)):
node = forward_nodes.pop(0)
adjacents = gen_adjacents(*node, n, m)
for adjacent in adjacents:
if adjacent in backward_visited:
return True
if adjacent not in forward_visited and table[adjacent[0]][adjacent[1]] == ".":
forward_visited.add(adjacent)
forward_nodes.append(adjacent)
for j in range(len(backward_nodes)):
node = backward_nodes.pop(0)
adjacents = gen_adjacents(*node, n, m)
for adjacent in adjacents:
if adjacent in forward_visited:
return True
if adjacent not in backward_visited and table[adjacent[0]][adjacent[1]] == ".":
backward_visited.add(adjacent)
backward_nodes.append(adjacent)
return False
def main():
n, m = map(int, input().split())
table = []
for _ in range(n):
row = input()
table.append(row)
r1, c1 = map(int, input().split())
r2, c2 = map(int, input().split())
r1 -= 1; c1 -= 1; r2 -= 1; c2 -= 1
if solve(table, r1, c1, r2, c2, n, m):
print("YES")
else:
print("NO")
main() | Title: Ice Cave
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice.
The level of the cave where you are is a rectangular square grid of *n* rows and *m* columns. Each cell consists either from intact or from cracked ice. From each cell you can move to cells that are side-adjacent with yours (due to some limitations of the game engine you cannot make jumps on the same place, i.e. jump from a cell to itself). If you move to the cell with cracked ice, then your character falls down through it and if you move to the cell with intact ice, then the ice on this cell becomes cracked.
Let's number the rows with integers from 1 to *n* from top to bottom and the columns with integers from 1 to *m* from left to right. Let's denote a cell on the intersection of the *r*-th row and the *c*-th column as (*r*,<=*c*).
You are staying in the cell (*r*1,<=*c*1) and this cell is cracked because you've just fallen here from a higher level. You need to fall down through the cell (*r*2,<=*c*2) since the exit to the next level is there. Can you do this?
Input Specification:
The first line contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=500) — the number of rows and columns in the cave description.
Each of the next *n* lines describes the initial state of the level of the cave, each line consists of *m* characters "." (that is, intact ice) and "X" (cracked ice).
The next line contains two integers, *r*1 and *c*1 (1<=≤<=*r*1<=≤<=*n*,<=1<=≤<=*c*1<=≤<=*m*) — your initial coordinates. It is guaranteed that the description of the cave contains character 'X' in cell (*r*1,<=*c*1), that is, the ice on the starting cell is initially cracked.
The next line contains two integers *r*2 and *c*2 (1<=≤<=*r*2<=≤<=*n*,<=1<=≤<=*c*2<=≤<=*m*) — the coordinates of the cell through which you need to fall. The final cell may coincide with the starting one.
Output Specification:
If you can reach the destination, print 'YES', otherwise print 'NO'.
Demo Input:
['4 6\nX...XX\n...XX.\n.X..X.\n......\n1 6\n2 2\n', '5 4\n.X..\n...X\nX.X.\n....\n.XX.\n5 3\n1 1\n', '4 7\n..X.XX.\n.XX..X.\nX...X..\nX......\n2 2\n1 6\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample test one possible path is:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c61f56de718beea14935ccdc85ae2c4ad45c1454.png" style="max-width: 100.0%;max-height: 100.0%;"/>
After the first visit of cell (2, 2) the ice on it cracks and when you step there for the second time, your character falls through the ice as intended. | ```python
def is_valid_tile(r, c, n, m):
return r >= 0 and c >= 0 and r < n and c < m
def gen_adjacents(r, c, n, m):
adjs = []
possibles = [(r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)]
for possible in possibles:
if is_valid_tile(*possible, n, m):
adjs.append(possible)
return adjs
def solve(table, r1, c1, r2, c2, n, m):
if r1 == r2 and c1 == c2:
adjacents = gen_adjacents(r1, c1, n, m)
for adjacent in adjacents:
if table[adjacent[0]][adjacent[1]] == "X":
return False
return True
if table[r2][c2] == ".":
adjacents = gen_adjacents(r2, c2, n, m)
solid = 0
for adj in adjacents:
if table[adj[0]][adj[1]] == ".":
solid += 1
if solid < 2:
return False
forward_visited = set([(r1, c1)]); backward_visited = set([(r2, c2)])
forward_nodes = [(r1, c1)]; backward_nodes = [(r2, c2)]
while forward_nodes and backward_nodes:
for i in range(len(forward_nodes)):
node = forward_nodes.pop(0)
adjacents = gen_adjacents(*node, n, m)
for adjacent in adjacents:
if adjacent in backward_visited:
return True
if adjacent not in forward_visited and table[adjacent[0]][adjacent[1]] == ".":
forward_visited.add(adjacent)
forward_nodes.append(adjacent)
for j in range(len(backward_nodes)):
node = backward_nodes.pop(0)
adjacents = gen_adjacents(*node, n, m)
for adjacent in adjacents:
if adjacent in forward_visited:
return True
if adjacent not in backward_visited and table[adjacent[0]][adjacent[1]] == ".":
backward_visited.add(adjacent)
backward_nodes.append(adjacent)
return False
def main():
n, m = map(int, input().split())
table = []
for _ in range(n):
row = input()
table.append(row)
r1, c1 = map(int, input().split())
r2, c2 = map(int, input().split())
r1 -= 1; c1 -= 1; r2 -= 1; c2 -= 1
if solve(table, r1, c1, r2, c2, n, m):
print("YES")
else:
print("NO")
main()
``` | 0 | |
743 | B | Chloe and the sequence | PROGRAMMING | 1,200 | [
"binary search",
"bitmasks",
"constructive algorithms",
"implementation"
] | null | null | Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem! | The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1). | Print single integer — the integer at the *k*-th position in the obtained sequence. | [
"3 2\n",
"4 8\n"
] | [
"2",
"4"
] | In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4. | 1,000 | [
{
"input": "3 2",
"output": "2"
},
{
"input": "4 8",
"output": "4"
},
{
"input": "5 27",
"output": "1"
},
{
"input": "7 44",
"output": "3"
},
{
"input": "15 18432",
"output": "12"
},
{
"input": "20 259676",
"output": "3"
},
{
"input": "30 6... | 1,632,881,408 | 2,147,483,647 | PyPy 3 | OK | TESTS | 39 | 108 | 20,172,800 | n, k = map(int, input().split())
cnt = 1
for i in range(n-1):
cnt = cnt*2 + 1
l = 1
r = cnt
p = n
while l <= r:
mid = (l + r) // 2
if mid == k:
print(n)
break
elif mid < k:
l = mid + 1
n -= 1
else:
r = mid - 1
n -=1
| Title: Chloe and the sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem!
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).
Output Specification:
Print single integer — the integer at the *k*-th position in the obtained sequence.
Demo Input:
['3 2\n', '4 8\n']
Demo Output:
['2', '4']
Note:
In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4. | ```python
n, k = map(int, input().split())
cnt = 1
for i in range(n-1):
cnt = cnt*2 + 1
l = 1
r = cnt
p = n
while l <= r:
mid = (l + r) // 2
if mid == k:
print(n)
break
elif mid < k:
l = mid + 1
n -= 1
else:
r = mid - 1
n -=1
``` | 3 | |
622 | A | Infinite Sequence | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence. | The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. | Print the element in the *n*-th position of the sequence (the elements are numerated from one). | [
"3\n",
"5\n",
"10\n",
"55\n",
"56\n"
] | [
"2\n",
"2\n",
"4\n",
"10\n",
"1\n"
] | none | 0 | [
{
"input": "3",
"output": "2"
},
{
"input": "5",
"output": "2"
},
{
"input": "10",
"output": "4"
},
{
"input": "55",
"output": "10"
},
{
"input": "56",
"output": "1"
},
{
"input": "1000000000000",
"output": "88209"
},
{
"input": "8471941278... | 1,683,031,698 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 10 | 46 | 0 | # LUOGU_RID: 109426988
l=1;r=10**7
end=int(input())
while l<r:
mid=(l+r)//2
if (mid**2+mid)//2 <= end:
l=mid+1
else:
r=mid
maxi=(l**2+l)//2
mini=((l-1)**2+l-1)//2
deta=end-mini
if deta==0:
print(l-1)
else:
print(deta)
| Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence.
Input Specification:
The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output Specification:
Print the element in the *n*-th position of the sequence (the elements are numerated from one).
Demo Input:
['3\n', '5\n', '10\n', '55\n', '56\n']
Demo Output:
['2\n', '2\n', '4\n', '10\n', '1\n']
Note:
none | ```python
# LUOGU_RID: 109426988
l=1;r=10**7
end=int(input())
while l<r:
mid=(l+r)//2
if (mid**2+mid)//2 <= end:
l=mid+1
else:
r=mid
maxi=(l**2+l)//2
mini=((l-1)**2+l-1)//2
deta=end-mini
if deta==0:
print(l-1)
else:
print(deta)
``` | 0 | |
784 | F | Crunching Numbers Just for You | PROGRAMMING | 1,900 | [
"*special",
"implementation"
] | null | null | You are developing a new feature for the website which sells airline tickets: being able to sort tickets by price! You have already extracted the tickets' prices, so there's just the last step to be done...
You are given an array of integers. Sort it in non-descending order. | The input consists of a single line of space-separated integers. The first number is *n* (1<=≤<=*n*<=≤<=10) — the size of the array. The following *n* numbers are the elements of the array (1<=≤<=*a**i*<=≤<=100). | Output space-separated elements of the sorted array. | [
"3 3 1 2\n"
] | [
"1 2 3 \n"
] | Remember, this is a very important feature, and you have to make sure the customers appreciate it! | 0 | [
{
"input": "3 3 1 2",
"output": "1 2 3 "
},
{
"input": "10 54 100 27 1 33 27 80 49 27 6",
"output": "1 6 27 27 27 33 49 54 80 100 "
}
] | 1,691,719,893 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 31 | 0 |
print(*sorted(list(set(list(map(int,input().split())))))) | Title: Crunching Numbers Just for You
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are developing a new feature for the website which sells airline tickets: being able to sort tickets by price! You have already extracted the tickets' prices, so there's just the last step to be done...
You are given an array of integers. Sort it in non-descending order.
Input Specification:
The input consists of a single line of space-separated integers. The first number is *n* (1<=≤<=*n*<=≤<=10) — the size of the array. The following *n* numbers are the elements of the array (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output space-separated elements of the sorted array.
Demo Input:
['3 3 1 2\n']
Demo Output:
['1 2 3 \n']
Note:
Remember, this is a very important feature, and you have to make sure the customers appreciate it! | ```python
print(*sorted(list(set(list(map(int,input().split()))))))
``` | 0 | |
133 | A | HQ9+ | PROGRAMMING | 900 | [
"implementation"
] | null | null | HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output. | The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive. | Output "YES", if executing the program will produce any output, and "NO" otherwise. | [
"Hi!\n",
"Codeforces\n"
] | [
"YES\n",
"NO\n"
] | In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions. | 500 | [
{
"input": "Hi!",
"output": "YES"
},
{
"input": "Codeforces",
"output": "NO"
},
{
"input": "a+b=c",
"output": "NO"
},
{
"input": "hq-lowercase",
"output": "NO"
},
{
"input": "Q",
"output": "YES"
},
{
"input": "9",
"output": "YES"
},
{
"inpu... | 1,684,323,727 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 92 | 0 | l = ["H", "Q", "9", "+"]
s = input()
check = False
for i in range(len(s)):
if(s[i] in l):
check = True
break
if(check == False):
print("NO")
else:
print("YES") | Title: HQ9+
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input Specification:
The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output Specification:
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Demo Input:
['Hi!\n', 'Codeforces\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions. | ```python
l = ["H", "Q", "9", "+"]
s = input()
check = False
for i in range(len(s)):
if(s[i] in l):
check = True
break
if(check == False):
print("NO")
else:
print("YES")
``` | 0 | |
698 | A | Vacations | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. | Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days. | [
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | 500 | [
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 ... | 1,686,921,382 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 5 | 46 | 0 | def f(v, prev):
c = 0
for tmp in range(len(v)):
if v[tmp] == 0:
c += 1
prev = 0
elif v[tmp] == 1 and prev != 1:
prev = 1
elif v[tmp] == 2 and prev != 2:
prev = 2
elif v[tmp] == 3 and prev != 0:
prev = 2 if prev == 1 else 1
elif v[tmp] == 3 and prev == 0:
c += min(f(v[tmp + 1:], 1), f(v[tmp + 1:], 2))
return c
else:
c += 1
return c
n = int(input())
v = list(map(int, input().split()))
print(f(v, 0))
| Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | ```python
def f(v, prev):
c = 0
for tmp in range(len(v)):
if v[tmp] == 0:
c += 1
prev = 0
elif v[tmp] == 1 and prev != 1:
prev = 1
elif v[tmp] == 2 and prev != 2:
prev = 2
elif v[tmp] == 3 and prev != 0:
prev = 2 if prev == 1 else 1
elif v[tmp] == 3 and prev == 0:
c += min(f(v[tmp + 1:], 1), f(v[tmp + 1:], 2))
return c
else:
c += 1
return c
n = int(input())
v = list(map(int, input().split()))
print(f(v, 0))
``` | 0 | |
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... | 1,639,861,157 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 78 | 0 | import sys
def main() :
n,m = map(int,input().split())
array = list(map(int,input().strip().split()))
max_ = 0
ind = -1
for i in range(len(array)):
a = array[i]//m
if a >= max_ :
max_ = a
ind = i+1
print(ind)
main() | 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
import sys
def main() :
n,m = map(int,input().split())
array = list(map(int,input().strip().split()))
max_ = 0
ind = -1
for i in range(len(array)):
a = array[i]//m
if a >= max_ :
max_ = a
ind = i+1
print(ind)
main()
``` | 0 | |
234 | A | Lefthanders and Righthanders | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | One fine October day a mathematics teacher Vasily Petrov went to a class and saw there *n* pupils who sat at the desks, two people at each desk. Vasily quickly realized that number *n* is even. Like all true mathematicians, Vasily has all students numbered from 1 to *n*.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists. | The first input line contains a single even integer *n* (4<=≤<=*n*<=≤<=100) — the number of students in the class. The second line contains exactly *n* capital English letters "L" and "R". If the *i*-th letter at the second line equals "L", then the student number *i* is a lefthander, otherwise he is a righthander. | Print integer pairs, one pair per line. In the *i*-th line print the numbers of students that will sit at the *i*-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them. | [
"6\nLLRLLL\n",
"4\nRRLL\n"
] | [
"1 4\n2 5\n6 3\n",
"3 1\n4 2\n"
] | none | 0 | [
{
"input": "6\nLLRLLL",
"output": "1 4\n2 5\n6 3"
},
{
"input": "4\nRRLL",
"output": "3 1\n4 2"
},
{
"input": "4\nLLRR",
"output": "1 3\n2 4"
},
{
"input": "6\nRLLRRL",
"output": "1 4\n2 5\n3 6"
},
{
"input": "8\nLRLRLLLR",
"output": "1 5\n6 2\n3 7\n4 8"
},
... | 1,675,268,095 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | n = int(input())
x = list(input())
output = ""
for i in range(n//2):
p1 = x[i]
p2 = x[i+(n//2)]
if p1 == p2 or p1 == 'L':
output += f"{i+1} {i+(n//2)+1}\n"
else:
output += f"{i+(n//2)+1} {i+1}\n"
print(output)
#DHER BARA
| Title: Lefthanders and Righthanders
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there *n* pupils who sat at the desks, two people at each desk. Vasily quickly realized that number *n* is even. Like all true mathematicians, Vasily has all students numbered from 1 to *n*.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
Input Specification:
The first input line contains a single even integer *n* (4<=≤<=*n*<=≤<=100) — the number of students in the class. The second line contains exactly *n* capital English letters "L" and "R". If the *i*-th letter at the second line equals "L", then the student number *i* is a lefthander, otherwise he is a righthander.
Output Specification:
Print integer pairs, one pair per line. In the *i*-th line print the numbers of students that will sit at the *i*-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
Demo Input:
['6\nLLRLLL\n', '4\nRRLL\n']
Demo Output:
['1 4\n2 5\n6 3\n', '3 1\n4 2\n']
Note:
none | ```python
n = int(input())
x = list(input())
output = ""
for i in range(n//2):
p1 = x[i]
p2 = x[i+(n//2)]
if p1 == p2 or p1 == 'L':
output += f"{i+1} {i+(n//2)+1}\n"
else:
output += f"{i+(n//2)+1} {i+1}\n"
print(output)
#DHER BARA
``` | -1 | |
854 | A | Fraction | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"math"
] | null | null | Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem. | In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction. | Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum. | [
"3\n",
"4\n",
"12\n"
] | [
"1 2\n",
"1 3\n",
"5 7\n"
] | none | 500 | [
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"ou... | 1,630,526,714 | 2,147,483,647 | PyPy 3 | OK | TESTS | 26 | 109 | 20,172,800 | from math import gcd
n = int(input())
k = n // 2
for i in range(k, n):
if gcd(i, n - i) == 1:
print(*sorted([n - i, i]))
break | Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none | ```python
from math import gcd
n = int(input())
k = n // 2
for i in range(k, n):
if gcd(i, n - i) == 1:
print(*sorted([n - i, i]))
break
``` | 3 | |
382 | A | Ksenia and Pan Scales | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Ksenia has ordinary pan scales and several weights of an equal mass. Ksenia has already put some weights on the scales, while other weights are untouched. Ksenia is now wondering whether it is possible to put all the remaining weights on the scales so that the scales were in equilibrium.
The scales is in equilibrium if the total sum of weights on the left pan is equal to the total sum of weights on the right pan. | The first line has a non-empty sequence of characters describing the scales. In this sequence, an uppercase English letter indicates a weight, and the symbol "|" indicates the delimiter (the character occurs in the sequence exactly once). All weights that are recorded in the sequence before the delimiter are initially on the left pan of the scale. All weights that are recorded in the sequence after the delimiter are initially on the right pan of the scale.
The second line contains a non-empty sequence containing uppercase English letters. Each letter indicates a weight which is not used yet.
It is guaranteed that all the English letters in the input data are different. It is guaranteed that the input does not contain any extra characters. | If you cannot put all the weights on the scales so that the scales were in equilibrium, print string "Impossible". Otherwise, print the description of the resulting scales, copy the format of the input.
If there are multiple answers, print any of them. | [
"AC|T\nL\n",
"|ABC\nXYZ\n",
"W|T\nF\n",
"ABC|\nD\n"
] | [
"AC|TL\n",
"XYZ|ABC\n",
"Impossible\n",
"Impossible\n"
] | none | 500 | [
{
"input": "AC|T\nL",
"output": "AC|TL"
},
{
"input": "|ABC\nXYZ",
"output": "XYZ|ABC"
},
{
"input": "W|T\nF",
"output": "Impossible"
},
{
"input": "ABC|\nD",
"output": "Impossible"
},
{
"input": "A|BC\nDEF",
"output": "ADF|BCE"
},
{
"input": "|\nABC",... | 1,662,484,981 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | # Ksenia and Pan Scales
left, right = list(map(str, input().split("|")))
w_add = input()
if (len(left) == len(right) + len(w_add)):
print(left, "|", right + w_add, sep="")
elif(len(left) + len(w_add) == len(right)):
print(left + w_add, "|", right, sep="")
else:
print("Impossible") | Title: Ksenia and Pan Scales
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ksenia has ordinary pan scales and several weights of an equal mass. Ksenia has already put some weights on the scales, while other weights are untouched. Ksenia is now wondering whether it is possible to put all the remaining weights on the scales so that the scales were in equilibrium.
The scales is in equilibrium if the total sum of weights on the left pan is equal to the total sum of weights on the right pan.
Input Specification:
The first line has a non-empty sequence of characters describing the scales. In this sequence, an uppercase English letter indicates a weight, and the symbol "|" indicates the delimiter (the character occurs in the sequence exactly once). All weights that are recorded in the sequence before the delimiter are initially on the left pan of the scale. All weights that are recorded in the sequence after the delimiter are initially on the right pan of the scale.
The second line contains a non-empty sequence containing uppercase English letters. Each letter indicates a weight which is not used yet.
It is guaranteed that all the English letters in the input data are different. It is guaranteed that the input does not contain any extra characters.
Output Specification:
If you cannot put all the weights on the scales so that the scales were in equilibrium, print string "Impossible". Otherwise, print the description of the resulting scales, copy the format of the input.
If there are multiple answers, print any of them.
Demo Input:
['AC|T\nL\n', '|ABC\nXYZ\n', 'W|T\nF\n', 'ABC|\nD\n']
Demo Output:
['AC|TL\n', 'XYZ|ABC\n', 'Impossible\n', 'Impossible\n']
Note:
none | ```python
# Ksenia and Pan Scales
left, right = list(map(str, input().split("|")))
w_add = input()
if (len(left) == len(right) + len(w_add)):
print(left, "|", right + w_add, sep="")
elif(len(left) + len(w_add) == len(right)):
print(left + w_add, "|", right, sep="")
else:
print("Impossible")
``` | 0 | |
821 | C | Okabe and Boxes | PROGRAMMING | 1,500 | [
"data structures",
"greedy",
"trees"
] | null | null | Okabe and Super Hacker Daru are stacking and removing boxes. There are *n* boxes numbered from 1 to *n*. Initially there are no boxes on the stack.
Okabe, being a control freak, gives Daru 2*n* commands: *n* of which are to add a box to the top of the stack, and *n* of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to *n*. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.
That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.
Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed. | The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of boxes.
Each of the next 2*n* lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer *x* (1<=≤<=*x*<=≤<=*n*) follows, indicating that Daru should add the box with number *x* to the top of the stack.
It is guaranteed that exactly *n* lines contain "add" operations, all the boxes added are distinct, and *n* lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed. | Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands. | [
"3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove\n",
"7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove\n"
] | [
"1\n",
"2\n"
] | In the first sample, Daru should reorder the boxes after adding box 3 to the stack.
In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack. | 1,500 | [
{
"input": "3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove",
"output": "1"
},
{
"input": "7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove",
"output": "2"
},
{
"input": "4\nadd 1\nadd 3\nremove\nadd 4\nadd 2\nremove\nremove\nremov... | 1,607,503,567 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define all(c) (c).begin(), (c).end()
#define data ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pb push_back
#define REP(i,a,b) for (int i = a; i < b; i++)
#define range(name, start,end,step) (int name = start; name < end; name+= step)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ar array;
int pow(int a, int b, int m){
int ans = 1;
while(b){
if(b&1) ans=(ans*a)%m;
b/=2;
a=(a*a)%m;
}
return ans;
}
int main(){
data;
int n, cur = 1, res = 0, x, len;
string s;
cin >> n;
vector<int> stack;
for (int i = 0; i< 2*n; i++){
cin >> s;
if (s == "add"){
cin >> x;
stack.push_back(x);
}else{
len = stack.size();
if (stack.at(len-1) != cur){
sort(stack.begin(), stack.end(), greater<int>());
res++;
}
stack.pop_back();
cur ++;
}
}
cout << res;
return 0;
} | Title: Okabe and Boxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Okabe and Super Hacker Daru are stacking and removing boxes. There are *n* boxes numbered from 1 to *n*. Initially there are no boxes on the stack.
Okabe, being a control freak, gives Daru 2*n* commands: *n* of which are to add a box to the top of the stack, and *n* of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to *n*. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.
That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.
Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed.
Input Specification:
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of boxes.
Each of the next 2*n* lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer *x* (1<=≤<=*x*<=≤<=*n*) follows, indicating that Daru should add the box with number *x* to the top of the stack.
It is guaranteed that exactly *n* lines contain "add" operations, all the boxes added are distinct, and *n* lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed.
Output Specification:
Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands.
Demo Input:
['3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove\n', '7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove\n']
Demo Output:
['1\n', '2\n']
Note:
In the first sample, Daru should reorder the boxes after adding box 3 to the stack.
In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack. | ```python
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define all(c) (c).begin(), (c).end()
#define data ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pb push_back
#define REP(i,a,b) for (int i = a; i < b; i++)
#define range(name, start,end,step) (int name = start; name < end; name+= step)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ar array;
int pow(int a, int b, int m){
int ans = 1;
while(b){
if(b&1) ans=(ans*a)%m;
b/=2;
a=(a*a)%m;
}
return ans;
}
int main(){
data;
int n, cur = 1, res = 0, x, len;
string s;
cin >> n;
vector<int> stack;
for (int i = 0; i< 2*n; i++){
cin >> s;
if (s == "add"){
cin >> x;
stack.push_back(x);
}else{
len = stack.size();
if (stack.at(len-1) != cur){
sort(stack.begin(), stack.end(), greater<int>());
res++;
}
stack.pop_back();
cur ++;
}
}
cout << res;
return 0;
}
``` | -1 | |
583 | A | Asphalting Roads | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct. | In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. | [
"2\n1 1\n1 2\n2 1\n2 2\n",
"1\n1 1\n"
] | [
"1 4 \n",
"1 \n"
] | In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | 500 | [
{
"input": "2\n1 1\n1 2\n2 1\n2 2",
"output": "1 4 "
},
{
"input": "1\n1 1",
"output": "1 "
},
{
"input": "2\n1 1\n2 2\n1 2\n2 1",
"output": "1 2 "
},
{
"input": "2\n1 2\n2 2\n2 1\n1 1",
"output": "1 3 "
},
{
"input": "3\n2 2\n1 2\n3 2\n3 3\n1 1\n2 3\n1 3\n3 1\n2 ... | 1,451,023,186 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | n = int(input())
inter = [[False]*n]*2
for i in range(n*n):
h,v = input().split()
hnum = int(h)-1
vnum = int(v)-1
if not inter[0][hnum] and not inter[1][vnum]:
inter[0][hnum] = True
inter[1][vnum] = True
print(i+1,end=' ') | Title: Asphalting Roads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
Output Specification:
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
Demo Input:
['2\n1 1\n1 2\n2 1\n2 2\n', '1\n1 1\n']
Demo Output:
['1 4 \n', '1 \n']
Note:
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | ```python
n = int(input())
inter = [[False]*n]*2
for i in range(n*n):
h,v = input().split()
hnum = int(h)-1
vnum = int(v)-1
if not inter[0][hnum] and not inter[1][vnum]:
inter[0][hnum] = True
inter[1][vnum] = True
print(i+1,end=' ')
``` | 0 | |
768 | A | Oath of the Night's Watch | PROGRAMMING | 900 | [
"constructive algorithms",
"sortings"
] | null | null | "Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support? | First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow.
Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards. | Output a single integer representing the number of stewards which Jon will feed. | [
"2\n1 5\n",
"3\n1 2 5\n"
] | [
"0",
"1"
] | In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | 500 | [
{
"input": "2\n1 5",
"output": "0"
},
{
"input": "3\n1 2 5",
"output": "1"
},
{
"input": "4\n1 2 3 4",
"output": "2"
},
{
"input": "8\n7 8 9 4 5 6 1 2",
"output": "6"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n100",
"output": "0"
},
... | 1,684,410,238 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 15 | 0 | a=int(input())
b=list(map(int,input().split()))
s=0
k=sorted(b)
for i in b:
if i<k[-1] and i > k[0]:
s+=1
if s ==0:
print(0)
else:
print(s-1)
| Title: Oath of the Night's Watch
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input Specification:
First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow.
Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards.
Output Specification:
Output a single integer representing the number of stewards which Jon will feed.
Demo Input:
['2\n1 5\n', '3\n1 2 5\n']
Demo Output:
['0', '1']
Note:
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | ```python
a=int(input())
b=list(map(int,input().split()))
s=0
k=sorted(b)
for i in b:
if i<k[-1] and i > k[0]:
s+=1
if s ==0:
print(0)
else:
print(s-1)
``` | 0 | |
149 | A | Business trip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters. | The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100). | Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1. | [
"5\n1 1 1 1 2 2 3 2 2 1 1 1\n",
"0\n0 0 0 0 0 0 0 1 1 2 3 0\n",
"11\n1 1 4 1 1 5 1 1 4 1 1 1\n"
] | [
"2\n",
"0\n",
"3\n"
] | Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all. | 500 | [
{
"input": "5\n1 1 1 1 2 2 3 2 2 1 1 1",
"output": "2"
},
{
"input": "0\n0 0 0 0 0 0 0 1 1 2 3 0",
"output": "0"
},
{
"input": "11\n1 1 4 1 1 5 1 1 4 1 1 1",
"output": "3"
},
{
"input": "15\n20 1 1 1 1 2 2 1 2 2 1 1",
"output": "1"
},
{
"input": "7\n8 9 100 12 14 ... | 1,640,940,099 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 92 | 0 | k = int(input())
lis = sorted(list(map(int, input().split())))[::-1]
s =0
m =0
flag=0
if k==0:
print(0)
else:
for i in lis:
s+=i
m+=1
if s>=k:
flag=1
break
if flag==1:
print(m)
else:
print(-1)
| Title: Business trip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
Input Specification:
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
Output Specification:
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
Demo Input:
['5\n1 1 1 1 2 2 3 2 2 1 1 1\n', '0\n0 0 0 0 0 0 0 1 1 2 3 0\n', '11\n1 1 4 1 1 5 1 1 4 1 1 1\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all. | ```python
k = int(input())
lis = sorted(list(map(int, input().split())))[::-1]
s =0
m =0
flag=0
if k==0:
print(0)
else:
for i in lis:
s+=i
m+=1
if s>=k:
flag=1
break
if flag==1:
print(m)
else:
print(-1)
``` | 3 | |
31 | C | Schedule | PROGRAMMING | 1,700 | [
"implementation"
] | C. Schedule | 2 | 256 | At the beginning of the new semester there is new schedule in the Berland State University. According to this schedule, *n* groups have lessons at the room 31. For each group the starting time of the lesson and the finishing time of the lesson are known. It has turned out that it is impossible to hold all lessons, because for some groups periods of their lessons intersect. If at some moment of time one groups finishes it's lesson, and the other group starts the lesson, their lessons don't intersect.
The dean wants to cancel the lesson in one group so that no two time periods of lessons of the remaining groups intersect. You are to find all ways to do that. | The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — amount of groups, which have lessons in the room 31. Then *n* lines follow, each of them contains two integers *l**i* *r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=106) — starting and finishing times of lesson of the *i*-th group. It is possible that initially no two lessons intersect (see sample 1). | Output integer *k* — amount of ways to cancel the lesson in exactly one group so that no two time periods of lessons of the remaining groups intersect. In the second line output *k* numbers — indexes of groups, where it is possible to cancel the lesson. Groups are numbered starting from 1 in the order that they were given in the input. Output the numbers in increasing order. | [
"3\n3 10\n20 30\n1 3\n",
"4\n3 10\n20 30\n1 3\n1 39\n",
"3\n1 5\n2 6\n3 7\n"
] | [
"3\n1 2 3 ",
"1\n4 ",
"0\n"
] | none | 1,500 | [
{
"input": "3\n3 10\n20 30\n1 3",
"output": "3\n1 2 3 "
},
{
"input": "4\n3 10\n20 30\n1 3\n1 39",
"output": "1\n4 "
},
{
"input": "3\n1 5\n2 6\n3 7",
"output": "0"
},
{
"input": "4\n1 5\n5 7\n6 9\n9 10",
"output": "2\n2 3 "
},
{
"input": "11\n717170 795210\n86642... | 1,522,078,581 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 122 | 6,963,200 | groups = []
overlapping = set()
for notused in range(int(input())):
groups.append(list(map(int, input().split(" "))))
for i in range(len(groups)-1):
if groups[-1][0] > groups[i][0] and groups[-1][0] < groups[i][1] or groups[-1][1] > groups[i][0] and groups[-1][1] < groups[i][1]:
overlapping.add(i)
overlapping.add(len(groups)-1)
if groups[i][0] > groups[-1][0] and groups[i][0] < groups[-1][1] or groups[i][1] > groups[-1][0] and groups[i][1] < groups[-1][1]:
overlapping.add(i)
overlapping.add(len(groups)-1)
if len(overlapping) == 0:
print(len(groups))
for i in range(len(groups)): print(i, end = " ")
print()
else:
# we now have a non-empty set of overlapping elements in the list.
possible = []
for item in overlapping:
newoverlaping = overlapping - {item}
ranges = []
overlap = False
for z in newoverlaping:
x = groups[z]
for y in ranges:
if x[0] > y[0] and x[0] < y[1] or x[1] > y[0] and x[1] < y[1]:
overlap = True
break;
if y[0] > x[0] and y[0] < x[1] or y[1] > x[0] and y[1] < x[1]:
overlap = True
break;
ranges.append(x)
if overlap: break;
if not overlap: possible.append(item+1)
print(len(possible))
for item in possible: print(item, end=" ")
if len(possible) != 0:print() | Title: Schedule
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
At the beginning of the new semester there is new schedule in the Berland State University. According to this schedule, *n* groups have lessons at the room 31. For each group the starting time of the lesson and the finishing time of the lesson are known. It has turned out that it is impossible to hold all lessons, because for some groups periods of their lessons intersect. If at some moment of time one groups finishes it's lesson, and the other group starts the lesson, their lessons don't intersect.
The dean wants to cancel the lesson in one group so that no two time periods of lessons of the remaining groups intersect. You are to find all ways to do that.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — amount of groups, which have lessons in the room 31. Then *n* lines follow, each of them contains two integers *l**i* *r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=106) — starting and finishing times of lesson of the *i*-th group. It is possible that initially no two lessons intersect (see sample 1).
Output Specification:
Output integer *k* — amount of ways to cancel the lesson in exactly one group so that no two time periods of lessons of the remaining groups intersect. In the second line output *k* numbers — indexes of groups, where it is possible to cancel the lesson. Groups are numbered starting from 1 in the order that they were given in the input. Output the numbers in increasing order.
Demo Input:
['3\n3 10\n20 30\n1 3\n', '4\n3 10\n20 30\n1 3\n1 39\n', '3\n1 5\n2 6\n3 7\n']
Demo Output:
['3\n1 2 3 ', '1\n4 ', '0\n']
Note:
none | ```python
groups = []
overlapping = set()
for notused in range(int(input())):
groups.append(list(map(int, input().split(" "))))
for i in range(len(groups)-1):
if groups[-1][0] > groups[i][0] and groups[-1][0] < groups[i][1] or groups[-1][1] > groups[i][0] and groups[-1][1] < groups[i][1]:
overlapping.add(i)
overlapping.add(len(groups)-1)
if groups[i][0] > groups[-1][0] and groups[i][0] < groups[-1][1] or groups[i][1] > groups[-1][0] and groups[i][1] < groups[-1][1]:
overlapping.add(i)
overlapping.add(len(groups)-1)
if len(overlapping) == 0:
print(len(groups))
for i in range(len(groups)): print(i, end = " ")
print()
else:
# we now have a non-empty set of overlapping elements in the list.
possible = []
for item in overlapping:
newoverlaping = overlapping - {item}
ranges = []
overlap = False
for z in newoverlaping:
x = groups[z]
for y in ranges:
if x[0] > y[0] and x[0] < y[1] or x[1] > y[0] and x[1] < y[1]:
overlap = True
break;
if y[0] > x[0] and y[0] < x[1] or y[1] > x[0] and y[1] < x[1]:
overlap = True
break;
ranges.append(x)
if overlap: break;
if not overlap: possible.append(item+1)
print(len(possible))
for item in possible: print(item, end=" ")
if len(possible) != 0:print()
``` | 0 |
873 | A | Chores | PROGRAMMING | 800 | [
"implementation"
] | null | null | Luba has to do *n* chores today. *i*-th chore takes *a**i* units of time to complete. It is guaranteed that for every the condition *a**i*<=≥<=*a**i*<=-<=1 is met, so the sequence is sorted.
Also Luba can work really hard on some chores. She can choose not more than *k* any chores and do each of them in *x* units of time instead of *a**i* ().
Luba is very responsible, so she has to do all *n* chores, and now she wants to know the minimum time she needs to do everything. Luba cannot do two chores simultaneously. | The first line contains three integers *n*,<=*k*,<=*x* (1<=≤<=*k*<=≤<=*n*<=≤<=100,<=1<=≤<=*x*<=≤<=99) — the number of chores Luba has to do, the number of chores she can do in *x* units of time, and the number *x* itself.
The second line contains *n* integer numbers *a**i* (2<=≤<=*a**i*<=≤<=100) — the time Luba has to spend to do *i*-th chore.
It is guaranteed that , and for each *a**i*<=≥<=*a**i*<=-<=1. | Print one number — minimum time Luba needs to do all *n* chores. | [
"4 2 2\n3 6 7 10\n",
"5 2 1\n100 100 100 100 100\n"
] | [
"13\n",
"302\n"
] | In the first example the best option would be to do the third and the fourth chore, spending *x* = 2 time on each instead of *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, respectively. Then the answer is 3 + 6 + 2 + 2 = 13.
In the second example Luba can choose any two chores to spend *x* time on them instead of *a*<sub class="lower-index">*i*</sub>. So the answer is 100·3 + 2·1 = 302. | 0 | [
{
"input": "4 2 2\n3 6 7 10",
"output": "13"
},
{
"input": "5 2 1\n100 100 100 100 100",
"output": "302"
},
{
"input": "1 1 1\n100",
"output": "1"
},
{
"input": "100 1 99\n100 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 ... | 1,636,384,035 | 2,147,483,647 | Python 3 | OK | TESTS | 11 | 46 | 4,300,800 | n,k,x=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
print(sum(a)+k*x-sum(a[n-k:])) | Title: Chores
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba has to do *n* chores today. *i*-th chore takes *a**i* units of time to complete. It is guaranteed that for every the condition *a**i*<=≥<=*a**i*<=-<=1 is met, so the sequence is sorted.
Also Luba can work really hard on some chores. She can choose not more than *k* any chores and do each of them in *x* units of time instead of *a**i* ().
Luba is very responsible, so she has to do all *n* chores, and now she wants to know the minimum time she needs to do everything. Luba cannot do two chores simultaneously.
Input Specification:
The first line contains three integers *n*,<=*k*,<=*x* (1<=≤<=*k*<=≤<=*n*<=≤<=100,<=1<=≤<=*x*<=≤<=99) — the number of chores Luba has to do, the number of chores she can do in *x* units of time, and the number *x* itself.
The second line contains *n* integer numbers *a**i* (2<=≤<=*a**i*<=≤<=100) — the time Luba has to spend to do *i*-th chore.
It is guaranteed that , and for each *a**i*<=≥<=*a**i*<=-<=1.
Output Specification:
Print one number — minimum time Luba needs to do all *n* chores.
Demo Input:
['4 2 2\n3 6 7 10\n', '5 2 1\n100 100 100 100 100\n']
Demo Output:
['13\n', '302\n']
Note:
In the first example the best option would be to do the third and the fourth chore, spending *x* = 2 time on each instead of *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, respectively. Then the answer is 3 + 6 + 2 + 2 = 13.
In the second example Luba can choose any two chores to spend *x* time on them instead of *a*<sub class="lower-index">*i*</sub>. So the answer is 100·3 + 2·1 = 302. | ```python
n,k,x=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
print(sum(a)+k*x-sum(a[n-k:]))
``` | 3 | |
136 | A | Presents | PROGRAMMING | 800 | [
"implementation"
] | null | null | Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves. | Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*. | [
"4\n2 3 4 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] | [
"4 1 2 3\n",
"1 3 2\n",
"1 2\n"
] | none | 500 | [
{
"input": "4\n2 3 4 1",
"output": "4 1 2 3"
},
{
"input": "3\n1 3 2",
"output": "1 3 2"
},
{
"input": "2\n1 2",
"output": "1 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 3 2 6 4 5 7 9 8 10",
"output": "1 3 2 5 6 4 7 9 8 10"
},
{
"input"... | 1,695,300,617 | 2,147,483,647 | Python 3 | OK | TESTS | 101 | 122 | 0 | n=int(input())
my_str=input()
my_list=[]
a=my_str.split()
h=1
while h<=n:
index=a.index(str(h))
my_list.append(str(index+1))
h +=1
result=' '.join(my_list)
print(result) | Title: Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Output Specification:
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
Demo Input:
['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['4 1 2 3\n', '1 3 2\n', '1 2\n']
Note:
none | ```python
n=int(input())
my_str=input()
my_list=[]
a=my_str.split()
h=1
while h<=n:
index=a.index(str(h))
my_list.append(str(index+1))
h +=1
result=' '.join(my_list)
print(result)
``` | 3 | |
20 | C | Dijkstra? | PROGRAMMING | 1,900 | [
"graphs",
"shortest paths"
] | C. Dijkstra? | 1 | 64 | You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | [
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] | [
"1 4 3 5 ",
"1 4 3 5 "
] | none | 1,500 | [
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input... | 1,599,946,878 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 27 | 1,000 | 26,521,600 | import heapq
from collections import defaultdict
def distances(graph, start,leng):
d = {vertex: float('infinity') for vertex in range(1,leng+1)}
prev = {vertex: 1 for vertex in graph}
d[start] = 0
pq = [(0, start)]
while len(pq) > 0:
dis, ver = heapq.heappop(pq)
if dis > d[ver]:
continue
for nei, weight in graph[ver].items():
alt = dis + weight
if alt < d[nei]:
d[nei] = alt
heapq.heappush(pq, (alt, nei))
prev[nei] = ver
return prev, d
def backtrace(parent, start, end):
path = [end]
while path[-1] != start:
path.append(parent[path[-1]])
path.reverse()
return path
g = defaultdict(dict)
n, m = map(int, input().split())
for i in range(m):
u,v,w = map(int,input().split())
g[u][v] = w
g[v][u] = w
prev,d = distances(g,1,n)
print((' '.join(str(i) for i in backtrace(prev,1,n))) if d[n] < float('infinity') else -1) | Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none | ```python
import heapq
from collections import defaultdict
def distances(graph, start,leng):
d = {vertex: float('infinity') for vertex in range(1,leng+1)}
prev = {vertex: 1 for vertex in graph}
d[start] = 0
pq = [(0, start)]
while len(pq) > 0:
dis, ver = heapq.heappop(pq)
if dis > d[ver]:
continue
for nei, weight in graph[ver].items():
alt = dis + weight
if alt < d[nei]:
d[nei] = alt
heapq.heappush(pq, (alt, nei))
prev[nei] = ver
return prev, d
def backtrace(parent, start, end):
path = [end]
while path[-1] != start:
path.append(parent[path[-1]])
path.reverse()
return path
g = defaultdict(dict)
n, m = map(int, input().split())
for i in range(m):
u,v,w = map(int,input().split())
g[u][v] = w
g[v][u] = w
prev,d = distances(g,1,n)
print((' '.join(str(i) for i in backtrace(prev,1,n))) if d[n] < float('infinity') else -1)
``` | 0 |
785 | A | Anton and Polyhedrons | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number! | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron. | Output one number — the total number of faces in all the polyhedrons in Anton's collection. | [
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] | [
"42\n",
"28\n"
] | In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | 500 | [
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosa... | 1,674,398,724 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 30 | 1,091 | 9,216,000 | d = {"Tetrahedron":4, "Cube":6, "Octahedron":8, "Dodecahedron":12, "Icosahedron":20}
n = int(input())
c=0
for i in range(n):
c += d[str(input())]
print(c) | Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | ```python
d = {"Tetrahedron":4, "Cube":6, "Octahedron":8, "Dodecahedron":12, "Icosahedron":20}
n = int(input())
c=0
for i in range(n):
c += d[str(input())]
print(c)
``` | 3 | |
527 | C | Glass Carving | PROGRAMMING | 1,500 | [
"binary search",
"data structures",
"implementation"
] | null | null | Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree? | The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts. | After each cut print on a single line the area of the maximum available glass fragment in mm2. | [
"4 3 4\nH 2\nV 2\nV 3\nV 1\n",
"7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n"
] | [
"8\n4\n4\n2\n",
"28\n16\n12\n6\n4\n"
] | Picture for the first sample test: | 1,500 | [
{
"input": "4 3 4\nH 2\nV 2\nV 3\nV 1",
"output": "8\n4\n4\n2"
},
{
"input": "7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1",
"output": "28\n16\n12\n6\n4"
},
{
"input": "2 2 1\nV 1",
"output": "2"
},
{
"input": "2 2 1\nH 1",
"output": "2"
},
{
"input": "2 2 2\nV 1\nH 1",
"ou... | 1,568,708,618 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 93 | 307,200 | __author__ = 'ruckus'
class Split:
def __init__(self, order, offset) -> None:
super().__init__()
self.order = order
self.offset = int(offset)
class Glass:
h_splits = [0]
w_splits = [0]
max_p = 0
max_p_h_offset = 0
max_p_w_offset = 0
def __init__(self, h, w) -> None:
super().__init__()
self.h_splits.append(h)
self.w_splits.append(w)
self.max_p = h * w
def split(self, split: Split):
if split.order == 'H':
self.h_splits.append(split.offset)
self.h_splits = self.h_splits.sort()
else:
self.w_splits.append(split.offset)
self.w_splits = self.w_splits.sort()
max_p = 0
max_p_h_offset = 0
max_p_w_offset = 0
for i in range(1,len(self.h_splits)):
for j in range(1,len(self.w_splits)):
p = (self.h_splits[i] - self.h_splits[i - 1]) * (self.w_splits[j] - self.w_splits[j - 1])
if p > max_p:
max_p = p
max_p_h_offset = self.h_splits[i - 1]
max_p_w_offset = self.w_splits[j - 1]
self.max_p = max_p
self.max_p_h_offset = max_p_h_offset
self.max_p_w_offset = max_p_w_offset
print(max_p)
w, h, n = map(int, input().split())
glass = Glass(h, w)
for i in range(n):
order, offset = input().split()
glass.split(Split(order, offset)) | Title: Glass Carving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?
Input Specification:
The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
Output Specification:
After each cut print on a single line the area of the maximum available glass fragment in mm2.
Demo Input:
['4 3 4\nH 2\nV 2\nV 3\nV 1\n', '7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n']
Demo Output:
['8\n4\n4\n2\n', '28\n16\n12\n6\n4\n']
Note:
Picture for the first sample test: | ```python
__author__ = 'ruckus'
class Split:
def __init__(self, order, offset) -> None:
super().__init__()
self.order = order
self.offset = int(offset)
class Glass:
h_splits = [0]
w_splits = [0]
max_p = 0
max_p_h_offset = 0
max_p_w_offset = 0
def __init__(self, h, w) -> None:
super().__init__()
self.h_splits.append(h)
self.w_splits.append(w)
self.max_p = h * w
def split(self, split: Split):
if split.order == 'H':
self.h_splits.append(split.offset)
self.h_splits = self.h_splits.sort()
else:
self.w_splits.append(split.offset)
self.w_splits = self.w_splits.sort()
max_p = 0
max_p_h_offset = 0
max_p_w_offset = 0
for i in range(1,len(self.h_splits)):
for j in range(1,len(self.w_splits)):
p = (self.h_splits[i] - self.h_splits[i - 1]) * (self.w_splits[j] - self.w_splits[j - 1])
if p > max_p:
max_p = p
max_p_h_offset = self.h_splits[i - 1]
max_p_w_offset = self.w_splits[j - 1]
self.max_p = max_p
self.max_p_h_offset = max_p_h_offset
self.max_p_w_offset = max_p_w_offset
print(max_p)
w, h, n = map(int, input().split())
glass = Glass(h, w)
for i in range(n):
order, offset = input().split()
glass.split(Split(order, offset))
``` | -1 | |
104 | A | Blackjack | PROGRAMMING | 800 | [
"implementation"
] | A. Blackjack | 2 | 256 | One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. | The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. | Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. | [
"12\n",
"20\n",
"10\n"
] | [
"4",
"15",
"0"
] | In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | 500 | [
{
"input": "12",
"output": "4"
},
{
"input": "20",
"output": "15"
},
{
"input": "10",
"output": "0"
},
{
"input": "11",
"output": "4"
},
{
"input": "15",
"output": "4"
},
{
"input": "18",
"output": "4"
},
{
"input": "25",
"output": "0"
... | 1,671,963,750 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 92 | 0 | n = int(input())
req = abs(n - 10)
if req == 1 or req == 11:
print(4)
elif req >= 2 and req <= 9:
print(4)
elif req == 10:
print(15)
else:
print(0) | Title: Blackjack
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.
Input Specification:
The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.
Output Specification:
Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.
Demo Input:
['12\n', '20\n', '10\n']
Demo Output:
['4', '15', '0']
Note:
In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | ```python
n = int(input())
req = abs(n - 10)
if req == 1 or req == 11:
print(4)
elif req >= 2 and req <= 9:
print(4)
elif req == 10:
print(15)
else:
print(0)
``` | 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... | 1,510,507,621 | 4,921 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 4 | 2,000 | 14,233,600 | a = int(input())
b = list(map(int, input().split()))
minnum = 0
k = a - 1
keys = []
while k >= 0:
if b[k] not in keys:
keys.append(b[k])
min = k
minnum = b[k]
else:
if b[k] == minnum:
if k - min > 1:
k = min - 1
min = k
minnum = b[k]
else:
min = k
minnum = b[k]
k -= 1
print(minnum) | 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
a = int(input())
b = list(map(int, input().split()))
minnum = 0
k = a - 1
keys = []
while k >= 0:
if b[k] not in keys:
keys.append(b[k])
min = k
minnum = b[k]
else:
if b[k] == minnum:
if k - min > 1:
k = min - 1
min = k
minnum = b[k]
else:
min = k
minnum = b[k]
k -= 1
print(minnum)
``` | 0 | |
617 | A | Elephant | PROGRAMMING | 800 | [
"math"
] | null | null | An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house. | The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house. | Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*. | [
"5\n",
"12\n"
] | [
"1\n",
"3\n"
] | In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves. | 500 | [
{
"input": "5",
"output": "1"
},
{
"input": "12",
"output": "3"
},
{
"input": "999999",
"output": "200000"
},
{
"input": "41",
"output": "9"
},
{
"input": "1000000",
"output": "200000"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
... | 1,699,022,901 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 61 | 0 | x=int(input())
if x%5==1: print(x//5)
else : print(x//5+1) | Title: Elephant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
Input Specification:
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Output Specification:
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
Demo Input:
['5\n', '12\n']
Demo Output:
['1\n', '3\n']
Note:
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves. | ```python
x=int(input())
if x%5==1: print(x//5)
else : print(x//5+1)
``` | 0 | |
610 | A | Pasha and Stick | PROGRAMMING | 1,000 | [
"combinatorics",
"math"
] | null | null | Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way. | The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick. | The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square. | [
"6\n",
"20\n"
] | [
"1\n",
"4\n"
] | There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work. | 500 | [
{
"input": "6",
"output": "1"
},
{
"input": "20",
"output": "4"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "2000000000",
"output": "4... | 1,581,761,997 | 2,147,483,647 | Python 3 | OK | TESTS | 76 | 109 | 307,200 | a = int(input())
if a % 2 != 0:
print(0)
exit()
print((a // 2 - 1) // 2) | Title: Pasha and Stick
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick.
Output Specification:
The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.
Demo Input:
['6\n', '20\n']
Demo Output:
['1\n', '4\n']
Note:
There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work. | ```python
a = int(input())
if a % 2 != 0:
print(0)
exit()
print((a // 2 - 1) // 2)
``` | 3 | |
224 | B | Array | PROGRAMMING | 1,500 | [
"bitmasks",
"implementation",
"two pointers"
] | null | null | You've got an array *a*, consisting of *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. Your task is to find a minimal by inclusion segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) such, that among numbers *a**l*,<= *a**l*<=+<=1,<= ...,<= *a**r* there are exactly *k* distinct numbers.
Segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; *l*,<=*r* are integers) of length *m*<==<=*r*<=-<=*l*<=+<=1, satisfying the given property, is called minimal by inclusion, if there is no segment [*x*,<=*y*] satisfying the property and less then *m* in length, such that 1<=≤<=*l*<=≤<=*x*<=≤<=*y*<=≤<=*r*<=≤<=*n*. Note that the segment [*l*,<=*r*] doesn't have to be minimal in length among all segments, satisfying the given property. | The first line contains two space-separated integers: *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* — elements of the array *a* (1<=≤<=*a**i*<=≤<=105). | Print a space-separated pair of integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) such, that the segment [*l*,<=*r*] is the answer to the problem. If the sought segment does not exist, print "-1 -1" without the quotes. If there are multiple correct answers, print any of them. | [
"4 2\n1 2 2 3\n",
"8 3\n1 1 2 2 3 3 4 5\n",
"7 4\n4 7 7 4 7 4 7\n"
] | [
"1 2\n",
"2 5\n",
"-1 -1\n"
] | In the first sample among numbers *a*<sub class="lower-index">1</sub> and *a*<sub class="lower-index">2</sub> there are exactly two distinct numbers.
In the second sample segment [2, 5] is a minimal by inclusion segment with three distinct numbers, but it is not minimal in length among such segments.
In the third sample there is no segment with four distinct numbers. | 1,000 | [
{
"input": "4 2\n1 2 2 3",
"output": "1 2"
},
{
"input": "8 3\n1 1 2 2 3 3 4 5",
"output": "2 5"
},
{
"input": "7 4\n4 7 7 4 7 4 7",
"output": "-1 -1"
},
{
"input": "5 1\n1 7 2 3 2",
"output": "1 1"
},
{
"input": "1 2\n666",
"output": "-1 -1"
},
{
"inp... | 1,590,317,378 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 342 | 7,680,000 | #####################
#1: Array,
#R -> ++Count[]
#L -> --Count[]
line1 = input().split(" ")
arr = list(map(int,input().split()))
n = int(line1[0])
k = int(line1[1])
counts = {}
l = 0
r = 0
count = 0
for i in range(len(arr)):
if arr[i] in counts:
counts[arr[i]] += 1
else:
count += 1
counts[arr[i]] = 1
if count >= k:
r = i
break
for i in range(len(arr)):
l = i
counts[arr[i]] -= 1
if counts[arr[i]] == 0:
l = i
break
if count < k:
print("-1 -1")
else:
print(l + 1, r + 1)
| Title: Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got an array *a*, consisting of *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. Your task is to find a minimal by inclusion segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) such, that among numbers *a**l*,<= *a**l*<=+<=1,<= ...,<= *a**r* there are exactly *k* distinct numbers.
Segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; *l*,<=*r* are integers) of length *m*<==<=*r*<=-<=*l*<=+<=1, satisfying the given property, is called minimal by inclusion, if there is no segment [*x*,<=*y*] satisfying the property and less then *m* in length, such that 1<=≤<=*l*<=≤<=*x*<=≤<=*y*<=≤<=*r*<=≤<=*n*. Note that the segment [*l*,<=*r*] doesn't have to be minimal in length among all segments, satisfying the given property.
Input Specification:
The first line contains two space-separated integers: *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* — elements of the array *a* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Print a space-separated pair of integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) such, that the segment [*l*,<=*r*] is the answer to the problem. If the sought segment does not exist, print "-1 -1" without the quotes. If there are multiple correct answers, print any of them.
Demo Input:
['4 2\n1 2 2 3\n', '8 3\n1 1 2 2 3 3 4 5\n', '7 4\n4 7 7 4 7 4 7\n']
Demo Output:
['1 2\n', '2 5\n', '-1 -1\n']
Note:
In the first sample among numbers *a*<sub class="lower-index">1</sub> and *a*<sub class="lower-index">2</sub> there are exactly two distinct numbers.
In the second sample segment [2, 5] is a minimal by inclusion segment with three distinct numbers, but it is not minimal in length among such segments.
In the third sample there is no segment with four distinct numbers. | ```python
#####################
#1: Array,
#R -> ++Count[]
#L -> --Count[]
line1 = input().split(" ")
arr = list(map(int,input().split()))
n = int(line1[0])
k = int(line1[1])
counts = {}
l = 0
r = 0
count = 0
for i in range(len(arr)):
if arr[i] in counts:
counts[arr[i]] += 1
else:
count += 1
counts[arr[i]] = 1
if count >= k:
r = i
break
for i in range(len(arr)):
l = i
counts[arr[i]] -= 1
if counts[arr[i]] == 0:
l = i
break
if count < k:
print("-1 -1")
else:
print(l + 1, r + 1)
``` | 3 | |
675 | A | Infinite Sequence | PROGRAMMING | 1,100 | [
"math"
] | null | null | Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help. | The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively. | If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes). | [
"1 7 3\n",
"10 10 0\n",
"1 -4 5\n",
"0 60 50\n"
] | [
"YES\n",
"YES\n",
"NO\n",
"NO\n"
] | In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer. | 500 | [
{
"input": "1 7 3",
"output": "YES"
},
{
"input": "10 10 0",
"output": "YES"
},
{
"input": "1 -4 5",
"output": "NO"
},
{
"input": "0 60 50",
"output": "NO"
},
{
"input": "1 -4 -5",
"output": "YES"
},
{
"input": "0 1 0",
"output": "NO"
},
{
... | 1,598,621,782 | 2,147,483,647 | PyPy 3 | OK | TESTS | 178 | 156 | 0 | def f(l):
a,b,c = l
if c==0:
return a==b
return (b-a)%c==0 and (b-a)//c>=0
l = list(map(int,input().split()))
print('YES' if f(l) else 'NO')
| Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help.
Input Specification:
The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.
Output Specification:
If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes).
Demo Input:
['1 7 3\n', '10 10 0\n', '1 -4 5\n', '0 60 50\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'NO\n']
Note:
In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer. | ```python
def f(l):
a,b,c = l
if c==0:
return a==b
return (b-a)%c==0 and (b-a)//c>=0
l = list(map(int,input().split()))
print('YES' if f(l) else 'NO')
``` | 3 | |
34 | B | Sale | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | B. Sale | 2 | 256 | Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets. | Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets. | [
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] | [
"8\n",
"7\n"
] | none | 1,000 | [
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "... | 1,623,589,393 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 186 | 0 | n, m = list(map(int, input().split()))
prices = list(map(int, input().split()))
negative_prices = []
for price in prices:
if price < 0:
negative_prices.append(price)
negative_prices.sort()
result = 0
for i in range(len(negative_prices)):
result -= negative_prices[i]
print(result) | Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none | ```python
n, m = list(map(int, input().split()))
prices = list(map(int, input().split()))
negative_prices = []
for price in prices:
if price < 0:
negative_prices.append(price)
negative_prices.sort()
result = 0
for i in range(len(negative_prices)):
result -= negative_prices[i]
print(result)
``` | 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... | 1,697,377,920 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 92 | 0 | cur = 0
ans = 0
for _ in range(int(input())):
a, b = list(map(int, input().split()))
cur -= a
cur += b
ans = max(ans, cur)
print(ans) | 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
cur = 0
ans = 0
for _ in range(int(input())):
a, b = list(map(int, input().split()))
cur -= a
cur += b
ans = max(ans, cur)
print(ans)
``` | 3 | |
762 | A | k-th divisor | PROGRAMMING | 1,400 | [
"math",
"number theory"
] | null | null | You are given two integers *n* and *k*. Find *k*-th smallest divisor of *n*, or report that it doesn't exist.
Divisor of *n* is any such natural number, that *n* can be divided by it without remainder. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1015, 1<=≤<=*k*<=≤<=109). | If *n* has less than *k* divisors, output -1.
Otherwise, output the *k*-th smallest divisor of *n*. | [
"4 2\n",
"5 3\n",
"12 5\n"
] | [
"2\n",
"-1\n",
"6\n"
] | In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | 0 | [
{
"input": "4 2",
"output": "2"
},
{
"input": "5 3",
"output": "-1"
},
{
"input": "12 5",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "866421317361600 26880",
"output": "866421317361600"
},
{
"input": "866421317361600 26881",
"ou... | 1,584,365,565 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 124 | 0 | # You
# Dont read my code
a,b = map(int,input().split())
div = [1]
if a % 2 != 0:
for i in range(3, a,2):
if a % i == 0:
div.append(i)
div.append(a)
else:
for i in range(2,a,3):
if a % i == 0:
div.append(i)
div.append(a)
if b <= len(div):
print(div[b-1])
else:
print(-1)
| Title: k-th divisor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two integers *n* and *k*. Find *k*-th smallest divisor of *n*, or report that it doesn't exist.
Divisor of *n* is any such natural number, that *n* can be divided by it without remainder.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1015, 1<=≤<=*k*<=≤<=109).
Output Specification:
If *n* has less than *k* divisors, output -1.
Otherwise, output the *k*-th smallest divisor of *n*.
Demo Input:
['4 2\n', '5 3\n', '12 5\n']
Demo Output:
['2\n', '-1\n', '6\n']
Note:
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | ```python
# You
# Dont read my code
a,b = map(int,input().split())
div = [1]
if a % 2 != 0:
for i in range(3, a,2):
if a % i == 0:
div.append(i)
div.append(a)
else:
for i in range(2,a,3):
if a % i == 0:
div.append(i)
div.append(a)
if b <= len(div):
print(div[b-1])
else:
print(-1)
``` | 0 | |
3 | B | Lorry | PROGRAMMING | 1,900 | [
"greedy",
"sortings"
] | B. Lorry | 2 | 64 | A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body. | The first line contains a pair of integer numbers *n* and *v* (1<=≤<=*n*<=≤<=105; 1<=≤<=*v*<=≤<=109), where *n* is the number of waterborne vehicles in the boat depot, and *v* is the truck body volume of the lorry in cubic metres. The following *n* lines contain the information about the waterborne vehicles, that is a pair of numbers *t**i*,<=*p**i* (1<=≤<=*t**i*<=≤<=2; 1<=≤<=*p**i*<=≤<=104), where *t**i* is the vehicle type (1 – a kayak, 2 – a catamaran), and *p**i* is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file. | In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them. | [
"3 2\n1 2\n2 7\n1 3\n"
] | [
"7\n2\n"
] | none | 0 | [
{
"input": "3 2\n1 2\n2 7\n1 3",
"output": "7\n2"
},
{
"input": "5 3\n1 9\n2 9\n1 9\n2 10\n1 6",
"output": "24\n3 1 5"
},
{
"input": "10 10\n1 14\n2 15\n2 11\n2 12\n2 9\n1 14\n2 15\n1 9\n2 11\n2 6",
"output": "81\n6 1 7 2 4 9"
},
{
"input": "20 19\n2 47\n1 37\n1 48\n2 42\n2 4... | 1,566,461,067 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 17,100,800 | import functools
def cmp(item1, item2):
if item1.carry / item1.typ > item2.carry / item2.typ:
return -1
else:
return 1
class Vehicles:
def __init__(self, idx, typ, carry):
self.idx = idx
self.typ = typ
self.carry = carry
n, v = map(int, input().split())
a = []
for i in range(n):
typ, carry = map(int, input().split())
a.append(Vehicles(i + 1, typ, carry))
a.sort(key=functools.cmp_to_key(cmp))
res = [0]
path = []
idx = 0
for i in range(n):
if v >= a[i].typ:
v -= a[i].typ
res[0] += a[i].carry
path.append(a[i].idx)
idx = i
if v == 1 and idx < n:
flag = False
tmp = [0, 0]
for i in range(idx + 1, n):
if a[i].typ == 2:
tmp = [a[i].carry, str(a[i].idx)]
flag = True
break
if flag:
path = []
for i in range(idx, -1, -1):
if flag and a[i].typ == 1 and a[i].carry < tmp[0]:
flag = False
res[0] += tmp[0] - a[i].carry
path.append(tmp[1])
else:
path.append(a[i].idx)
print(res[0])
l = len(path)
for i in range(l):
if i == 0:
print(str(path[i]), end='')
else:
print(' ' + str(path[i]), end='') | Title: Lorry
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body.
Input Specification:
The first line contains a pair of integer numbers *n* and *v* (1<=≤<=*n*<=≤<=105; 1<=≤<=*v*<=≤<=109), where *n* is the number of waterborne vehicles in the boat depot, and *v* is the truck body volume of the lorry in cubic metres. The following *n* lines contain the information about the waterborne vehicles, that is a pair of numbers *t**i*,<=*p**i* (1<=≤<=*t**i*<=≤<=2; 1<=≤<=*p**i*<=≤<=104), where *t**i* is the vehicle type (1 – a kayak, 2 – a catamaran), and *p**i* is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file.
Output Specification:
In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them.
Demo Input:
['3 2\n1 2\n2 7\n1 3\n']
Demo Output:
['7\n2\n']
Note:
none | ```python
import functools
def cmp(item1, item2):
if item1.carry / item1.typ > item2.carry / item2.typ:
return -1
else:
return 1
class Vehicles:
def __init__(self, idx, typ, carry):
self.idx = idx
self.typ = typ
self.carry = carry
n, v = map(int, input().split())
a = []
for i in range(n):
typ, carry = map(int, input().split())
a.append(Vehicles(i + 1, typ, carry))
a.sort(key=functools.cmp_to_key(cmp))
res = [0]
path = []
idx = 0
for i in range(n):
if v >= a[i].typ:
v -= a[i].typ
res[0] += a[i].carry
path.append(a[i].idx)
idx = i
if v == 1 and idx < n:
flag = False
tmp = [0, 0]
for i in range(idx + 1, n):
if a[i].typ == 2:
tmp = [a[i].carry, str(a[i].idx)]
flag = True
break
if flag:
path = []
for i in range(idx, -1, -1):
if flag and a[i].typ == 1 and a[i].carry < tmp[0]:
flag = False
res[0] += tmp[0] - a[i].carry
path.append(tmp[1])
else:
path.append(a[i].idx)
print(res[0])
l = len(path)
for i in range(l):
if i == 0:
print(str(path[i]), end='')
else:
print(' ' + str(path[i]), end='')
``` | 0 |
832 | A | Sasha and Sticks | PROGRAMMING | 800 | [
"games",
"math"
] | null | null | It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him. | The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn. | If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower). | [
"1 1\n",
"10 4\n"
] | [
"YES\n",
"NO\n"
] | In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win. | 500 | [
{
"input": "1 1",
"output": "YES"
},
{
"input": "10 4",
"output": "NO"
},
{
"input": "251656215122324104 164397544865601257",
"output": "YES"
},
{
"input": "963577813436662285 206326039287271924",
"output": "NO"
},
{
"input": "1000000000000000000 1",
"output":... | 1,628,001,929 | 2,147,483,647 | PyPy 3 | OK | TESTS | 62 | 93 | 20,172,800 | n,k = map(int,input().split())
if n < k:
print('YES')
else:
diff = n//k
if diff % 2 == 0:
print('NO')
if diff % 2 != 0:
print('YES') | Title: Sasha and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
Output Specification:
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
Demo Input:
['1 1\n', '10 4\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win. | ```python
n,k = map(int,input().split())
if n < k:
print('YES')
else:
diff = n//k
if diff % 2 == 0:
print('NO')
if diff % 2 != 0:
print('YES')
``` | 3 | |
274 | A | k-Multiple Free Set | PROGRAMMING | 1,500 | [
"binary search",
"greedy",
"sortings"
] | null | null | A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset. | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces. | On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}. | [
"6 2\n2 3 6 5 4 10\n"
] | [
"3\n"
] | In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | 500 | [
{
"input": "6 2\n2 3 6 5 4 10",
"output": "3"
},
{
"input": "10 2\n1 2 3 4 5 6 7 8 9 10",
"output": "6"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "100 2\n191 17 61 40 77 95 128 88 26 69 79 10 131 106 142 152 68 39 182 53 83 81 6 89 65 148 33 22 5 47 107 121 52 163 1... | 1,618,941,400 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 8,192,000 | n,k=map(int,input().split())
l=list(map(int,input().split()))
u=[]
for i in range(len(l)):
f=0
for g in u:
if 2*g[-1]==l[i]:
g.append(l[i])
f=1
break
if f==0:
u.append([l[i]])
ct=0
for i in u:
if len(i)==1:
ct+=1
else:
if len(i)%2==0:
ct+=len(i)//2
else:
ct+=(len(i)//2)
ct+=1
print(ct)
| Title: k-Multiple Free Set
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces.
Output Specification:
On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}.
Demo Input:
['6 2\n2 3 6 5 4 10\n']
Demo Output:
['3\n']
Note:
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | ```python
n,k=map(int,input().split())
l=list(map(int,input().split()))
u=[]
for i in range(len(l)):
f=0
for g in u:
if 2*g[-1]==l[i]:
g.append(l[i])
f=1
break
if f==0:
u.append([l[i]])
ct=0
for i in u:
if len(i)==1:
ct+=1
else:
if len(i)%2==0:
ct+=len(i)//2
else:
ct+=(len(i)//2)
ct+=1
print(ct)
``` | 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",... | 1,658,406,807 | 2,147,483,647 | PyPy 3 | OK | TESTS | 52 | 93 | 0 | x, y = map(int, input().split())
print('Yes' if (x - y) % 2 == 1 and y - x <= 1 and y > 0 and (x == 0 or y > 1) else '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 = map(int, input().split())
print('Yes' if (x - y) % 2 == 1 and y - x <= 1 and y > 0 and (x == 0 or y > 1) else 'No')
``` | 3 | |
660 | C | Hard Process | PROGRAMMING | 1,600 | [
"binary search",
"dp",
"two pointers"
] | null | null | You are given an array *a* with *n* elements. Each element of *a* is either 0 or 1.
Let's denote the length of the longest subsegment of consecutive elements in *a*, consisting of only numbers one, as *f*(*a*). You can change no more than *k* zeroes to ones to maximize *f*(*a*). | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=3·105,<=0<=≤<=*k*<=≤<=*n*) — the number of elements in *a* and the parameter *k*.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1) — the elements of *a*. | On the first line print a non-negative integer *z* — the maximal value of *f*(*a*) after no more than *k* changes of zeroes to ones.
On the second line print *n* integers *a**j* — the elements of the array *a* after the changes.
If there are multiple answers, you can print any one of them. | [
"7 1\n1 0 0 1 1 0 1\n",
"10 2\n1 0 0 1 0 1 0 1 0 1\n"
] | [
"4\n1 0 0 1 1 1 1\n",
"5\n1 0 0 1 1 1 1 1 0 1\n"
] | none | 0 | [
{
"input": "7 1\n1 0 0 1 1 0 1",
"output": "4\n1 0 0 1 1 1 1"
},
{
"input": "10 2\n1 0 0 1 0 1 0 1 0 1",
"output": "5\n1 0 0 1 1 1 1 1 0 1"
},
{
"input": "1 0\n0",
"output": "0\n0"
},
{
"input": "1 0\n0",
"output": "0\n0"
},
{
"input": "7 0\n0 1 0 0 0 1 0",
"o... | 1,623,649,676 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 5 | 92 | 0 | n,k=map(int,input().split())
a=list(map(int,input().split()))
i=0
j=0
countzero=0
ss=0
ee=0
aa=0
while i<n-1:
while countzero<=k:
if a[i]==0:
countzero+=1
i+=1
while countzero>k:
if a[j]==0:
countzero-=1
j+=1
if (i-j+1)>aa:
ss=i
ee=j
aa=(i-j+1)
print(aa)
for k in range(j,i):
a[k]=1
print(*a)
| Title: Hard Process
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a* with *n* elements. Each element of *a* is either 0 or 1.
Let's denote the length of the longest subsegment of consecutive elements in *a*, consisting of only numbers one, as *f*(*a*). You can change no more than *k* zeroes to ones to maximize *f*(*a*).
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=3·105,<=0<=≤<=*k*<=≤<=*n*) — the number of elements in *a* and the parameter *k*.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1) — the elements of *a*.
Output Specification:
On the first line print a non-negative integer *z* — the maximal value of *f*(*a*) after no more than *k* changes of zeroes to ones.
On the second line print *n* integers *a**j* — the elements of the array *a* after the changes.
If there are multiple answers, you can print any one of them.
Demo Input:
['7 1\n1 0 0 1 1 0 1\n', '10 2\n1 0 0 1 0 1 0 1 0 1\n']
Demo Output:
['4\n1 0 0 1 1 1 1\n', '5\n1 0 0 1 1 1 1 1 0 1\n']
Note:
none | ```python
n,k=map(int,input().split())
a=list(map(int,input().split()))
i=0
j=0
countzero=0
ss=0
ee=0
aa=0
while i<n-1:
while countzero<=k:
if a[i]==0:
countzero+=1
i+=1
while countzero>k:
if a[j]==0:
countzero-=1
j+=1
if (i-j+1)>aa:
ss=i
ee=j
aa=(i-j+1)
print(aa)
for k in range(j,i):
a[k]=1
print(*a)
``` | 0 | |
840 | B | Leha and another game about graph | PROGRAMMING | 2,100 | [
"constructive algorithms",
"data structures",
"dfs and similar",
"dp",
"graphs"
] | null | null | Leha plays a computer game, where is on each level is given a connected graph with *n* vertices and *m* edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer *d**i*, which can be equal to 0, 1 or <=-<=1. To pass the level, he needs to find a «good» subset of edges of the graph or say, that it doesn't exist. Subset is called «good», if by by leaving only edges from this subset in the original graph, we obtain the following: for every vertex i, *d**i*<==<=<=-<=1 or it's degree modulo 2 is equal to *d**i*. Leha wants to pass the game as soon as possible and ask you to help him. In case of multiple correct answers, print any of them. | The first line contains two integers *n*, *m* (1<=≤<=*n*<=≤<=3·105, *n*<=-<=1<=≤<=*m*<=≤<=3·105) — number of vertices and edges.
The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (<=-<=1<=≤<=*d**i*<=≤<=1) — numbers on the vertices.
Each of the next *m* lines contains two integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*) — edges. It's guaranteed, that graph in the input is connected. | Print <=-<=1 in a single line, if solution doesn't exist. Otherwise in the first line *k* — number of edges in a subset. In the next *k* lines indexes of edges. Edges are numerated in order as they are given in the input, starting from 1. | [
"1 0\n1\n",
"4 5\n0 0 0 -1\n1 2\n2 3\n3 4\n1 4\n2 4\n",
"2 1\n1 1\n1 2\n",
"3 3\n0 -1 1\n1 2\n2 3\n1 3\n"
] | [
"-1\n",
"0\n",
"1\n1\n",
"1\n2\n"
] | In the first sample we have single vertex without edges. It's degree is 0 and we can not get 1. | 1,000 | [
{
"input": "1 0\n1",
"output": "-1"
},
{
"input": "4 5\n0 0 0 -1\n1 2\n2 3\n3 4\n1 4\n2 4",
"output": "0"
},
{
"input": "2 1\n1 1\n1 2",
"output": "1\n1"
},
{
"input": "3 3\n0 -1 1\n1 2\n2 3\n1 3",
"output": "1\n2"
},
{
"input": "10 10\n-1 -1 -1 -1 -1 -1 -1 -1 -1 ... | 1,689,249,608 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1689249607.7267241")# 1689249607.726764 | Title: Leha and another game about graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leha plays a computer game, where is on each level is given a connected graph with *n* vertices and *m* edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer *d**i*, which can be equal to 0, 1 or <=-<=1. To pass the level, he needs to find a «good» subset of edges of the graph or say, that it doesn't exist. Subset is called «good», if by by leaving only edges from this subset in the original graph, we obtain the following: for every vertex i, *d**i*<==<=<=-<=1 or it's degree modulo 2 is equal to *d**i*. Leha wants to pass the game as soon as possible and ask you to help him. In case of multiple correct answers, print any of them.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*<=≤<=3·105, *n*<=-<=1<=≤<=*m*<=≤<=3·105) — number of vertices and edges.
The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (<=-<=1<=≤<=*d**i*<=≤<=1) — numbers on the vertices.
Each of the next *m* lines contains two integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*) — edges. It's guaranteed, that graph in the input is connected.
Output Specification:
Print <=-<=1 in a single line, if solution doesn't exist. Otherwise in the first line *k* — number of edges in a subset. In the next *k* lines indexes of edges. Edges are numerated in order as they are given in the input, starting from 1.
Demo Input:
['1 0\n1\n', '4 5\n0 0 0 -1\n1 2\n2 3\n3 4\n1 4\n2 4\n', '2 1\n1 1\n1 2\n', '3 3\n0 -1 1\n1 2\n2 3\n1 3\n']
Demo Output:
['-1\n', '0\n', '1\n1\n', '1\n2\n']
Note:
In the first sample we have single vertex without edges. It's degree is 0 and we can not get 1. | ```python
print("_RANDOM_GUESS_1689249607.7267241")# 1689249607.726764
``` | 0 | |
667 | B | Coat of Anticubism | PROGRAMMING | 1,100 | [
"constructive algorithms",
"geometry"
] | null | null | As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.
A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-dimensional objects through three-dimensional objects by using his magnificent sculptures. And his new project is connected with this. Cicasso wants to make a coat for the haters of anticubism. To do this, he wants to create a sculpture depicting a well-known geometric primitive — convex polygon.
Cicasso prepared for this a few blanks, which are rods with integer lengths, and now he wants to bring them together. The *i*-th rod is a segment of length *l**i*.
The sculptor plans to make a convex polygon with a nonzero area, using all rods he has as its sides. Each rod should be used as a side to its full length. It is forbidden to cut, break or bend rods. However, two sides may form a straight angle .
Cicasso knows that it is impossible to make a convex polygon with a nonzero area out of the rods with the lengths which he had chosen. Cicasso does not want to leave the unused rods, so the sculptor decides to make another rod-blank with an integer length so that his problem is solvable. Of course, he wants to make it as short as possible, because the materials are expensive, and it is improper deed to spend money for nothing.
Help sculptor! | The first line contains an integer *n* (3<=≤<=*n*<=≤<=105) — a number of rod-blanks.
The second line contains *n* integers *l**i* (1<=≤<=*l**i*<=≤<=109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with *n* vertices and nonzero area using the rods Cicasso already has. | Print the only integer *z* — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with (*n*<=+<=1) vertices and nonzero area from all of the rods. | [
"3\n1 2 1\n",
"5\n20 4 3 2 1\n"
] | [
"1\n",
"11\n"
] | In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.
In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}. | 1,000 | [
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "5\n20 4 3 2 1",
"output": "11"
},
{
"input": "7\n77486105 317474713 89523018 332007362 7897847 949616701 54820086",
"output": "70407571"
},
{
"input": "14\n245638694 2941428 4673577 12468 991349408 44735727 14046308 60637707... | 1,677,322,571 | 2,147,483,647 | Python 3 | OK | TESTS | 51 | 77 | 3,174,400 | def main():
param = 0
sum = 0
geom = []
maximum = 0
index = 0
n = int(input())
geom = list(map(int, input().split()))
for i in range(0, n):
if geom[i] > maximum:
maximum = geom[i]
index = i
geom = geom[:index] + geom[index+1:]
for i in range(0, n - 1):
sum += geom[i]
param = maximum - sum + 1
print(param)
if __name__ == '__main__':
main()
| Title: Coat of Anticubism
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.
A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-dimensional objects through three-dimensional objects by using his magnificent sculptures. And his new project is connected with this. Cicasso wants to make a coat for the haters of anticubism. To do this, he wants to create a sculpture depicting a well-known geometric primitive — convex polygon.
Cicasso prepared for this a few blanks, which are rods with integer lengths, and now he wants to bring them together. The *i*-th rod is a segment of length *l**i*.
The sculptor plans to make a convex polygon with a nonzero area, using all rods he has as its sides. Each rod should be used as a side to its full length. It is forbidden to cut, break or bend rods. However, two sides may form a straight angle .
Cicasso knows that it is impossible to make a convex polygon with a nonzero area out of the rods with the lengths which he had chosen. Cicasso does not want to leave the unused rods, so the sculptor decides to make another rod-blank with an integer length so that his problem is solvable. Of course, he wants to make it as short as possible, because the materials are expensive, and it is improper deed to spend money for nothing.
Help sculptor!
Input Specification:
The first line contains an integer *n* (3<=≤<=*n*<=≤<=105) — a number of rod-blanks.
The second line contains *n* integers *l**i* (1<=≤<=*l**i*<=≤<=109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with *n* vertices and nonzero area using the rods Cicasso already has.
Output Specification:
Print the only integer *z* — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with (*n*<=+<=1) vertices and nonzero area from all of the rods.
Demo Input:
['3\n1 2 1\n', '5\n20 4 3 2 1\n']
Demo Output:
['1\n', '11\n']
Note:
In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.
In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}. | ```python
def main():
param = 0
sum = 0
geom = []
maximum = 0
index = 0
n = int(input())
geom = list(map(int, input().split()))
for i in range(0, n):
if geom[i] > maximum:
maximum = geom[i]
index = i
geom = geom[:index] + geom[index+1:]
for i in range(0, n - 1):
sum += geom[i]
param = maximum - sum + 1
print(param)
if __name__ == '__main__':
main()
``` | 3 | |
380 | C | Sereja and Brackets | PROGRAMMING | 2,000 | [
"data structures",
"schedules"
] | null | null | Sereja has a bracket sequence *s*1,<=*s*2,<=...,<=*s**n*, or, in other words, a string *s* of length *n*, consisting of characters "(" and ")".
Sereja needs to answer *m* queries, each of them is described by two integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). The answer to the *i*-th query is the length of the maximum correct bracket subsequence of sequence *s**l**i*,<=*s**l**i*<=+<=1,<=...,<=*s**r**i*. Help Sereja answer all queries.
You can find the definitions for a subsequence and a correct bracket sequence in the notes. | The first line contains a sequence of characters *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*n*<=≤<=106) without any spaces. Each character is either a "(" or a ")". The second line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains a pair of integers. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) — the description of the *i*-th query. | Print the answer to each question on a single line. Print the answers in the order they go in the input. | [
"())(())(())(\n7\n1 1\n2 3\n1 2\n1 12\n8 12\n5 11\n2 10\n"
] | [
"0\n0\n2\n10\n4\n6\n6\n"
] | A subsequence of length |*x*| of string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (where |*s*| is the length of string *s*) is string *x* = *s*<sub class="lower-index">*k*<sub class="lower-index">1</sub></sub>*s*<sub class="lower-index">*k*<sub class="lower-index">2</sub></sub>... *s*<sub class="lower-index">*k*<sub class="lower-index">|*x*|</sub></sub> (1 ≤ *k*<sub class="lower-index">1</sub> < *k*<sub class="lower-index">2</sub> < ... < *k*<sub class="lower-index">|*x*|</sub> ≤ |*s*|).
A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1" and "+" between the characters of the string. For example, bracket sequences "()()", "(())" are correct (the resulting expressions "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
For the third query required sequence will be «()».
For the fourth query required sequence will be «()(())(())». | 1,500 | [
{
"input": "())(())(())(\n7\n1 1\n2 3\n1 2\n1 12\n8 12\n5 11\n2 10",
"output": "0\n0\n2\n10\n4\n6\n6"
},
{
"input": "(((((()((((((((((()((()(((((\n1\n8 15",
"output": "0"
},
{
"input": "((()((())(((((((((()(()(()(((((((((((((((()(()((((((((((((((()(((((((((((((((((((()(((\n39\n28 56\n39 ... | 1,683,753,323 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 61 | 0 | import sys, math
def merge(left, right):
ans = (
left[0] + right[0] - min(left[0], right[1]),
left[1] + right[1] - min(left[0], right[1]),
left[2] + right[2] + min(left[0], right[1])
)
return ans
def build_iter(tree):
n = len(arr)
# print(n, len(tree))
for i in range(n):
tree[n+i] = (
int(arr[i] == "("),
int(arr[i] == ")"),
0
)
for i in range(n - 1, 0, -1):
tree[i] = merge(tree[2*i], tree[2*i+1])
def query_iter(l, r):
ans = None
n = len(arr)
l += n
r += n
while l <= r:
# print(l, r)
if l % 2 == 1:
# print("\t",'before', l, seg_tree[l], ans)
ans = seg_tree[l] if not ans else merge(ans, seg_tree[l])
# print("\t", "left-aft",ans)
if r % 2 == 0:
# print("\t",'rgt-bef', r, seg_tree[r], ans)
ans = seg_tree[r] if not ans else merge(ans, seg_tree[r])
# print("\t", "rgt-aft",ans)
l = (l+1)//2
r = (r-1)//2
# print(ans)
return ans
def nearestPowerOf2(N):
# Calculate log2 of N
a = int(math.log2(N))
# If 2^a is equal to N, return N
if 2**a == N:
return N
# Return 2^(a + 1)
return 2**(a + 1)
st = sys.stdin.readline()
arr = [*st][:-1]
n = int(sys.stdin.readline())
# print(n,arr)
ln = len(arr)
pow_2 = nearestPowerOf2(ln)
if ln < pow_2:
temp = [None] * (pow_2 - ln)
arr.extend(temp)
seg_tree = [(0, 0, 0)] * (2 * pow_2)
build_iter(seg_tree)
# for i, val in enumerate(seg_tree):
# print(i,val, end="\n")
# print()
for _ in range(n):
l,r = map(int, sys.stdin.readline().split())
# print(l,r)
# print(seg_tree)
ans = query_iter(l - 1, r - 1)
print(ans[2] * 2) | Title: Sereja and Brackets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja has a bracket sequence *s*1,<=*s*2,<=...,<=*s**n*, or, in other words, a string *s* of length *n*, consisting of characters "(" and ")".
Sereja needs to answer *m* queries, each of them is described by two integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). The answer to the *i*-th query is the length of the maximum correct bracket subsequence of sequence *s**l**i*,<=*s**l**i*<=+<=1,<=...,<=*s**r**i*. Help Sereja answer all queries.
You can find the definitions for a subsequence and a correct bracket sequence in the notes.
Input Specification:
The first line contains a sequence of characters *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*n*<=≤<=106) without any spaces. Each character is either a "(" or a ")". The second line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains a pair of integers. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) — the description of the *i*-th query.
Output Specification:
Print the answer to each question on a single line. Print the answers in the order they go in the input.
Demo Input:
['())(())(())(\n7\n1 1\n2 3\n1 2\n1 12\n8 12\n5 11\n2 10\n']
Demo Output:
['0\n0\n2\n10\n4\n6\n6\n']
Note:
A subsequence of length |*x*| of string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (where |*s*| is the length of string *s*) is string *x* = *s*<sub class="lower-index">*k*<sub class="lower-index">1</sub></sub>*s*<sub class="lower-index">*k*<sub class="lower-index">2</sub></sub>... *s*<sub class="lower-index">*k*<sub class="lower-index">|*x*|</sub></sub> (1 ≤ *k*<sub class="lower-index">1</sub> < *k*<sub class="lower-index">2</sub> < ... < *k*<sub class="lower-index">|*x*|</sub> ≤ |*s*|).
A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1" and "+" between the characters of the string. For example, bracket sequences "()()", "(())" are correct (the resulting expressions "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
For the third query required sequence will be «()».
For the fourth query required sequence will be «()(())(())». | ```python
import sys, math
def merge(left, right):
ans = (
left[0] + right[0] - min(left[0], right[1]),
left[1] + right[1] - min(left[0], right[1]),
left[2] + right[2] + min(left[0], right[1])
)
return ans
def build_iter(tree):
n = len(arr)
# print(n, len(tree))
for i in range(n):
tree[n+i] = (
int(arr[i] == "("),
int(arr[i] == ")"),
0
)
for i in range(n - 1, 0, -1):
tree[i] = merge(tree[2*i], tree[2*i+1])
def query_iter(l, r):
ans = None
n = len(arr)
l += n
r += n
while l <= r:
# print(l, r)
if l % 2 == 1:
# print("\t",'before', l, seg_tree[l], ans)
ans = seg_tree[l] if not ans else merge(ans, seg_tree[l])
# print("\t", "left-aft",ans)
if r % 2 == 0:
# print("\t",'rgt-bef', r, seg_tree[r], ans)
ans = seg_tree[r] if not ans else merge(ans, seg_tree[r])
# print("\t", "rgt-aft",ans)
l = (l+1)//2
r = (r-1)//2
# print(ans)
return ans
def nearestPowerOf2(N):
# Calculate log2 of N
a = int(math.log2(N))
# If 2^a is equal to N, return N
if 2**a == N:
return N
# Return 2^(a + 1)
return 2**(a + 1)
st = sys.stdin.readline()
arr = [*st][:-1]
n = int(sys.stdin.readline())
# print(n,arr)
ln = len(arr)
pow_2 = nearestPowerOf2(ln)
if ln < pow_2:
temp = [None] * (pow_2 - ln)
arr.extend(temp)
seg_tree = [(0, 0, 0)] * (2 * pow_2)
build_iter(seg_tree)
# for i, val in enumerate(seg_tree):
# print(i,val, end="\n")
# print()
for _ in range(n):
l,r = map(int, sys.stdin.readline().split())
# print(l,r)
# print(seg_tree)
ans = query_iter(l - 1, r - 1)
print(ans[2] * 2)
``` | 0 | |
259 | B | Little Elephant and Magic Square | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Little Elephant loves magic squares very much.
A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes. | The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105. | Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions. | [
"0 1 1\n1 0 1\n1 1 0\n",
"0 3 6\n5 0 5\n4 7 0\n"
] | [
"1 1 1\n1 1 1\n1 1 1\n",
"6 3 6\n5 5 5\n4 7 4\n"
] | none | 1,000 | [
{
"input": "0 1 1\n1 0 1\n1 1 0",
"output": "1 1 1\n1 1 1\n1 1 1"
},
{
"input": "0 3 6\n5 0 5\n4 7 0",
"output": "6 3 6\n5 5 5\n4 7 4"
},
{
"input": "0 4 4\n4 0 4\n4 4 0",
"output": "4 4 4\n4 4 4\n4 4 4"
},
{
"input": "0 54 48\n36 0 78\n66 60 0",
"output": "69 54 48\n36 5... | 1,686,481,476 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 24 | 124 | 1,843,200 | r1=list(map(int,input().split()))
r2=list(map(int,input().split()))
r3=list(map(int,input().split()))
s1=sum(r1)
s2=sum(r2)
s3=sum(r3)
for i in range(1,10**5+1):
r1[0]=i
s1+=i
r2[1]=s1-s2
s2+=r2[1]
r3[2]=s1-s3
s3+=r3[2]
if r2[1]>0 and r3[2]>0 and s1==s2==s3==(r1[0]+r2[0]+r3[0])==(r1[1]+r2[1]+r3[1])==(r1[2]+r2[2]+r3[2])==(r1[0]+r2[1]+r3[2])==(r1[2]+r2[1]+r3[0]):
print(*r1)
print(*r2)
print(*r3)
break
r1[0]=0
s1-=i
s2-=r2[1]
r2[1]=0
s3-=r3[2]
r3[2]=0 | Title: Little Elephant and Magic Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Elephant loves magic squares very much.
A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes.
Input Specification:
The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.
Output Specification:
Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions.
Demo Input:
['0 1 1\n1 0 1\n1 1 0\n', '0 3 6\n5 0 5\n4 7 0\n']
Demo Output:
['1 1 1\n1 1 1\n1 1 1\n', '6 3 6\n5 5 5\n4 7 4\n']
Note:
none | ```python
r1=list(map(int,input().split()))
r2=list(map(int,input().split()))
r3=list(map(int,input().split()))
s1=sum(r1)
s2=sum(r2)
s3=sum(r3)
for i in range(1,10**5+1):
r1[0]=i
s1+=i
r2[1]=s1-s2
s2+=r2[1]
r3[2]=s1-s3
s3+=r3[2]
if r2[1]>0 and r3[2]>0 and s1==s2==s3==(r1[0]+r2[0]+r3[0])==(r1[1]+r2[1]+r3[1])==(r1[2]+r2[2]+r3[2])==(r1[0]+r2[1]+r3[2])==(r1[2]+r2[1]+r3[0]):
print(*r1)
print(*r2)
print(*r3)
break
r1[0]=0
s1-=i
s2-=r2[1]
r2[1]=0
s3-=r3[2]
r3[2]=0
``` | 3 | |
888 | D | Almost Identity Permutations | PROGRAMMING | 1,600 | [
"combinatorics",
"dp",
"math"
] | null | null | A permutation *p* of size *n* is an array such that every integer from 1 to *n* occurs exactly once in this array.
Let's call a permutation an almost identity permutation iff there exist at least *n*<=-<=*k* indices *i* (1<=≤<=*i*<=≤<=*n*) such that *p**i*<==<=*i*.
Your task is to count the number of almost identity permutations for given numbers *n* and *k*. | The first line contains two integers *n* and *k* (4<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=4). | Print the number of almost identity permutations for given *n* and *k*. | [
"4 1\n",
"4 2\n",
"5 3\n",
"5 4\n"
] | [
"1\n",
"7\n",
"31\n",
"76\n"
] | none | 0 | [
{
"input": "4 1",
"output": "1"
},
{
"input": "4 2",
"output": "7"
},
{
"input": "5 3",
"output": "31"
},
{
"input": "5 4",
"output": "76"
},
{
"input": "200 1",
"output": "1"
},
{
"input": "200 2",
"output": "19901"
},
{
"input": "200 3",
... | 1,660,810,027 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | from math import factorial as f
n, k = map(int, input().split())
d = {1: 0, 2: 1, 3: 2, 4: 11}
ans = 1
for i in range(2, k + 1):
ans += f(n) // f(n - i) // f(i) * d[i]
print(ans)
| Title: Almost Identity Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation *p* of size *n* is an array such that every integer from 1 to *n* occurs exactly once in this array.
Let's call a permutation an almost identity permutation iff there exist at least *n*<=-<=*k* indices *i* (1<=≤<=*i*<=≤<=*n*) such that *p**i*<==<=*i*.
Your task is to count the number of almost identity permutations for given numbers *n* and *k*.
Input Specification:
The first line contains two integers *n* and *k* (4<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=4).
Output Specification:
Print the number of almost identity permutations for given *n* and *k*.
Demo Input:
['4 1\n', '4 2\n', '5 3\n', '5 4\n']
Demo Output:
['1\n', '7\n', '31\n', '76\n']
Note:
none | ```python
from math import factorial as f
n, k = map(int, input().split())
d = {1: 0, 2: 1, 3: 2, 4: 11}
ans = 1
for i in range(2, k + 1):
ans += f(n) // f(n - i) // f(i) * d[i]
print(ans)
``` | 0 | |
404 | A | Valera and X | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him. | The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper. | Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes. | [
"5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n",
"3\nwsw\nsws\nwsw\n",
"3\nxpx\npxp\nxpe\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nwsw\nsws\nwsw",
"output": "YES"
},
{
"input": "3\nxpx\npxp\nxpe",
"output": "NO"
},
{
"input": "5\nliiil\nilili\niilii\nilili\nliiil",
"output": "YES"
},
{
"input": "7\nbwccccb\nck... | 1,642,804,611 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | a=int(input())
f=True
for i in range(a):
f=True if (f and ((b:=input())[i]==b[-i-1] and b.count(b[i])==2) or (i==a//2 and b.count(b[i])==1)) else False
print("YES" if f else "NO") | Title: Valera and X
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper.
Output Specification:
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
Demo Input:
['5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n', '3\nwsw\nsws\nwsw\n', '3\nxpx\npxp\nxpe\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
a=int(input())
f=True
for i in range(a):
f=True if (f and ((b:=input())[i]==b[-i-1] and b.count(b[i])==2) or (i==a//2 and b.count(b[i])==1)) else False
print("YES" if f else "NO")
``` | 0 | |
884 | D | Boxes And Balls | PROGRAMMING | 2,300 | [
"data structures",
"greedy"
] | null | null | Ivan has *n* different boxes. The first of them contains some balls of *n* different colors.
Ivan wants to play a strange game. He wants to distribute the balls into boxes in such a way that for every *i* (1<=≤<=*i*<=≤<=*n*) *i*-th box will contain all balls with color *i*.
In order to do this, Ivan will make some turns. Each turn he does the following:
1. Ivan chooses any non-empty box and takes all balls from this box; 1. Then Ivan chooses any *k* empty boxes (the box from the first step becomes empty, and Ivan is allowed to choose it), separates the balls he took on the previous step into *k* non-empty groups and puts each group into one of the boxes. He should put each group into a separate box. He can choose either *k*<==<=2 or *k*<==<=3.
The penalty of the turn is the number of balls Ivan takes from the box during the first step of the turn. And penalty of the game is the total penalty of turns made by Ivan until he distributes all balls to corresponding boxes.
Help Ivan to determine the minimum possible penalty of the game! | The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200000) — the number of boxes and colors.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the number of balls with color *i*. | Print one number — the minimum possible penalty of the game. | [
"3\n1 2 3\n",
"4\n2 3 4 5\n"
] | [
"6\n",
"19\n"
] | In the first example you take all the balls from the first box, choose *k* = 3 and sort all colors to corresponding boxes. Penalty is 6.
In the second example you make two turns:
1. Take all the balls from the first box, choose *k* = 3, put balls of color 3 to the third box, of color 4 — to the fourth box and the rest put back into the first box. Penalty is 14; 1. Take all the balls from the first box, choose *k* = 2, put balls of color 1 to the first box, of color 2 — to the second box. Penalty is 5.
Total penalty is 19. | 0 | [
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "4\n2 3 4 5",
"output": "19"
},
{
"input": "6\n1 4 4 4 4 4",
"output": "38"
},
{
"input": "8\n821407370 380061316 428719552 90851747 825473738 704702117 845629927 245820158",
"output": "8176373828"
},
{
"input": "... | 1,509,644,165 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 15,257,600 | class HeapMin:
def __init__(self):
self.heap = []
self.length = 0
def insert(self,valor):
self.heap.append(valor)
self.up(self.length)
self.length+=1
def remove(self):
if self.length > 0:
ret = self.heap[0]
self.length-=1
self.heap[0] = self.heap[self.length]
self.down(0)
del(self.heap[self.length])
return ret
def up(self,i):
j = (i-1)//2
if(j >= 0):
if(self.heap[i]<self.heap[j]):
aux = self.heap[i]
self.heap[i] = self.heap[j]
self.heap[j] = aux
self.up(j)
def down(self,i):
j = (i+1)*2 - 1
if(j<=self.length):
if(j<self.length):
if(self.heap[j+1] < self.heap[j]):
j+=1
if(self.heap[i] > self.heap[j]):
aux = self.heap[i]
self.heap[i] = self.heap[j]
self.heap[j] = aux
self.down(j)
n = int(input())
boxes = HeapMin()
resp=0
for i in input().split():
boxes.insert(int(i))
if(n%2==0):
boxes.insert(0)
n+=1
while n!=1:
c = boxes.remove() + boxes.remove() + boxes.remove()
resp+=c
boxes.insert(c)
n-=2
print(resp)
| Title: Boxes And Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan has *n* different boxes. The first of them contains some balls of *n* different colors.
Ivan wants to play a strange game. He wants to distribute the balls into boxes in such a way that for every *i* (1<=≤<=*i*<=≤<=*n*) *i*-th box will contain all balls with color *i*.
In order to do this, Ivan will make some turns. Each turn he does the following:
1. Ivan chooses any non-empty box and takes all balls from this box; 1. Then Ivan chooses any *k* empty boxes (the box from the first step becomes empty, and Ivan is allowed to choose it), separates the balls he took on the previous step into *k* non-empty groups and puts each group into one of the boxes. He should put each group into a separate box. He can choose either *k*<==<=2 or *k*<==<=3.
The penalty of the turn is the number of balls Ivan takes from the box during the first step of the turn. And penalty of the game is the total penalty of turns made by Ivan until he distributes all balls to corresponding boxes.
Help Ivan to determine the minimum possible penalty of the game!
Input Specification:
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200000) — the number of boxes and colors.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the number of balls with color *i*.
Output Specification:
Print one number — the minimum possible penalty of the game.
Demo Input:
['3\n1 2 3\n', '4\n2 3 4 5\n']
Demo Output:
['6\n', '19\n']
Note:
In the first example you take all the balls from the first box, choose *k* = 3 and sort all colors to corresponding boxes. Penalty is 6.
In the second example you make two turns:
1. Take all the balls from the first box, choose *k* = 3, put balls of color 3 to the third box, of color 4 — to the fourth box and the rest put back into the first box. Penalty is 14; 1. Take all the balls from the first box, choose *k* = 2, put balls of color 1 to the first box, of color 2 — to the second box. Penalty is 5.
Total penalty is 19. | ```python
class HeapMin:
def __init__(self):
self.heap = []
self.length = 0
def insert(self,valor):
self.heap.append(valor)
self.up(self.length)
self.length+=1
def remove(self):
if self.length > 0:
ret = self.heap[0]
self.length-=1
self.heap[0] = self.heap[self.length]
self.down(0)
del(self.heap[self.length])
return ret
def up(self,i):
j = (i-1)//2
if(j >= 0):
if(self.heap[i]<self.heap[j]):
aux = self.heap[i]
self.heap[i] = self.heap[j]
self.heap[j] = aux
self.up(j)
def down(self,i):
j = (i+1)*2 - 1
if(j<=self.length):
if(j<self.length):
if(self.heap[j+1] < self.heap[j]):
j+=1
if(self.heap[i] > self.heap[j]):
aux = self.heap[i]
self.heap[i] = self.heap[j]
self.heap[j] = aux
self.down(j)
n = int(input())
boxes = HeapMin()
resp=0
for i in input().split():
boxes.insert(int(i))
if(n%2==0):
boxes.insert(0)
n+=1
while n!=1:
c = boxes.remove() + boxes.remove() + boxes.remove()
resp+=c
boxes.insert(c)
n-=2
print(resp)
``` | 0 | |
393 | A | Nineteen | PROGRAMMING | 0 | [] | null | null | Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string. | The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100. | Print a single integer — the maximum number of "nineteen"s that she can get in her string. | [
"nniinneetteeeenn\n",
"nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n",
"nineteenineteen\n"
] | [
"2",
"2",
"2"
] | none | 500 | [
{
"input": "nniinneetteeeenn",
"output": "2"
},
{
"input": "nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii",
"output": "2"
},
{
"input": "nineteenineteen",
"output": "2"
},
{
"input": "nssemsnnsitjtihtthij",
"output": "0"
},
{
"input": "eehihnttehtherjsihihn... | 1,554,135,137 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 109 | 0 | n=input().split()
x=n[0].count("n")%3
y=n[0].count("n")/3
if(x==0 and n[0].count("i")==y and n[0].count("e")==3*y and n[0].count("t")==y):
print(y)
else:
print("NOt")
| Title: Nineteen
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string.
Input Specification:
The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100.
Output Specification:
Print a single integer — the maximum number of "nineteen"s that she can get in her string.
Demo Input:
['nniinneetteeeenn\n', 'nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n', 'nineteenineteen\n']
Demo Output:
['2', '2', '2']
Note:
none | ```python
n=input().split()
x=n[0].count("n")%3
y=n[0].count("n")/3
if(x==0 and n[0].count("i")==y and n[0].count("e")==3*y and n[0].count("t")==y):
print(y)
else:
print("NOt")
``` | 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": "... | 1,686,667,647 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 1,000 | 102,400 | q=int(input())
a=q
s=''
while a>0:
s=s+str(a%10)
a=a//10
s=str(q)+s
print(s) | 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
q=int(input())
a=q
s=''
while a>0:
s=s+str(a%10)
a=a//10
s=str(q)+s
print(s)
``` | 0 | |
755 | E | PolandBall and White-Red graph | PROGRAMMING | 2,400 | [
"constructive algorithms",
"graphs",
"shortest paths"
] | null | null | PolandBall has an undirected simple graph consisting of *n* vertices. Unfortunately, it has no edges. The graph is very sad because of that. PolandBall wanted to make it happier, adding some red edges. Then, he will add white edges in every remaining place. Therefore, the final graph will be a clique in two colors: white and red.
Colorfulness of the graph is a value *min*(*d**r*,<=*d**w*), where *d**r* is the diameter of the red subgraph and *d**w* is the diameter of white subgraph. The diameter of a graph is a largest value *d* such that shortest path between some pair of vertices in it is equal to *d*. If the graph is not connected, we consider its diameter to be -1.
PolandBall wants the final graph to be as neat as possible. He wants the final colorfulness to be equal to *k*. Can you help him and find any graph which satisfies PolandBall's requests? | The only one input line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=1000), representing graph's size and sought colorfulness. | If it's impossible to find a suitable graph, print -1.
Otherwise, you can output any graph which fulfills PolandBall's requirements. First, output *m* — the number of red edges in your graph. Then, you should output *m* lines, each containing two integers *a**i* and *b**i*, (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) which means that there is an undirected red edge between vertices *a**i* and *b**i*. Every red edge should be printed exactly once, you can print the edges and the vertices of every edge in arbitrary order.
Remember that PolandBall's graph should remain simple, so no loops or multiple edges are allowed. | [
"4 1\n",
"5 2\n"
] | [
"-1\n",
"4\n1 2\n2 3\n3 4\n4 5\n"
] | In the first sample case, no graph can fulfill PolandBall's requirements.
In the second sample case, red graph is a path from 1 to 5. Its diameter is 4. However, white graph has diameter 2, because it consists of edges 1-3, 1-4, 1-5, 2-4, 2-5, 3-5. | 2,500 | [
{
"input": "4 1",
"output": "-1"
},
{
"input": "5 2",
"output": "4\n1 2\n2 3\n3 4\n4 5"
},
{
"input": "500 3",
"output": "123755\n1 2\n499 500\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 2... | 1,605,199,234 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 187 | 3,072,000 | l=input().split()
n=int(l[0])
k=int(l[1])
if(n<=3 or (k!=2 and k!=3)):
print(-1)
quit()
if(k==2):
print(n-1)
for i in range(n-1):
print(i+1,i+2)
else:
print(2*n-5)
print(1,2)
for i in range(n):
if(i==0 or i==2 or i==1):
continue
print(1,i+1)
for i in range(n):
if(i==0 or i==1 or i==3):
continue
print(2,i+1)
| Title: PolandBall and White-Red graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PolandBall has an undirected simple graph consisting of *n* vertices. Unfortunately, it has no edges. The graph is very sad because of that. PolandBall wanted to make it happier, adding some red edges. Then, he will add white edges in every remaining place. Therefore, the final graph will be a clique in two colors: white and red.
Colorfulness of the graph is a value *min*(*d**r*,<=*d**w*), where *d**r* is the diameter of the red subgraph and *d**w* is the diameter of white subgraph. The diameter of a graph is a largest value *d* such that shortest path between some pair of vertices in it is equal to *d*. If the graph is not connected, we consider its diameter to be -1.
PolandBall wants the final graph to be as neat as possible. He wants the final colorfulness to be equal to *k*. Can you help him and find any graph which satisfies PolandBall's requests?
Input Specification:
The only one input line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=1000), representing graph's size and sought colorfulness.
Output Specification:
If it's impossible to find a suitable graph, print -1.
Otherwise, you can output any graph which fulfills PolandBall's requirements. First, output *m* — the number of red edges in your graph. Then, you should output *m* lines, each containing two integers *a**i* and *b**i*, (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) which means that there is an undirected red edge between vertices *a**i* and *b**i*. Every red edge should be printed exactly once, you can print the edges and the vertices of every edge in arbitrary order.
Remember that PolandBall's graph should remain simple, so no loops or multiple edges are allowed.
Demo Input:
['4 1\n', '5 2\n']
Demo Output:
['-1\n', '4\n1 2\n2 3\n3 4\n4 5\n']
Note:
In the first sample case, no graph can fulfill PolandBall's requirements.
In the second sample case, red graph is a path from 1 to 5. Its diameter is 4. However, white graph has diameter 2, because it consists of edges 1-3, 1-4, 1-5, 2-4, 2-5, 3-5. | ```python
l=input().split()
n=int(l[0])
k=int(l[1])
if(n<=3 or (k!=2 and k!=3)):
print(-1)
quit()
if(k==2):
print(n-1)
for i in range(n-1):
print(i+1,i+2)
else:
print(2*n-5)
print(1,2)
for i in range(n):
if(i==0 or i==2 or i==1):
continue
print(1,i+1)
for i in range(n):
if(i==0 or i==1 or i==3):
continue
print(2,i+1)
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | A flower shop has got *n* bouquets, and the *i*-th bouquet consists of *a**i* flowers. Vasya, the manager of the shop, decided to make large bouquets from these bouquets.
Vasya thinks that a bouquet is large if it is made of two or more initial bouquets, and there is a constraint: the total number of flowers in a large bouquet should be odd. Each of the initial bouquets can be a part of at most one large bouquet. If an initial bouquet becomes a part of a large bouquet, all its flowers are included in the large bouquet.
Determine the maximum possible number of large bouquets Vasya can make. | The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of initial bouquets.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the number of flowers in each of the initial bouquets. | Print the maximum number of large bouquets Vasya can make. | [
"5\n2 3 4 2 7\n",
"6\n2 2 6 8 6 12\n",
"3\n11 4 10\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first example Vasya can make 2 large bouquets. For example, the first bouquet can contain the first and the fifth initial bouquets (the total number of flowers is then equal to 9), and the second bouquet can consist of the second and the third initial bouquets (the total number of flowers is then equal to 7). The fourth initial bouquet is unused in this scheme.
In the second example it is not possible to form a single bouquet with odd number of flowers.
In the third example Vasya can make one large bouquet. For example, he can make it using all three initial bouquets. The size of the large bouquet is then equal to 11 + 4 + 10 = 25. | 0 | [
{
"input": "5\n2 3 4 2 7",
"output": "2"
},
{
"input": "6\n2 2 6 8 6 12",
"output": "0"
},
{
"input": "3\n11 4 10",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n999999",
"output": "0"
},
... | 1,521,366,706 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 155 | 14,438,400 | n = int(input())
numbers = list(map(int, input().split()))
odd, even = 0, 0
for number in numbers:
if number%2==0:
even += 1
else:
odd += 1
ans = min(odd, even)
odd -= ans
even -= ans
ans += odd // 3
print(ans) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A flower shop has got *n* bouquets, and the *i*-th bouquet consists of *a**i* flowers. Vasya, the manager of the shop, decided to make large bouquets from these bouquets.
Vasya thinks that a bouquet is large if it is made of two or more initial bouquets, and there is a constraint: the total number of flowers in a large bouquet should be odd. Each of the initial bouquets can be a part of at most one large bouquet. If an initial bouquet becomes a part of a large bouquet, all its flowers are included in the large bouquet.
Determine the maximum possible number of large bouquets Vasya can make.
Input Specification:
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of initial bouquets.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the number of flowers in each of the initial bouquets.
Output Specification:
Print the maximum number of large bouquets Vasya can make.
Demo Input:
['5\n2 3 4 2 7\n', '6\n2 2 6 8 6 12\n', '3\n11 4 10\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first example Vasya can make 2 large bouquets. For example, the first bouquet can contain the first and the fifth initial bouquets (the total number of flowers is then equal to 9), and the second bouquet can consist of the second and the third initial bouquets (the total number of flowers is then equal to 7). The fourth initial bouquet is unused in this scheme.
In the second example it is not possible to form a single bouquet with odd number of flowers.
In the third example Vasya can make one large bouquet. For example, he can make it using all three initial bouquets. The size of the large bouquet is then equal to 11 + 4 + 10 = 25. | ```python
n = int(input())
numbers = list(map(int, input().split()))
odd, even = 0, 0
for number in numbers:
if number%2==0:
even += 1
else:
odd += 1
ans = min(odd, even)
odd -= ans
even -= ans
ans += odd // 3
print(ans)
``` | 3 | |
296 | A | Yaroslav and Permutations | PROGRAMMING | 1,100 | [
"greedy",
"math"
] | null | null | Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements. | In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise. | [
"1\n1\n",
"3\n1 1 2\n",
"4\n7 7 7 7\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs. | 500 | [
{
"input": "1\n1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "4\n7 7 7 7",
"output": "NO"
},
{
"input": "4\n479 170 465 146",
"output": "YES"
},
{
"input": "5\n996 437 605 996 293",
"output": "YES"
},
{
"input": "6\n727 53... | 1,567,661,453 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 280 | 409,600 | ######################################################################
# Write your code here
import sys
from math import *
input = sys.stdin.readline
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x10000000, resource.RLIM_INFINITY])
#sys.setrecursionlimit(0x100000)
# Write your code here
RI = lambda : [int(x) for x in sys.stdin.readline().strip().split()]
rw = lambda : input().strip().split()
ls = lambda : list(input().strip()) # for strings to list of char
from collections import defaultdict as df
import heapq
#heapq.heapify(li) heappush(li,4) heappop(li)
#import random
#random.shuffle(list)
infinite = float('inf')
#######################################################################
n=int(input())
l=RI()
d=df(int)
for i in l:
d[i]+=1
f=0
m=(n+1)//2
for i in d:
if(d[i]>m):
f=1
break
if(f==0):
print("YES")
else:
print("NO")
| Title: Yaroslav and Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
Output Specification:
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
Demo Input:
['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs. | ```python
######################################################################
# Write your code here
import sys
from math import *
input = sys.stdin.readline
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x10000000, resource.RLIM_INFINITY])
#sys.setrecursionlimit(0x100000)
# Write your code here
RI = lambda : [int(x) for x in sys.stdin.readline().strip().split()]
rw = lambda : input().strip().split()
ls = lambda : list(input().strip()) # for strings to list of char
from collections import defaultdict as df
import heapq
#heapq.heapify(li) heappush(li,4) heappop(li)
#import random
#random.shuffle(list)
infinite = float('inf')
#######################################################################
n=int(input())
l=RI()
d=df(int)
for i in l:
d[i]+=1
f=0
m=(n+1)//2
for i in d:
if(d[i]>m):
f=1
break
if(f==0):
print("YES")
else:
print("NO")
``` | 3 |
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