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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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",
... | 1,669,128,693 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 171 | 7,270,400 | n=int(input())
l=[0]*100005
for i in [int(j) for j in input().split()]:
l[i]+=1
dp=[0]*100005
for i in range(1,100001):
dp[i]=max( dp[i-1], dp[i-2] + i*l[i] )
print(max(dp)) | 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())
l=[0]*100005
for i in [int(j) for j in input().split()]:
l[i]+=1
dp=[0]*100005
for i in range(1,100001):
dp[i]=max( dp[i-1], dp[i-2] + i*l[i] )
print(max(dp))
``` | 3 | |
627 | B | Factory Repairs | PROGRAMMING | 1,700 | [
"data structures"
] | null | null | A factory produces thimbles in bulk. Typically, it can produce up to *a* thimbles a day. However, some of the machinery is defective, so it can currently only produce *b* thimbles each day. The factory intends to choose a *k*-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of *a* thimbles per day after the *k* days are complete.
Initially, no orders are pending. The factory receives updates of the form *d**i*, *a**i*, indicating that *a**i* new orders have been placed for the *d**i*-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.
As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day *p**i*. Help the owner answer his questions. | The first line contains five integers *n*, *k*, *a*, *b*, and *q* (1<=≤<=*k*<=≤<=*n*<=≤<=200<=000, 1<=≤<=*b*<=<<=*a*<=≤<=10 000, 1<=≤<=*q*<=≤<=200<=000) — the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.
The next *q* lines contain the descriptions of the queries. Each query is of one of the following two forms:
- 1 *d**i* *a**i* (1<=≤<=*d**i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=10 000), representing an update of *a**i* orders on day *d**i*, or - 2 *p**i* (1<=≤<=*p**i*<=≤<=*n*<=-<=*k*<=+<=1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day *p**i*?
It's guaranteed that the input will contain at least one query of the second type. | For each query of the second type, print a line containing a single integer — the maximum number of orders that the factory can fill over all *n* days. | [
"5 2 2 1 8\n1 1 2\n1 5 3\n1 2 1\n2 2\n1 4 2\n1 3 2\n2 1\n2 3\n",
"5 4 10 1 6\n1 1 5\n1 5 5\n1 3 2\n1 5 2\n2 1\n2 2\n"
] | [
"3\n6\n4\n",
"7\n1\n"
] | Consider the first sample.
We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.
For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.
For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders. | 1,000 | [
{
"input": "5 2 2 1 8\n1 1 2\n1 5 3\n1 2 1\n2 2\n1 4 2\n1 3 2\n2 1\n2 3",
"output": "3\n6\n4"
},
{
"input": "5 4 10 1 6\n1 1 5\n1 5 5\n1 3 2\n1 5 2\n2 1\n2 2",
"output": "7\n1"
},
{
"input": "1 1 2 1 1\n2 1",
"output": "0"
}
] | 1,584,038,294 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | //B - Factory Repairs:
//Assunto:
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define loop(i, a, b) for(int i = a; i < b; i++)
#define loopBack(i, a, b) for(int i = a; i >= b; i--)
#define INF INT_MAX
#define LINF LLONG_MAX
#define DINF DBL_MAX
#define MAX 10050
#define MOD 1000000007
using namespace std;
typedef long long int ll;
typedef pair<int, int> ii;
ll n, k, a, b, q, type, x, y, ans;
ll sum(ll k, vector<ll> &bit) {
int s = 0;
while(k > 0) {
s += bit[k];
k -= k & -k;
}
return s;
}
void add(ll val, ll k, vector<ll> &bit, ll aux) {
while(k <= n) {
bit[k] += min(aux, val);
k += k & -k;
}
}
int main() {
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
//freopen("in.txt", "r", stdin);
cin >> n >> k >> a >> b >> q;
vector<ll> bit1(n+1, 0), bit2(n+1, 0);
while(q--) {
cin >> type;
if(type == 1) {
cin >> x >> y;
add(y, x, bit1, b);
add(y, x, bit2, a);
} else {
cin >> x;
ans = sum(x-1, bit1);
ans += sum(n, bit2) - sum((x + k) - 1, bit2);
cout << ans << "\n";
}
}
}
| Title: Factory Repairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A factory produces thimbles in bulk. Typically, it can produce up to *a* thimbles a day. However, some of the machinery is defective, so it can currently only produce *b* thimbles each day. The factory intends to choose a *k*-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of *a* thimbles per day after the *k* days are complete.
Initially, no orders are pending. The factory receives updates of the form *d**i*, *a**i*, indicating that *a**i* new orders have been placed for the *d**i*-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.
As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day *p**i*. Help the owner answer his questions.
Input Specification:
The first line contains five integers *n*, *k*, *a*, *b*, and *q* (1<=≤<=*k*<=≤<=*n*<=≤<=200<=000, 1<=≤<=*b*<=<<=*a*<=≤<=10 000, 1<=≤<=*q*<=≤<=200<=000) — the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.
The next *q* lines contain the descriptions of the queries. Each query is of one of the following two forms:
- 1 *d**i* *a**i* (1<=≤<=*d**i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=10 000), representing an update of *a**i* orders on day *d**i*, or - 2 *p**i* (1<=≤<=*p**i*<=≤<=*n*<=-<=*k*<=+<=1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day *p**i*?
It's guaranteed that the input will contain at least one query of the second type.
Output Specification:
For each query of the second type, print a line containing a single integer — the maximum number of orders that the factory can fill over all *n* days.
Demo Input:
['5 2 2 1 8\n1 1 2\n1 5 3\n1 2 1\n2 2\n1 4 2\n1 3 2\n2 1\n2 3\n', '5 4 10 1 6\n1 1 5\n1 5 5\n1 3 2\n1 5 2\n2 1\n2 2\n']
Demo Output:
['3\n6\n4\n', '7\n1\n']
Note:
Consider the first sample.
We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.
For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.
For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders. | ```python
//B - Factory Repairs:
//Assunto:
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define loop(i, a, b) for(int i = a; i < b; i++)
#define loopBack(i, a, b) for(int i = a; i >= b; i--)
#define INF INT_MAX
#define LINF LLONG_MAX
#define DINF DBL_MAX
#define MAX 10050
#define MOD 1000000007
using namespace std;
typedef long long int ll;
typedef pair<int, int> ii;
ll n, k, a, b, q, type, x, y, ans;
ll sum(ll k, vector<ll> &bit) {
int s = 0;
while(k > 0) {
s += bit[k];
k -= k & -k;
}
return s;
}
void add(ll val, ll k, vector<ll> &bit, ll aux) {
while(k <= n) {
bit[k] += min(aux, val);
k += k & -k;
}
}
int main() {
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
//freopen("in.txt", "r", stdin);
cin >> n >> k >> a >> b >> q;
vector<ll> bit1(n+1, 0), bit2(n+1, 0);
while(q--) {
cin >> type;
if(type == 1) {
cin >> x >> y;
add(y, x, bit1, b);
add(y, x, bit2, a);
} else {
cin >> x;
ans = sum(x-1, bit1);
ans += sum(n, bit2) - sum((x + k) - 1, bit2);
cout << ans << "\n";
}
}
}
``` | -1 | |
721 | A | One-dimensional Japanese Crossword | PROGRAMMING | 800 | [
"implementation"
] | null | null | Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). | The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. | [
"3\nBBW\n",
"5\nBWBWB\n",
"4\nWWWW\n",
"4\nBBBB\n",
"13\nWBBBBWWBWBBBW\n"
] | [
"1\n2 ",
"3\n1 1 1 ",
"0\n",
"1\n4 ",
"3\n4 1 3 "
] | The last sample case correspond to the picture in the statement. | 500 | [
{
"input": "3\nBBW",
"output": "1\n2 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "4\nWWWW",
"output": "0"
},
{
"input": "4\nBBBB",
"output": "1\n4 "
},
{
"input": "13\nWBBBBWWBWBBBW",
"output": "3\n4 1 3 "
},
{
"input": "1\nB",
... | 1,664,794,493 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 46 | 0 | n = int(input())
s = input()
count = 0
k = 0
ans = []
for i in s:
if i=="B":
count+=1
else:
if count!=0:
k+=1
ans.append(count)
count=0
if count!=0:
ans.append(count)
k+=1
print(k)
print(*ans) | Title: One-dimensional Japanese Crossword
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output Specification:
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Demo Input:
['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n']
Demo Output:
['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 ']
Note:
The last sample case correspond to the picture in the statement. | ```python
n = int(input())
s = input()
count = 0
k = 0
ans = []
for i in s:
if i=="B":
count+=1
else:
if count!=0:
k+=1
ans.append(count)
count=0
if count!=0:
ans.append(count)
k+=1
print(k)
print(*ans)
``` | 3 | |
219 | A | k-String | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string. | The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*. | Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes). | [
"2\naazz\n",
"3\nabcabcabz\n"
] | [
"azaz\n",
"-1\n"
] | none | 500 | [
{
"input": "2\naazz",
"output": "azaz"
},
{
"input": "3\nabcabcabz",
"output": "-1"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "2\nabba",
"output": "abab"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "7\nabacaba",
"output": "-1"
}... | 1,679,653,685 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 92 | 0 | k = int(input())
st = input()
dic = {}
for i in st:
if(i in dic):
dic[i] = dic[i] +1
else :
dic[i] = 1
# print(dic)
e= 0
for i in dic:
if dic[i]%k !=0:
e=e+1
if e:
print(-1)
else :
ans = ''
for i in dic:
ans = ans+ i*(dic[i]//k)
temp = ans
for i in range(k-1):
ans = ans + temp
print(ans) | Title: k-String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
Input Specification:
The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*.
Output Specification:
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
Demo Input:
['2\naazz\n', '3\nabcabcabz\n']
Demo Output:
['azaz\n', '-1\n']
Note:
none | ```python
k = int(input())
st = input()
dic = {}
for i in st:
if(i in dic):
dic[i] = dic[i] +1
else :
dic[i] = 1
# print(dic)
e= 0
for i in dic:
if dic[i]%k !=0:
e=e+1
if e:
print(-1)
else :
ans = ''
for i in dic:
ans = ans+ i*(dic[i]//k)
temp = ans
for i in range(k-1):
ans = ans + temp
print(ans)
``` | 3 | |
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... | 1,664,282,944 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 31 | 0 | n,m = map(int,input().split())
numbers = [int(num) for num in input().split(" ", n-1)]
i = 1
j = n
for num in numbers:
if m<num:
j = i%n
k = (num-m)
numbers.append(k)
i = i+1
print(j) | 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
n,m = map(int,input().split())
numbers = [int(num) for num in input().split(" ", n-1)]
i = 1
j = n
for num in numbers:
if m<num:
j = i%n
k = (num-m)
numbers.append(k)
i = i+1
print(j)
``` | 0 | |
985 | D | Sand Fortress | PROGRAMMING | 2,100 | [
"binary search",
"constructive algorithms",
"math"
] | null | null | You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought *n* packs of sand with you. Let height *h**i* of the sand pillar on some spot *i* be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with *H* sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
- *h*1<=≤<=*H*: no sand from the leftmost spot should go over the fence; - For any |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen; - : you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold. | The only line contains two integer numbers *n* and *H* (1<=≤<=*n*,<=*H*<=≤<=1018) — the number of sand packs you have and the height of the fence, respectively. | Print the minimum number of spots you can occupy so the all the castle building conditions hold. | [
"5 2\n",
"6 8\n"
] | [
"3\n",
"3\n"
] | Here are the heights of some valid castles:
- *n* = 5, *H* = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...] - *n* = 6, *H* = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
- *n* = 5, *H* = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...] - *n* = 6, *H* = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | 0 | [
{
"input": "5 2",
"output": "3"
},
{
"input": "6 8",
"output": "3"
},
{
"input": "20 4",
"output": "7"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1414213562"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 10000000000000000... | 1,527,318,936 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 2,000 | 307,200 | n, H = tuple(map(int, input().split(' ')))
answer = 0
def stackUp(mid):
sum = 0
for i in range(mid):
sum += (i + 1)
return sum
spots = 0
low = 0
high = H
while low < high-1:
mid = (low + high)//2
sandBags = stackUp(mid)
if sandBags >= n:
high = mid
elif sandBags < n:
low = mid
sandBags = stackUp(high)
remain = n-sandBags
if remain == 0:
answer = high
else:
answer = high
while remain > 0:
if remain > H:
remain -= answer
answer += 1
elif remain <= H and remain > answer:
remain -= (answer + 1)
answer += 1
elif remain <= answer:
answer += 1
remain = 0
print(answer)
| Title: Sand Fortress
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought *n* packs of sand with you. Let height *h**i* of the sand pillar on some spot *i* be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with *H* sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
- *h*1<=≤<=*H*: no sand from the leftmost spot should go over the fence; - For any |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen; - : you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input Specification:
The only line contains two integer numbers *n* and *H* (1<=≤<=*n*,<=*H*<=≤<=1018) — the number of sand packs you have and the height of the fence, respectively.
Output Specification:
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Demo Input:
['5 2\n', '6 8\n']
Demo Output:
['3\n', '3\n']
Note:
Here are the heights of some valid castles:
- *n* = 5, *H* = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...] - *n* = 6, *H* = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
- *n* = 5, *H* = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...] - *n* = 6, *H* = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | ```python
n, H = tuple(map(int, input().split(' ')))
answer = 0
def stackUp(mid):
sum = 0
for i in range(mid):
sum += (i + 1)
return sum
spots = 0
low = 0
high = H
while low < high-1:
mid = (low + high)//2
sandBags = stackUp(mid)
if sandBags >= n:
high = mid
elif sandBags < n:
low = mid
sandBags = stackUp(high)
remain = n-sandBags
if remain == 0:
answer = high
else:
answer = high
while remain > 0:
if remain > H:
remain -= answer
answer += 1
elif remain <= H and remain > answer:
remain -= (answer + 1)
answer += 1
elif remain <= answer:
answer += 1
remain = 0
print(answer)
``` | 0 | |
165 | A | Supercentral Point | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | [
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] | [
"2\n",
"1\n"
] | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | 500 | [
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n... | 1,658,232,761 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 124 | 0 | n = int(input())
s = []
for i in range(n):
x, y = map(int, input().split())
s.append([x, y])
kol = 0
for elem in s:
u = 0
d = 0
r = 0
l = 0
for ilem in s:
if elem != ilem:
if ilem[0] == elem[0] and ilem[1] > elem[1]:
u += 1
elif ilem[0] == elem[0] and ilem[1] < elem[1]:
d += 1
elif ilem[0] > elem[0] and ilem[1] == elem[1]:
r += 1
elif ilem[0] < elem[0] and ilem[1] == elem[1]:
l += 1
if u * d * r * l > 0:
kol += 1
print(kol)
| Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | ```python
n = int(input())
s = []
for i in range(n):
x, y = map(int, input().split())
s.append([x, y])
kol = 0
for elem in s:
u = 0
d = 0
r = 0
l = 0
for ilem in s:
if elem != ilem:
if ilem[0] == elem[0] and ilem[1] > elem[1]:
u += 1
elif ilem[0] == elem[0] and ilem[1] < elem[1]:
d += 1
elif ilem[0] > elem[0] and ilem[1] == elem[1]:
r += 1
elif ilem[0] < elem[0] and ilem[1] == elem[1]:
l += 1
if u * d * r * l > 0:
kol += 1
print(kol)
``` | 3 | |
957 | A | Tritonic Iridescence | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them. | The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas.
The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one). | If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower). | [
"5\nCY??Y\n",
"5\nC?C?Y\n",
"5\n?CYC?\n",
"5\nC??MM\n",
"3\nMMY\n"
] | [
"Yes\n",
"Yes\n",
"Yes\n",
"No\n",
"No\n"
] | For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | 500 | [
{
"input": "5\nCY??Y",
"output": "Yes"
},
{
"input": "5\nC?C?Y",
"output": "Yes"
},
{
"input": "5\n?CYC?",
"output": "Yes"
},
{
"input": "5\nC??MM",
"output": "No"
},
{
"input": "3\nMMY",
"output": "No"
},
{
"input": "15\n??YYYYYY??YYYY?",
"output"... | 1,522,157,883 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 7,065,600 | n = int(input())
s = input()
flag=0
for i in range(n):
if i==0 and s[0] == '?':
print('Yes')
flag=1
break
elif i==n-1 and s[n-1] == '?':
print('Yes')
flag=1
break
else:
if s[i-1]=='?' or s[i+1]=='?':
print('Yes')
flag=1
break
elif s[i-1] == s[i+1]:
print('Yes')
flag=1
break
if flag==0:
print('No') | Title: Tritonic Iridescence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input Specification:
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas.
The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output Specification:
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Demo Input:
['5\nCY??Y\n', '5\nC?C?Y\n', '5\n?CYC?\n', '5\nC??MM\n', '3\nMMY\n']
Demo Output:
['Yes\n', 'Yes\n', 'Yes\n', 'No\n', 'No\n']
Note:
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | ```python
n = int(input())
s = input()
flag=0
for i in range(n):
if i==0 and s[0] == '?':
print('Yes')
flag=1
break
elif i==n-1 and s[n-1] == '?':
print('Yes')
flag=1
break
else:
if s[i-1]=='?' or s[i+1]=='?':
print('Yes')
flag=1
break
elif s[i-1] == s[i+1]:
print('Yes')
flag=1
break
if flag==0:
print('No')
``` | 0 | |
706 | B | Interesting drink | PROGRAMMING | 1,100 | [
"binary search",
"dp",
"implementation"
] | null | null | Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola". | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day. | Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day. | [
"5\n3 10 8 6 11\n4\n1\n10\n3\n11\n"
] | [
"0\n4\n1\n5\n"
] | On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | 1,000 | [
{
"input": "5\n3 10 8 6 11\n4\n1\n10\n3\n11",
"output": "0\n4\n1\n5"
},
{
"input": "5\n868 987 714 168 123\n10\n424\n192\n795\n873\n117\n914\n735\n158\n631\n471",
"output": "2\n2\n3\n4\n0\n4\n3\n1\n2\n2"
},
{
"input": "3\n435 482 309\n7\n245\n241\n909\n745\n980\n29\n521",
"output": "... | 1,695,809,542 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 6,860,800 | # https://codeforces.com/problemset/problem/706/B
from typing import List
def search(sorted_xs: int, left:int, right:int, m: int) -> int:
'''Returns the greatest index in ascending array.
`-1` if does not exists.'''
while left < right:
mid: int = left + ((right - left ) // 2)
if sorted_xs[mid] == m:
return mid
if sorted_xs[mid] < m:
left = mid + 1
elif sorted_xs[mid] > m:
right = mid - 1
return - 1
def solve(q:int, sorted_xs: List[int], ms:List[int]) -> None:
for m in ms:
print(search(sorted_xs,0, q, m) + 1);
def main() -> None:
n: int = int(input())
xs: List[int] = list(map(int, input().split(" ")))
xs.sort()
q:int = int(input())
ms:List[int] = []
for _ in range(q):
m = int(input())
ms.append(m)
solve(q, xs, ms)
if __name__ == "__main__":
main()
| Title: Interesting drink
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day.
Output Specification:
Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day.
Demo Input:
['5\n3 10 8 6 11\n4\n1\n10\n3\n11\n']
Demo Output:
['0\n4\n1\n5\n']
Note:
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | ```python
# https://codeforces.com/problemset/problem/706/B
from typing import List
def search(sorted_xs: int, left:int, right:int, m: int) -> int:
'''Returns the greatest index in ascending array.
`-1` if does not exists.'''
while left < right:
mid: int = left + ((right - left ) // 2)
if sorted_xs[mid] == m:
return mid
if sorted_xs[mid] < m:
left = mid + 1
elif sorted_xs[mid] > m:
right = mid - 1
return - 1
def solve(q:int, sorted_xs: List[int], ms:List[int]) -> None:
for m in ms:
print(search(sorted_xs,0, q, m) + 1);
def main() -> None:
n: int = int(input())
xs: List[int] = list(map(int, input().split(" ")))
xs.sort()
q:int = int(input())
ms:List[int] = []
for _ in range(q):
m = int(input())
ms.append(m)
solve(q, xs, ms)
if __name__ == "__main__":
main()
``` | 0 | |
711 | A | Bus to Udayland | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit? | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details. | If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them. | [
"6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n",
"5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n"
] | [
"YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"NO\n",
"YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n"
] | Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | 500 | [
{
"input": "6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX",
"output": "YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX"
},
{
"input": "4\nXO|OX\nXO|XX\nOX|OX\nXX|OX",
"output": "NO"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO"
... | 1,617,200,514 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 78 | 0 | ans=[]
flag=False
for _ in range(int(input())):
a=input().split('|')
if(not flag):
if(a[0]=='OO' or a[1]=='OO'):
if(a[0]=='OO'):
a[0]='++'
else:
a[1]="++"
flag=True
ans.append(a)
if(flag):
print("YES")
for i in ans:
print(i[0]+"|"+i[1])
else:
print("NO") | Title: Bus to Udayland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.
Output Specification:
If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them.
Demo Input:
['6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', '4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n', '5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n']
Demo Output:
['YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', 'NO\n', 'YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n']
Note:
Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | ```python
ans=[]
flag=False
for _ in range(int(input())):
a=input().split('|')
if(not flag):
if(a[0]=='OO' or a[1]=='OO'):
if(a[0]=='OO'):
a[0]='++'
else:
a[1]="++"
flag=True
ans.append(a)
if(flag):
print("YES")
for i in ans:
print(i[0]+"|"+i[1])
else:
print("NO")
``` | 3 | |
245 | E | Mishap in Club | PROGRAMMING | 1,400 | [
"greedy",
"implementation"
] | null | null | Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen.
On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended.
Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. | The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. | Print the sought minimum number of people | [
"+-+-+\n",
"---"
] | [
"1\n",
"3"
] | none | 0 | [
{
"input": "+-+-+",
"output": "1"
},
{
"input": "---",
"output": "3"
},
{
"input": "-",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "---",
"output": "3"
},
{
"input": "----",
"output": "4"
},
{
"input": "---+",
"output... | 1,367,837,349 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 93 | 102,400 | m,p,c=0,0,0
s=(input())
for i in s:
if i=='-':
c-=1
else:
c+=1
m=min(m,c)
p=max(p,c)
print (p-m)
| Title: Mishap in Club
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen.
On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended.
Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times.
Input Specification:
The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive.
Output Specification:
Print the sought minimum number of people
Demo Input:
['+-+-+\n', '---']
Demo Output:
['1\n', '3']
Note:
none | ```python
m,p,c=0,0,0
s=(input())
for i in s:
if i=='-':
c-=1
else:
c+=1
m=min(m,c)
p=max(p,c)
print (p-m)
``` | 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... | 1,588,259,129 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 216 | 0 | n = int(input())
numbers = (map(int, input().split()))
two_count = 0
ans_dict = dict()
for i in numbers:
ans_dict[i] = ans_dict.get(i, 0) + 1
for k, v in ans_dict.items():
if v > (n +1)/2:
print("NO")
break
else:
print("YES")
| 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
n = int(input())
numbers = (map(int, input().split()))
two_count = 0
ans_dict = dict()
for i in numbers:
ans_dict[i] = ans_dict.get(i, 0) + 1
for k, v in ans_dict.items():
if v > (n +1)/2:
print("NO")
break
else:
print("YES")
``` | 3 | |
248 | A | Cupboards | PROGRAMMING | 800 | [
"implementation"
] | null | null | One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*. | The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces. | In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs. | [
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 ... | 1,621,668,597 | 2,147,483,647 | PyPy 3 | OK | TESTS | 57 | 404 | 3,788,800 | l=[]
r=[]
for i in range(int(input())):
f=[int(i) for i in input().split()]
l.append(f[0])
r.append(f[1])
cl_1=l.count(1)
cl_0=l.count(0)
cr_1=r.count(1)
cr_0=r.count(0)
if cl_1<cl_0:
c=cl_1
else:
c=cl_0
if cr_1<cr_0:
c+=cr_1
else:
c+=cr_0
print(c)
| Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none | ```python
l=[]
r=[]
for i in range(int(input())):
f=[int(i) for i in input().split()]
l.append(f[0])
r.append(f[1])
cl_1=l.count(1)
cl_0=l.count(0)
cr_1=r.count(1)
cr_0=r.count(0)
if cl_1<cl_0:
c=cl_1
else:
c=cl_0
if cr_1<cr_0:
c+=cr_1
else:
c+=cr_0
print(c)
``` | 3 | |
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 ... | 1,588,830,482 | 2,147,483,647 | Python 3 | OK | TESTS | 88 | 109 | 6,656,000 | n=int(input())
a = [int(x) for x in input().split()]
c = 0
last = 0
for i in range(n):
if a[i]==0 or a[i]==last:
c+=1
last = 0
elif a[i]==3 :
if(last!=0):
last = 3 - last
else:
last=a[i]
print(c) | 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
n=int(input())
a = [int(x) for x in input().split()]
c = 0
last = 0
for i in range(n):
if a[i]==0 or a[i]==last:
c+=1
last = 0
elif a[i]==3 :
if(last!=0):
last = 3 - last
else:
last=a[i]
print(c)
``` | 3 | |
519 | C | A and B and Team Training | PROGRAMMING | 1,300 | [
"greedy",
"implementation",
"math",
"number theory"
] | null | null | A and B are preparing themselves for programming contests.
An important part of preparing for a competition is sharing programming knowledge from the experienced members to those who are just beginning to deal with the contests. Therefore, during the next team training A decided to make teams so that newbies are solving problems together with experienced participants.
A believes that the optimal team of three people should consist of one experienced participant and two newbies. Thus, each experienced participant can share the experience with a large number of people.
However, B believes that the optimal team should have two experienced members plus one newbie. Thus, each newbie can gain more knowledge and experience.
As a result, A and B have decided that all the teams during the training session should belong to one of the two types described above. Furthermore, they agree that the total number of teams should be as much as possible.
There are *n* experienced members and *m* newbies on the training session. Can you calculate what maximum number of teams can be formed? | The first line contains two integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=5·105) — the number of experienced participants and newbies that are present at the training session. | Print the maximum number of teams that can be formed. | [
"2 6\n",
"4 5\n"
] | [
"2\n",
"3\n"
] | Let's represent the experienced players as XP and newbies as NB.
In the first test the teams look as follows: (XP, NB, NB), (XP, NB, NB).
In the second test sample the teams look as follows: (XP, NB, NB), (XP, NB, NB), (XP, XP, NB). | 1,500 | [
{
"input": "2 6",
"output": "2"
},
{
"input": "4 5",
"output": "3"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "3 3",
"output": "2"
},
{
"input": "500000 500000",
"output": "333333"
},
{
"input": "70 100",
"output": "56"
},
{
"input": ... | 1,637,396,898 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 61 | 0 | a,b=map(int,input().split())
a,b=min(a,b),max(a,b)
if b>=2*a:
print(a)
else:
s=2*a-b
if s%3==2:
print(b-a+s//3*2+1)
else:
print(b-a+s//3*2) | Title: A and B and Team Training
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
An important part of preparing for a competition is sharing programming knowledge from the experienced members to those who are just beginning to deal with the contests. Therefore, during the next team training A decided to make teams so that newbies are solving problems together with experienced participants.
A believes that the optimal team of three people should consist of one experienced participant and two newbies. Thus, each experienced participant can share the experience with a large number of people.
However, B believes that the optimal team should have two experienced members plus one newbie. Thus, each newbie can gain more knowledge and experience.
As a result, A and B have decided that all the teams during the training session should belong to one of the two types described above. Furthermore, they agree that the total number of teams should be as much as possible.
There are *n* experienced members and *m* newbies on the training session. Can you calculate what maximum number of teams can be formed?
Input Specification:
The first line contains two integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=5·105) — the number of experienced participants and newbies that are present at the training session.
Output Specification:
Print the maximum number of teams that can be formed.
Demo Input:
['2 6\n', '4 5\n']
Demo Output:
['2\n', '3\n']
Note:
Let's represent the experienced players as XP and newbies as NB.
In the first test the teams look as follows: (XP, NB, NB), (XP, NB, NB).
In the second test sample the teams look as follows: (XP, NB, NB), (XP, NB, NB), (XP, XP, NB). | ```python
a,b=map(int,input().split())
a,b=min(a,b),max(a,b)
if b>=2*a:
print(a)
else:
s=2*a-b
if s%3==2:
print(b-a+s//3*2+1)
else:
print(b-a+s//3*2)
``` | 3 | |
780 | A | Andryusha and Socks | PROGRAMMING | 800 | [
"implementation"
] | null | null | Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time? | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair. | Print single integer — the maximum number of socks that were on the table at the same time. | [
"1\n1 1\n",
"3\n2 1 1 3 2 3\n"
] | [
"1\n",
"2\n"
] | In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe. | 500 | [
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n2 1 1 3 2 3",
"output": "2"
},
{
"input": "5\n5 1 3 2 4 3 1 2 4 5",
"output": "5"
},
{
"input": "10\n4 2 6 3 4 8 7 1 1 5 2 10 6 8 3 5 10 9 9 7",
"output": "6"
},
{
"input": "50\n30 47 31 38 37 50 36 43 9 23 2 2 ... | 1,598,703,637 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 217 | 13,312,000 | n = int(input())
socks = list(map(int, input().split()))
table = [0] * n
maximum = 0
temp = 0
for i in range(n):
if table[socks[i] - 1] == 0:
table[socks[i] - 1] = 1
temp += 1
else:
temp -= 1
maximum = max(maximum, temp)
print(maximum) | Title: Andryusha and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
Output Specification:
Print single integer — the maximum number of socks that were on the table at the same time.
Demo Input:
['1\n1 1\n', '3\n2 1 1 3 2 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe. | ```python
n = int(input())
socks = list(map(int, input().split()))
table = [0] * n
maximum = 0
temp = 0
for i in range(n):
if table[socks[i] - 1] == 0:
table[socks[i] - 1] = 1
temp += 1
else:
temp -= 1
maximum = max(maximum, temp)
print(maximum)
``` | 0 | |
454 | A | Little Pony and Crystal Mine | PROGRAMMING | 800 | [
"implementation"
] | null | null | Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=><=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it.
You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. | The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd). | Output a crystal of size *n*. | [
"3\n",
"5\n",
"7\n"
] | [
"*D*\nDDD\n*D*\n",
"**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n",
"***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n"
] | none | 500 | [
{
"input": "3",
"output": "*D*\nDDD\n*D*"
},
{
"input": "5",
"output": "**D**\n*DDD*\nDDDDD\n*DDD*\n**D**"
},
{
"input": "7",
"output": "***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***"
},
{
"input": "11",
"output": "*****D*****\n****DDD****\n***DDDDD***\n**... | 1,606,495,682 | 1,082 | Python 3 | OK | TESTS | 12 | 109 | 307,200 | import math
a=int(input())
arr=[a//2]
ans=[]
for i in range(math.ceil(a//2)+1):
trr=[]
for j in range(a):
if(j in arr):
trr.append('D')
else:
trr.append("*")
ans.append(trr)
arr.append(arr[0]-1)
arr.sort()
arr.append(arr[len(arr)-1]+1)
for i in ans:
print(''.join(i))
ans=ans[::-1]
ans.pop(0)
for i in ans:
print(''.join(i)) | Title: Little Pony and Crystal Mine
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=><=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it.
You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.
Input Specification:
The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd).
Output Specification:
Output a crystal of size *n*.
Demo Input:
['3\n', '5\n', '7\n']
Demo Output:
['*D*\nDDD\n*D*\n', '**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n', '***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n']
Note:
none | ```python
import math
a=int(input())
arr=[a//2]
ans=[]
for i in range(math.ceil(a//2)+1):
trr=[]
for j in range(a):
if(j in arr):
trr.append('D')
else:
trr.append("*")
ans.append(trr)
arr.append(arr[0]-1)
arr.sort()
arr.append(arr[len(arr)-1]+1)
for i in ans:
print(''.join(i))
ans=ans[::-1]
ans.pop(0)
for i in ans:
print(''.join(i))
``` | 3 | |
679 | A | Bear and Prime 100 | PROGRAMMING | 1,400 | [
"constructive algorithms",
"interactive",
"math"
] | null | null | This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2,<=100]. Your task is to say if the hidden number is prime or composite.
Integer *x*<=><=1 is called prime if it has exactly two distinct divisors, 1 and *x*. If integer *x*<=><=1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2,<=100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2,<=100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below). | After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise. | Up to 20 times you can ask a query — print an integer from interval [2,<=100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
- fflush(stdout) in C++; - System.out.flush() in Java; - stdout.flush() in Python; - flush(output) in Pascal; - See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2,<=100]. Of course, his/her solution won't be able to read the hidden number from the input. | [
"yes\nno\nyes\n",
"no\nyes\nno\nno\nno\n"
] | [
"2\n80\n5\ncomposite\n",
"58\n59\n78\n78\n2\nprime\n"
] | The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea790051c34ea7d2761cd9b096412ca7c647a173.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/35c6952617fa94ec3e0ea8e63aa1c3c49b3ba420.png" style="max-width: 100.0%;max-height: 100.0%;"/>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | 750 | [
{
"input": "30",
"output": "composite 4"
},
{
"input": "59",
"output": "prime 15"
},
{
"input": "2",
"output": "prime 16"
},
{
"input": "7",
"output": "prime 16"
},
{
"input": "9",
"output": "composite 3"
},
{
"input": "13",
"output": "prime 15"
... | 1,678,189,771 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | def main():
for i in range(2, 11):
print(i)
query = input()
if query == 'yes':
print('composite')
return
print('prime')
if __name__ == '__main__':
main() | Title: Bear and Prime 100
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2,<=100]. Your task is to say if the hidden number is prime or composite.
Integer *x*<=><=1 is called prime if it has exactly two distinct divisors, 1 and *x*. If integer *x*<=><=1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2,<=100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2,<=100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input Specification:
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output Specification:
Up to 20 times you can ask a query — print an integer from interval [2,<=100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
- fflush(stdout) in C++; - System.out.flush() in Java; - stdout.flush() in Python; - flush(output) in Pascal; - See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2,<=100]. Of course, his/her solution won't be able to read the hidden number from the input.
Demo Input:
['yes\nno\nyes\n', 'no\nyes\nno\nno\nno\n']
Demo Output:
['2\n80\n5\ncomposite\n', '58\n59\n78\n78\n2\nprime\n']
Note:
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea790051c34ea7d2761cd9b096412ca7c647a173.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/35c6952617fa94ec3e0ea8e63aa1c3c49b3ba420.png" style="max-width: 100.0%;max-height: 100.0%;"/>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | ```python
def main():
for i in range(2, 11):
print(i)
query = input()
if query == 'yes':
print('composite')
return
print('prime')
if __name__ == '__main__':
main()
``` | 0 | |
479 | A | Expression | PROGRAMMING | 1,000 | [
"brute force",
"math"
] | null | null | Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get. | The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10). | Print the maximum value of the expression that you can obtain. | [
"1\n2\n3\n",
"2\n10\n3\n"
] | [
"9\n",
"60\n"
] | none | 500 | [
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{... | 1,694,013,974 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 29 | 62 | 0 | a, b, c = map(int, [input() for i in range(3)])
result = max(a+b+c, a+b*c, a*b*c, a*b+c, a*(b+c), (a+b)*c)
print(result) | Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none | ```python
a, b, c = map(int, [input() for i in range(3)])
result = max(a+b+c, a+b*c, a*b*c, a*b+c, a*(b+c), (a+b)*c)
print(result)
``` | 3 | |
404 | C | Restore Graph | PROGRAMMING | 1,800 | [
"dfs and similar",
"graphs",
"sortings"
] | null | null | Valera had an undirected connected graph without self-loops and multiple edges consisting of *n* vertices. The graph had an interesting property: there were at most *k* edges adjacent to each of its vertices. For convenience, we will assume that the graph vertices were indexed by integers from 1 to *n*.
One day Valera counted the shortest distances from one of the graph vertices to all other ones and wrote them out in array *d*. Thus, element *d*[*i*] of the array shows the shortest distance from the vertex Valera chose to vertex number *i*.
Then something irreparable terrible happened. Valera lost the initial graph. However, he still has the array *d*. Help him restore the lost graph. | The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=<<=*n*<=≤<=105). Number *n* shows the number of vertices in the original graph. Number *k* shows that at most *k* edges were adjacent to each vertex in the original graph.
The second line contains space-separated integers *d*[1],<=*d*[2],<=...,<=*d*[*n*] (0<=≤<=*d*[*i*]<=<<=*n*). Number *d*[*i*] shows the shortest distance from the vertex Valera chose to the vertex number *i*. | If Valera made a mistake in his notes and the required graph doesn't exist, print in the first line number -1. Otherwise, in the first line print integer *m* (0<=≤<=*m*<=≤<=106) — the number of edges in the found graph.
In each of the next *m* lines print two space-separated integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*), denoting the edge that connects vertices with numbers *a**i* and *b**i*. The graph shouldn't contain self-loops and multiple edges. If there are multiple possible answers, print any of them. | [
"3 2\n0 1 1\n",
"4 2\n2 0 1 3\n",
"3 1\n0 0 0\n"
] | [
"3\n1 2\n1 3\n3 2\n",
"3\n1 3\n1 4\n2 3\n",
"-1\n"
] | none | 1,500 | [
{
"input": "3 2\n0 1 1",
"output": "2\n1 2\n1 3"
},
{
"input": "4 2\n2 0 1 3",
"output": "3\n1 3\n1 4\n2 3"
},
{
"input": "3 1\n0 0 0",
"output": "-1"
},
{
"input": "5 3\n0 2 1 2 1",
"output": "4\n1 3\n1 5\n2 5\n4 5"
},
{
"input": "7 3\n2 2 0 1 3 2 1",
"output... | 1,599,056,182 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 2 | 187 | 2,457,600 | #!/usr/bin/env python3
# created : 2020. 9. 1. 22:40
import os
from sys import stdin, stdout
from collections import deque
def solve(tc):
n, k = map(int, stdin.readline().split())
dseq = list(map(int, stdin.readline().split()))
idx = list(range(n))
idx = sorted(idx, key=lambda x: dseq[x])
i = 0
while dseq[idx[i]] == 0:
i += 1
if i == 0 or i > 1:
print(-1)
return
ans = []
ok = True
cnt = 0
queue = deque()
queue.append([idx[0], 0])
par = queue.popleft()
while i < n:
if dseq[idx[i]] <= par[1]:
ok = False
break
elif dseq[idx[i]] == par[1]+1:
ans.append([par[0]+1, idx[i]+1])
queue.append([idx[i], par[1]+1])
cnt += 1
if cnt == k:
if len(queue)==0:
ok = False
break
par = queue.popleft()
cnt = 0
else:
if len(queue):
par = queue.popleft()
continue
else:
ok = False
break
i += 1
if not ok:
print(-1)
return
print(len(ans))
for a, b in ans:
print(a,b)
tcs = 1
tc = 1
while tc <= tcs:
solve(tc)
tc += 1
| Title: Restore Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera had an undirected connected graph without self-loops and multiple edges consisting of *n* vertices. The graph had an interesting property: there were at most *k* edges adjacent to each of its vertices. For convenience, we will assume that the graph vertices were indexed by integers from 1 to *n*.
One day Valera counted the shortest distances from one of the graph vertices to all other ones and wrote them out in array *d*. Thus, element *d*[*i*] of the array shows the shortest distance from the vertex Valera chose to vertex number *i*.
Then something irreparable terrible happened. Valera lost the initial graph. However, he still has the array *d*. Help him restore the lost graph.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=<<=*n*<=≤<=105). Number *n* shows the number of vertices in the original graph. Number *k* shows that at most *k* edges were adjacent to each vertex in the original graph.
The second line contains space-separated integers *d*[1],<=*d*[2],<=...,<=*d*[*n*] (0<=≤<=*d*[*i*]<=<<=*n*). Number *d*[*i*] shows the shortest distance from the vertex Valera chose to the vertex number *i*.
Output Specification:
If Valera made a mistake in his notes and the required graph doesn't exist, print in the first line number -1. Otherwise, in the first line print integer *m* (0<=≤<=*m*<=≤<=106) — the number of edges in the found graph.
In each of the next *m* lines print two space-separated integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*), denoting the edge that connects vertices with numbers *a**i* and *b**i*. The graph shouldn't contain self-loops and multiple edges. If there are multiple possible answers, print any of them.
Demo Input:
['3 2\n0 1 1\n', '4 2\n2 0 1 3\n', '3 1\n0 0 0\n']
Demo Output:
['3\n1 2\n1 3\n3 2\n', '3\n1 3\n1 4\n2 3\n', '-1\n']
Note:
none | ```python
#!/usr/bin/env python3
# created : 2020. 9. 1. 22:40
import os
from sys import stdin, stdout
from collections import deque
def solve(tc):
n, k = map(int, stdin.readline().split())
dseq = list(map(int, stdin.readline().split()))
idx = list(range(n))
idx = sorted(idx, key=lambda x: dseq[x])
i = 0
while dseq[idx[i]] == 0:
i += 1
if i == 0 or i > 1:
print(-1)
return
ans = []
ok = True
cnt = 0
queue = deque()
queue.append([idx[0], 0])
par = queue.popleft()
while i < n:
if dseq[idx[i]] <= par[1]:
ok = False
break
elif dseq[idx[i]] == par[1]+1:
ans.append([par[0]+1, idx[i]+1])
queue.append([idx[i], par[1]+1])
cnt += 1
if cnt == k:
if len(queue)==0:
ok = False
break
par = queue.popleft()
cnt = 0
else:
if len(queue):
par = queue.popleft()
continue
else:
ok = False
break
i += 1
if not ok:
print(-1)
return
print(len(ans))
for a, b in ans:
print(a,b)
tcs = 1
tc = 1
while tc <= tcs:
solve(tc)
tc += 1
``` | -1 | |
625 | A | Guest From the Past | PROGRAMMING | 1,700 | [
"implementation",
"math"
] | null | null | Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help. | First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively. | Print the only integer — maximum number of liters of kefir, that Kolya can drink. | [
"10\n11\n9\n8\n",
"10\n5\n6\n1\n"
] | [
"2\n",
"2\n"
] | In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | 750 | [
{
"input": "10\n11\n9\n8",
"output": "2"
},
{
"input": "10\n5\n6\n1",
"output": "2"
},
{
"input": "2\n2\n2\n1",
"output": "1"
},
{
"input": "10\n3\n3\n1",
"output": "4"
},
{
"input": "10\n1\n2\n1",
"output": "10"
},
{
"input": "10\n2\n3\n1",
"outpu... | 1,594,014,117 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 52 | 155 | 20,172,800 | import math as ma
n=int(input())
a=int(input())
b=int(input())
c=int(input())
if a<=b-c:
print(n//a)
else:
d=(n-b)//(b-c)+1
n=n-d*(b-c)
d+=n//a
print(d) | Title: Guest From the Past
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output Specification:
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Demo Input:
['10\n11\n9\n8\n', '10\n5\n6\n1\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | ```python
import math as ma
n=int(input())
a=int(input())
b=int(input())
c=int(input())
if a<=b-c:
print(n//a)
else:
d=(n-b)//(b-c)+1
n=n-d*(b-c)
d+=n//a
print(d)
``` | 0 | |
381 | A | Sereja and Dima | PROGRAMMING | 800 | [
"greedy",
"implementation",
"two pointers"
] | null | null | Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. | On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. | [
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] | [
"12 5\n",
"16 12\n"
] | In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | 500 | [
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 ... | 1,672,546,868 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 0 | n = int(input())
li = list(map(int, input().split()))
s = 0
d = 0
i = 1
j = 0
k = n-1
while j <= k and i <= n:
if li[j] > li[k]:
if i%2==0:
d += li[j]
else:
s += li[j]
j += 1
else:
if i%2==0:
d += li[k]
else:
s += li[k]
k -= 1
i += 1
print(s, d)
| Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | ```python
n = int(input())
li = list(map(int, input().split()))
s = 0
d = 0
i = 1
j = 0
k = n-1
while j <= k and i <= n:
if li[j] > li[k]:
if i%2==0:
d += li[j]
else:
s += li[j]
j += 1
else:
if i%2==0:
d += li[k]
else:
s += li[k]
k -= 1
i += 1
print(s, d)
``` | 3 | |
383 | D | Antimatter | PROGRAMMING | 2,300 | [
"dp"
] | null | null | Iahub accidentally discovered a secret lab. He found there *n* devices ordered in a line, numbered from 1 to *n* from left to right. Each device *i* (1<=≤<=*i*<=≤<=*n*) can create either *a**i* units of matter or *a**i* units of antimatter.
Iahub wants to choose some contiguous subarray of devices in the lab, specify the production mode for each of them (produce matter or antimatter) and finally take a photo of it. However he will be successful only if the amounts of matter and antimatter produced in the selected subarray will be the same (otherwise there would be overflowing matter or antimatter in the photo).
You are requested to compute the number of different ways Iahub can successful take a photo. A photo is different than another if it represents another subarray, or if at least one device of the subarray is set to produce matter in one of the photos and antimatter in the other one. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000).
The sum *a*1<=+<=*a*2<=+<=...<=+<=*a**n* will be less than or equal to 10000. | Output a single integer, the number of ways Iahub can take a photo, modulo 1000000007 (109<=+<=7). | [
"4\n1 1 1 1\n"
] | [
"12\n"
] | The possible photos are [1+, 2-], [1-, 2+], [2+, 3-], [2-, 3+], [3+, 4-], [3-, 4+], [1+, 2+, 3-, 4-], [1+, 2-, 3+, 4-], [1+, 2-, 3-, 4+], [1-, 2+, 3+, 4-], [1-, 2+, 3-, 4+] and [1-, 2-, 3+, 4+], where "*i*+" means that the *i*-th element produces matter, and "*i*-" means that the *i*-th element produces antimatter. | 2,000 | [
{
"input": "4\n1 1 1 1",
"output": "12"
},
{
"input": "10\n16 9 9 11 10 12 9 6 10 8",
"output": "86"
},
{
"input": "50\n2 1 5 2 1 3 1 2 3 2 1 1 5 2 2 2 3 2 1 2 2 2 3 3 1 3 1 1 2 2 2 2 1 2 3 1 2 4 1 1 1 3 2 1 1 1 3 2 1 3",
"output": "115119382"
},
{
"input": "100\n8 3 3 7 3 6 ... | 1,538,597,403 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 1,000 | 18,432,000 | # http://codeforces.com/contest/383/problem/D
from sys import stdin
M = 10000
def read_input():
n = int(stdin.readline())
devices = [int(x) for x in stdin.readline().split()]
return n, devices
if __name__ == "__main__":
n, devices = read_input()
res = [{} for _ in range(n+1)]
for nb, val in enumerate(devices, 1):
for i in range(-M, M, 1):
if i - val >= -M:
res[nb][i-val] = res[nb].get(i-val,0) + res[nb-1].get(i, 0)
if i + val <= M:
res[nb][i+val] = res[nb].get(i+val,0)+ res[nb-1].get(i, 0)
res[nb][val] += 1
res[nb][-val] += 1
print(sum([el.get(0, 0) for el in res]))
| Title: Antimatter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub accidentally discovered a secret lab. He found there *n* devices ordered in a line, numbered from 1 to *n* from left to right. Each device *i* (1<=≤<=*i*<=≤<=*n*) can create either *a**i* units of matter or *a**i* units of antimatter.
Iahub wants to choose some contiguous subarray of devices in the lab, specify the production mode for each of them (produce matter or antimatter) and finally take a photo of it. However he will be successful only if the amounts of matter and antimatter produced in the selected subarray will be the same (otherwise there would be overflowing matter or antimatter in the photo).
You are requested to compute the number of different ways Iahub can successful take a photo. A photo is different than another if it represents another subarray, or if at least one device of the subarray is set to produce matter in one of the photos and antimatter in the other one.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000).
The sum *a*1<=+<=*a*2<=+<=...<=+<=*a**n* will be less than or equal to 10000.
Output Specification:
Output a single integer, the number of ways Iahub can take a photo, modulo 1000000007 (109<=+<=7).
Demo Input:
['4\n1 1 1 1\n']
Demo Output:
['12\n']
Note:
The possible photos are [1+, 2-], [1-, 2+], [2+, 3-], [2-, 3+], [3+, 4-], [3-, 4+], [1+, 2+, 3-, 4-], [1+, 2-, 3+, 4-], [1+, 2-, 3-, 4+], [1-, 2+, 3+, 4-], [1-, 2+, 3-, 4+] and [1-, 2-, 3+, 4+], where "*i*+" means that the *i*-th element produces matter, and "*i*-" means that the *i*-th element produces antimatter. | ```python
# http://codeforces.com/contest/383/problem/D
from sys import stdin
M = 10000
def read_input():
n = int(stdin.readline())
devices = [int(x) for x in stdin.readline().split()]
return n, devices
if __name__ == "__main__":
n, devices = read_input()
res = [{} for _ in range(n+1)]
for nb, val in enumerate(devices, 1):
for i in range(-M, M, 1):
if i - val >= -M:
res[nb][i-val] = res[nb].get(i-val,0) + res[nb-1].get(i, 0)
if i + val <= M:
res[nb][i+val] = res[nb].get(i+val,0)+ res[nb-1].get(i, 0)
res[nb][val] += 1
res[nb][-val] += 1
print(sum([el.get(0, 0) for el in res]))
``` | 0 | |
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 6... | 1,680,556,916 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 171 | 14,438,400 | n = int(input())
e1 = [int(e) for e in input().split()]
e2 = [int(e) for e in input().split()]
e3 = [int(e) for e in input().split()]
primeiro, segundo, terceiro = sum(e1), sum(e2), sum(e3)
print(abs(segundo - primeiro))
print(abs(terceiro - segundo))
| Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
n = int(input())
e1 = [int(e) for e in input().split()]
e2 = [int(e) for e in input().split()]
e3 = [int(e) for e in input().split()]
primeiro, segundo, terceiro = sum(e1), sum(e2), sum(e3)
print(abs(segundo - primeiro))
print(abs(terceiro - segundo))
``` | 3 | |
447 | A | DZY Loves Hash | PROGRAMMING | 800 | [
"implementation"
] | null | null | DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x* *mod* *p*. Operation *a* *mod* *b* denotes taking a remainder after division *a* by *b*.
However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1. | The first line contains two integers, *p* and *n* (2<=≤<=*p*,<=*n*<=≤<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=≤<=*x**i*<=≤<=109). | Output a single integer — the answer to the problem. | [
"10 5\n0\n21\n53\n41\n53\n",
"5 5\n0\n1\n2\n3\n4\n"
] | [
"4\n",
"-1\n"
] | none | 500 | [
{
"input": "10 5\n0\n21\n53\n41\n53",
"output": "4"
},
{
"input": "5 5\n0\n1\n2\n3\n4",
"output": "-1"
},
{
"input": "10 6\n811966798\n734823552\n790326404\n929189974\n414343256\n560346537",
"output": "4"
},
{
"input": "2 2\n788371161\n801743052",
"output": "-1"
},
{
... | 1,633,951,927 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | p,n = map(int, input().split())
x = [int(input()) for i in range(n)]
for j in range(1,n):
if x[j] in x[:j-1]:
print(j)
break
else:
pass
else:
print(-1) | Title: DZY Loves Hash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x* *mod* *p*. Operation *a* *mod* *b* denotes taking a remainder after division *a* by *b*.
However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1.
Input Specification:
The first line contains two integers, *p* and *n* (2<=≤<=*p*,<=*n*<=≤<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=≤<=*x**i*<=≤<=109).
Output Specification:
Output a single integer — the answer to the problem.
Demo Input:
['10 5\n0\n21\n53\n41\n53\n', '5 5\n0\n1\n2\n3\n4\n']
Demo Output:
['4\n', '-1\n']
Note:
none | ```python
p,n = map(int, input().split())
x = [int(input()) for i in range(n)]
for j in range(1,n):
if x[j] in x[:j-1]:
print(j)
break
else:
pass
else:
print(-1)
``` | 0 | |
433 | A | Kitahara Haruki's Gift | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends? | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple. | In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes). | [
"3\n100 200 100\n",
"4\n100 100 100 200\n"
] | [
"YES\n",
"NO\n"
] | In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | 500 | [
{
"input": "3\n100 200 100",
"output": "YES"
},
{
"input": "4\n100 100 100 200",
"output": "NO"
},
{
"input": "1\n100",
"output": "NO"
},
{
"input": "1\n200",
"output": "NO"
},
{
"input": "2\n100 100",
"output": "YES"
},
{
"input": "2\n200 200",
"o... | 1,661,856,616 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | n=int(input())
dic={}
for i in input().split(" "):
if i in list(dic.keys()):
dic[i]+=1
else:
dic[i]=1
if {"100":2*n/3,"200":n/3}==dic or {"200":2*n/3,"100":n/3}==dic:
print("YES")
else:
print("NO") | Title: Kitahara Haruki's Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
Output Specification:
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
Demo Input:
['3\n100 200 100\n', '4\n100 100 100 200\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | ```python
n=int(input())
dic={}
for i in input().split(" "):
if i in list(dic.keys()):
dic[i]+=1
else:
dic[i]=1
if {"100":2*n/3,"200":n/3}==dic or {"200":2*n/3,"100":n/3}==dic:
print("YES")
else:
print("NO")
``` | 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",
... | 1,627,271,791 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 154 | 6,758,400 | y,w = map(int,input().split())
y = max(y,w)
hm={0:"0/1",1:"1/6",2:"1/3",3:"1/2",4:"2/3",5:"5/6",6:"1/1"}
print(hm[6-y+1]) | 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
y,w = map(int,input().split())
y = max(y,w)
hm={0:"0/1",1:"1/6",2:"1/3",3:"1/2",4:"2/3",5:"5/6",6:"1/1"}
print(hm[6-y+1])
``` | 3.872646 |
387 | A | George and Sleep | PROGRAMMING | 900 | [
"implementation"
] | null | null | George woke up and saw the current time *s* on the digital clock. Besides, George knows that he has slept for time *t*.
Help George! Write a program that will, given time *s* and *t*, determine the time *p* when George went to bed. Note that George could have gone to bed yesterday relatively to the current time (see the second test sample). | The first line contains current time *s* as a string in the format "hh:mm". The second line contains time *t* in the format "hh:mm" — the duration of George's sleep. It is guaranteed that the input contains the correct time in the 24-hour format, that is, 00<=≤<=*hh*<=≤<=23, 00<=≤<=*mm*<=≤<=59. | In the single line print time *p* — the time George went to bed in the format similar to the format of the time in the input. | [
"05:50\n05:44\n",
"00:00\n01:00\n",
"00:01\n00:00\n"
] | [
"00:06\n",
"23:00\n",
"00:01\n"
] | In the first sample George went to bed at "00:06". Note that you should print the time only in the format "00:06". That's why answers "0:06", "00:6" and others will be considered incorrect.
In the second sample, George went to bed yesterday.
In the third sample, George didn't do to bed at all. | 500 | [
{
"input": "05:50\n05:44",
"output": "00:06"
},
{
"input": "00:00\n01:00",
"output": "23:00"
},
{
"input": "00:01\n00:00",
"output": "00:01"
},
{
"input": "23:59\n23:59",
"output": "00:00"
},
{
"input": "23:44\n23:55",
"output": "23:49"
},
{
"input": "... | 1,586,094,698 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 139 | 0 | a=input()
b=input()
p=int(a[0:2])-int(b[0:2])
if p<0:
p = 24 + p
m = int(a[3:])-int(b[3:])
if m<0:
m= 60+m
p-=1
if len(str(p))==1:
if p==0:
p = str(p)+'0'
else:
p= '0'+str(p)
if len(str(m))==1:
if m==0:
m= str(m)+'0'
else:
m= '0' + str(m)
print(str(p)+':'+str(m))
| Title: George and Sleep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George woke up and saw the current time *s* on the digital clock. Besides, George knows that he has slept for time *t*.
Help George! Write a program that will, given time *s* and *t*, determine the time *p* when George went to bed. Note that George could have gone to bed yesterday relatively to the current time (see the second test sample).
Input Specification:
The first line contains current time *s* as a string in the format "hh:mm". The second line contains time *t* in the format "hh:mm" — the duration of George's sleep. It is guaranteed that the input contains the correct time in the 24-hour format, that is, 00<=≤<=*hh*<=≤<=23, 00<=≤<=*mm*<=≤<=59.
Output Specification:
In the single line print time *p* — the time George went to bed in the format similar to the format of the time in the input.
Demo Input:
['05:50\n05:44\n', '00:00\n01:00\n', '00:01\n00:00\n']
Demo Output:
['00:06\n', '23:00\n', '00:01\n']
Note:
In the first sample George went to bed at "00:06". Note that you should print the time only in the format "00:06". That's why answers "0:06", "00:6" and others will be considered incorrect.
In the second sample, George went to bed yesterday.
In the third sample, George didn't do to bed at all. | ```python
a=input()
b=input()
p=int(a[0:2])-int(b[0:2])
if p<0:
p = 24 + p
m = int(a[3:])-int(b[3:])
if m<0:
m= 60+m
p-=1
if len(str(p))==1:
if p==0:
p = str(p)+'0'
else:
p= '0'+str(p)
if len(str(m))==1:
if m==0:
m= str(m)+'0'
else:
m= '0' + str(m)
print(str(p)+':'+str(m))
``` | 0 | |
118 | A | String Task | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task. | The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive. | Print the resulting string. It is guaranteed that this string is not empty. | [
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] | [
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] | none | 500 | [
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": "... | 1,698,323,045 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 14 | 122 | 0 | s=input().lower()
v=['a','e','i','o','u']
v=set(v)
for i in s:
if i not in v:
print(f".{i}",end="")
| Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none | ```python
s=input().lower()
v=['a','e','i','o','u']
v=set(v)
for i in s:
if i not in v:
print(f".{i}",end="")
``` | 0 | |
19 | B | Checkout Assistant | PROGRAMMING | 1,900 | [
"dp"
] | B. Checkout Assistant | 1 | 256 | Bob came to a cash & carry store, put *n* items into his trolley, and went to the checkout counter to pay. Each item is described by its price *c**i* and time *t**i* in seconds that a checkout assistant spends on this item. While the checkout assistant is occupied with some item, Bob can steal some other items from his trolley. To steal one item Bob needs exactly 1 second. What is the minimum amount of money that Bob will have to pay to the checkout assistant? Remember, please, that it is Bob, who determines the order of items for the checkout assistant. | The first input line contains number *n* (1<=≤<=*n*<=≤<=2000). In each of the following *n* lines each item is described by a pair of numbers *t**i*, *c**i* (0<=≤<=*t**i*<=≤<=2000,<=1<=≤<=*c**i*<=≤<=109). If *t**i* is 0, Bob won't be able to steal anything, while the checkout assistant is occupied with item *i*. | Output one number — answer to the problem: what is the minimum amount of money that Bob will have to pay. | [
"4\n2 10\n0 20\n1 5\n1 3\n",
"3\n0 1\n0 10\n0 100\n"
] | [
"8\n",
"111\n"
] | none | 0 | [
{
"input": "4\n2 10\n0 20\n1 5\n1 3",
"output": "8"
},
{
"input": "3\n0 1\n0 10\n0 100",
"output": "111"
},
{
"input": "2\n0 635254032\n0 75159864",
"output": "710413896"
},
{
"input": "2\n0 861438648\n1 469893784",
"output": "469893784"
},
{
"input": "2\n2 876232... | 1,658,308,342 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 34 | 140 | 5,734,400 | n=int(input())
ar=[float('inf')]*(n+1)
ar[0]=0
for i in range(n):
t,c=map(int,input().split())
for j in range(n-1,-1,-1):
w=min(j+t+1,n)
ar[w]=min(ar[w],ar[j]+c)
print(ar[n])
| Title: Checkout Assistant
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob came to a cash & carry store, put *n* items into his trolley, and went to the checkout counter to pay. Each item is described by its price *c**i* and time *t**i* in seconds that a checkout assistant spends on this item. While the checkout assistant is occupied with some item, Bob can steal some other items from his trolley. To steal one item Bob needs exactly 1 second. What is the minimum amount of money that Bob will have to pay to the checkout assistant? Remember, please, that it is Bob, who determines the order of items for the checkout assistant.
Input Specification:
The first input line contains number *n* (1<=≤<=*n*<=≤<=2000). In each of the following *n* lines each item is described by a pair of numbers *t**i*, *c**i* (0<=≤<=*t**i*<=≤<=2000,<=1<=≤<=*c**i*<=≤<=109). If *t**i* is 0, Bob won't be able to steal anything, while the checkout assistant is occupied with item *i*.
Output Specification:
Output one number — answer to the problem: what is the minimum amount of money that Bob will have to pay.
Demo Input:
['4\n2 10\n0 20\n1 5\n1 3\n', '3\n0 1\n0 10\n0 100\n']
Demo Output:
['8\n', '111\n']
Note:
none | ```python
n=int(input())
ar=[float('inf')]*(n+1)
ar[0]=0
for i in range(n):
t,c=map(int,input().split())
for j in range(n-1,-1,-1):
w=min(j+t+1,n)
ar[w]=min(ar[w],ar[j]+c)
print(ar[n])
``` | 3.919319 |
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"... | 1,691,503,599 | 2,147,483,647 | PyPy 3 | OK | TESTS | 20 | 93 | 0 | from math import ceil
n,m,a = map(int,input().split())
x = int(ceil(n/a)*ceil(m/a))
print(x) | 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
from math import ceil
n,m,a = map(int,input().split())
x = int(ceil(n/a)*ceil(m/a))
print(x)
``` | 3.9535 |
234 | D | Cinema | PROGRAMMING | 1,600 | [
"implementation"
] | null | null | Overall there are *m* actors in Berland. Each actor has a personal identifier — an integer from 1 to *m* (distinct actors have distinct identifiers). Vasya likes to watch Berland movies with Berland actors, and he has *k* favorite actors. He watched the movie trailers for the next month and wrote the following information for every movie: the movie title, the number of actors who starred in it, and the identifiers of these actors. Besides, he managed to copy the movie titles and how many actors starred there, but he didn't manage to write down the identifiers of some actors. Vasya looks at his records and wonders which movies may be his favourite, and which ones may not be. Once Vasya learns the exact cast of all movies, his favorite movies will be determined as follows: a movie becomes favorite movie, if no other movie from Vasya's list has more favorite actors.
Help the boy to determine the following for each movie:
- whether it surely will be his favourite movie;- whether it surely won't be his favourite movie; - can either be favourite or not. | The first line of the input contains two integers *m* and *k* (1<=≤<=*m*<=≤<=100,<=1<=≤<=*k*<=≤<=*m*) — the number of actors in Berland and the number of Vasya's favourite actors.
The second line contains *k* distinct integers *a**i* (1<=≤<=*a**i*<=≤<=*m*) — the identifiers of Vasya's favourite actors.
The third line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of movies in Vasya's list.
Then follow *n* blocks of lines, each block contains a movie's description. The *i*-th movie's description contains three lines:
- the first line contains string *s**i* (*s**i* consists of lowercase English letters and can have the length of from 1 to 10 characters, inclusive) — the movie's title, - the second line contains a non-negative integer *d**i* (1<=≤<=*d**i*<=≤<=*m*) — the number of actors who starred in this movie,- the third line has *d**i* integers *b**i*,<=*j* (0<=≤<=*b**i*,<=*j*<=≤<=*m*) — the identifiers of the actors who star in this movie. If *b**i*,<=*j*<==<=0, than Vasya doesn't remember the identifier of the *j*-th actor. It is guaranteed that the list of actors for a movie doesn't contain the same actors.
All movies have distinct names. The numbers on the lines are separated by single spaces. | Print *n* lines in the output. In the *i*-th line print:
- 0, if the *i*-th movie will surely be the favourite; - 1, if the *i*-th movie won't surely be the favourite; - 2, if the *i*-th movie can either be favourite, or not favourite. | [
"5 3\n1 2 3\n6\nfirstfilm\n3\n0 0 0\nsecondfilm\n4\n0 0 4 5\nthirdfilm\n1\n2\nfourthfilm\n1\n5\nfifthfilm\n1\n4\nsixthfilm\n2\n1 0\n",
"5 3\n1 3 5\n4\njumanji\n3\n0 0 0\ntheeagle\n5\n1 2 3 4 0\nmatrix\n3\n2 4 0\nsourcecode\n2\n2 4\n"
] | [
"2\n2\n1\n1\n1\n2\n",
"2\n0\n1\n1\n"
] | Note to the second sample:
- Movie jumanji can theoretically have from 1 to 3 Vasya's favourite actors. - Movie theeagle has all three favourite actors, as the actor Vasya failed to remember, can only have identifier 5. - Movie matrix can have exactly one favourite actor. - Movie sourcecode doesn't have any favourite actors.
Thus, movie theeagle will surely be favourite, movies matrix and sourcecode won't surely be favourite, and movie jumanji can be either favourite (if it has all three favourite actors), or not favourite. | 0 | [
{
"input": "5 3\n1 2 3\n6\nfirstfilm\n3\n0 0 0\nsecondfilm\n4\n0 0 4 5\nthirdfilm\n1\n2\nfourthfilm\n1\n5\nfifthfilm\n1\n4\nsixthfilm\n2\n1 0",
"output": "2\n2\n1\n1\n1\n2"
},
{
"input": "5 3\n1 3 5\n4\njumanji\n3\n0 0 0\ntheeagle\n5\n1 2 3 4 0\nmatrix\n3\n2 4 0\nsourcecode\n2\n2 4",
"output": "... | 1,620,453,923 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 124 | 6,963,200 | import sys
try:
sys.stdin = open('input.txt')
sys.stdout = open('output.txt', 'w')
except:
pass
def compl(n, s):
return set(filter(lambda x: x not in s, range(1, n + 1)))
m, k = list(map(int, input().split()))
id = list(map(int, input().split()))
n = int(input())
favorite = set(id)
if n == 1:
print(0)
exit()
best = []
guaranteed = []
for i in range(n):
name = input()
cnt = int(input())
tmp = list(map(int, input().split()))
s = set()
cnt_zero = 0
for elem in tmp:
if elem != 0:
s.add(elem)
else:
cnt_zero += 1
inter = s & favorite
comp = compl(m, favorite | s)
pred = max(0, cnt_zero - len(comp))
g = len(inter) + pred
guaranteed.append(g)
cnt_zero -= pred
b = g + min(cnt_zero, len(favorite - s) - pred)
best.append(b)
max_g = max(guaranteed)
for i in range(n):
max_b = 0
for j in range(n):
if i == j:
continue
max_b = max(max_b, best[j])
if guaranteed[i] >= max_b:
print(0)
elif best[i] < max_g:
print(1)
else:
print(2)
| Title: Cinema
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Overall there are *m* actors in Berland. Each actor has a personal identifier — an integer from 1 to *m* (distinct actors have distinct identifiers). Vasya likes to watch Berland movies with Berland actors, and he has *k* favorite actors. He watched the movie trailers for the next month and wrote the following information for every movie: the movie title, the number of actors who starred in it, and the identifiers of these actors. Besides, he managed to copy the movie titles and how many actors starred there, but he didn't manage to write down the identifiers of some actors. Vasya looks at his records and wonders which movies may be his favourite, and which ones may not be. Once Vasya learns the exact cast of all movies, his favorite movies will be determined as follows: a movie becomes favorite movie, if no other movie from Vasya's list has more favorite actors.
Help the boy to determine the following for each movie:
- whether it surely will be his favourite movie;- whether it surely won't be his favourite movie; - can either be favourite or not.
Input Specification:
The first line of the input contains two integers *m* and *k* (1<=≤<=*m*<=≤<=100,<=1<=≤<=*k*<=≤<=*m*) — the number of actors in Berland and the number of Vasya's favourite actors.
The second line contains *k* distinct integers *a**i* (1<=≤<=*a**i*<=≤<=*m*) — the identifiers of Vasya's favourite actors.
The third line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of movies in Vasya's list.
Then follow *n* blocks of lines, each block contains a movie's description. The *i*-th movie's description contains three lines:
- the first line contains string *s**i* (*s**i* consists of lowercase English letters and can have the length of from 1 to 10 characters, inclusive) — the movie's title, - the second line contains a non-negative integer *d**i* (1<=≤<=*d**i*<=≤<=*m*) — the number of actors who starred in this movie,- the third line has *d**i* integers *b**i*,<=*j* (0<=≤<=*b**i*,<=*j*<=≤<=*m*) — the identifiers of the actors who star in this movie. If *b**i*,<=*j*<==<=0, than Vasya doesn't remember the identifier of the *j*-th actor. It is guaranteed that the list of actors for a movie doesn't contain the same actors.
All movies have distinct names. The numbers on the lines are separated by single spaces.
Output Specification:
Print *n* lines in the output. In the *i*-th line print:
- 0, if the *i*-th movie will surely be the favourite; - 1, if the *i*-th movie won't surely be the favourite; - 2, if the *i*-th movie can either be favourite, or not favourite.
Demo Input:
['5 3\n1 2 3\n6\nfirstfilm\n3\n0 0 0\nsecondfilm\n4\n0 0 4 5\nthirdfilm\n1\n2\nfourthfilm\n1\n5\nfifthfilm\n1\n4\nsixthfilm\n2\n1 0\n', '5 3\n1 3 5\n4\njumanji\n3\n0 0 0\ntheeagle\n5\n1 2 3 4 0\nmatrix\n3\n2 4 0\nsourcecode\n2\n2 4\n']
Demo Output:
['2\n2\n1\n1\n1\n2\n', '2\n0\n1\n1\n']
Note:
Note to the second sample:
- Movie jumanji can theoretically have from 1 to 3 Vasya's favourite actors. - Movie theeagle has all three favourite actors, as the actor Vasya failed to remember, can only have identifier 5. - Movie matrix can have exactly one favourite actor. - Movie sourcecode doesn't have any favourite actors.
Thus, movie theeagle will surely be favourite, movies matrix and sourcecode won't surely be favourite, and movie jumanji can be either favourite (if it has all three favourite actors), or not favourite. | ```python
import sys
try:
sys.stdin = open('input.txt')
sys.stdout = open('output.txt', 'w')
except:
pass
def compl(n, s):
return set(filter(lambda x: x not in s, range(1, n + 1)))
m, k = list(map(int, input().split()))
id = list(map(int, input().split()))
n = int(input())
favorite = set(id)
if n == 1:
print(0)
exit()
best = []
guaranteed = []
for i in range(n):
name = input()
cnt = int(input())
tmp = list(map(int, input().split()))
s = set()
cnt_zero = 0
for elem in tmp:
if elem != 0:
s.add(elem)
else:
cnt_zero += 1
inter = s & favorite
comp = compl(m, favorite | s)
pred = max(0, cnt_zero - len(comp))
g = len(inter) + pred
guaranteed.append(g)
cnt_zero -= pred
b = g + min(cnt_zero, len(favorite - s) - pred)
best.append(b)
max_g = max(guaranteed)
for i in range(n):
max_b = 0
for j in range(n):
if i == j:
continue
max_b = max(max_b, best[j])
if guaranteed[i] >= max_b:
print(0)
elif best[i] < max_g:
print(1)
else:
print(2)
``` | 3 | |
892 | A | Greed | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not! | The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans. | Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower). | [
"2\n3 5\n3 6\n",
"3\n6 8 9\n6 10 12\n",
"5\n0 0 5 0 0\n1 1 8 10 5\n",
"4\n4 1 0 3\n5 2 2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"YES\n"
] | In the first sample, there are already 2 cans, so the answer is "YES". | 500 | [
{
"input": "2\n3 5\n3 6",
"output": "YES"
},
{
"input": "3\n6 8 9\n6 10 12",
"output": "NO"
},
{
"input": "5\n0 0 5 0 0\n1 1 8 10 5",
"output": "YES"
},
{
"input": "4\n4 1 0 3\n5 2 2 3",
"output": "YES"
},
{
"input": "10\n9 10 24 11 1 7 8 3 28 14\n86 20 34 11 22 9... | 1,678,368,072 | 2,147,483,647 | PyPy 3 | OK | TESTS | 39 | 280 | 13,414,400 | n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
largest_can = 0
second_largest_can = 0
for i in b:
if i >= largest_can:
second_largest_can = largest_can
largest_can = i
elif i > second_largest_can:
second_largest_can = i
remaining = largest_can + second_largest_can
flag = "YES"
for i in range(len(a)):
remaining -= a[i]
if remaining < 0:
flag = "NO"
break
print(flag) | Title: Greed
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input Specification:
The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans.
Output Specification:
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Demo Input:
['2\n3 5\n3 6\n', '3\n6 8 9\n6 10 12\n', '5\n0 0 5 0 0\n1 1 8 10 5\n', '4\n4 1 0 3\n5 2 2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'YES\n']
Note:
In the first sample, there are already 2 cans, so the answer is "YES". | ```python
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
largest_can = 0
second_largest_can = 0
for i in b:
if i >= largest_can:
second_largest_can = largest_can
largest_can = i
elif i > second_largest_can:
second_largest_can = i
remaining = largest_can + second_largest_can
flag = "YES"
for i in range(len(a)):
remaining -= a[i]
if remaining < 0:
flag = "NO"
break
print(flag)
``` | 3 | |
989 | A | A Blend of Springtime | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | "What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order. | The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively. | Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower). | [
".BAC.\n",
"AA..CB\n"
] | [
"Yes\n",
"No\n"
] | In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | 500 | [
{
"input": ".BAC.",
"output": "Yes"
},
{
"input": "AA..CB",
"output": "No"
},
{
"input": ".",
"output": "No"
},
{
"input": "ACB.AAAAAA",
"output": "Yes"
},
{
"input": "B.BC.BBBCA",
"output": "Yes"
},
{
"input": "BA..CAB..B",
"output": "Yes"
},
... | 1,618,490,635 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 62 | 0 | s = input()
flag = 0
for i in range(len(s)-2):
if s[i:i+3] == 'BAC' or s[i:i+3] == 'ABC' or s[i:i+3] == 'CAB' or s[i:i+3] == 'ACB' or s[i:i+3] == 'BCA' or s[i:i+3] == 'CBA':
flag = 1
print("Yes")
break
if flag != 1:
print('No') | Title: A Blend of Springtime
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input Specification:
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output Specification:
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['.BAC.\n', 'AA..CB\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | ```python
s = input()
flag = 0
for i in range(len(s)-2):
if s[i:i+3] == 'BAC' or s[i:i+3] == 'ABC' or s[i:i+3] == 'CAB' or s[i:i+3] == 'ACB' or s[i:i+3] == 'BCA' or s[i:i+3] == 'CBA':
flag = 1
print("Yes")
break
if flag != 1:
print('No')
``` | 3 | |
977 | D | Divide by three, multiply by two | PROGRAMMING | 1,400 | [
"dfs and similar",
"math",
"sortings"
] | null | null | Polycarp likes to play with numbers. He takes some integer number $x$, writes it down on the board, and then performs with it $n - 1$ operations of the two kinds:
- divide the number $x$ by $3$ ($x$ must be divisible by $3$); - multiply the number $x$ by $2$.
After each operation, Polycarp writes down the result on the board and replaces $x$ by the result. So there will be $n$ numbers on the board after all.
You are given a sequence of length $n$ — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.
Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.
It is guaranteed that the answer exists. | The first line of the input contatins an integer number $n$ ($2 \le n \le 100$) — the number of the elements in the sequence. The second line of the input contains $n$ integer numbers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 3 \cdot 10^{18}$) — rearranged (reordered) sequence that Polycarp can wrote down on the board. | Print $n$ integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.
It is guaranteed that the answer exists. | [
"6\n4 8 6 3 12 9\n",
"4\n42 28 84 126\n",
"2\n1000000000000000000 3000000000000000000\n"
] | [
"9 3 6 12 4 8 \n",
"126 42 84 28 \n",
"3000000000000000000 1000000000000000000 \n"
] | In the first example the given sequence can be rearranged in the following way: $[9, 3, 6, 12, 4, 8]$. It can match possible Polycarp's game which started with $x = 9$. | 0 | [
{
"input": "6\n4 8 6 3 12 9",
"output": "9 3 6 12 4 8 "
},
{
"input": "4\n42 28 84 126",
"output": "126 42 84 28 "
},
{
"input": "2\n1000000000000000000 3000000000000000000",
"output": "3000000000000000000 1000000000000000000 "
},
{
"input": "19\n46875000000000000 73242187500... | 1,695,773,719 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 46 | 32,358,400 | ln = int(input())
arr = list(map(int,input().split()))
dc = {}
ans = []
dcans = {}
def check(num,cnt,anss):
global dc,ln,ans,dcans
if num in dc and dc[num]:
if cnt==ln:
ans = anss
dcans = dc.copy()
return
dc[num]-=1
else:
return
if cnt>ln:
return
if num%3==0:
check(int(num/3),cnt+1,anss+[num])
check(num*2,cnt+1,anss+[num])
for i in arr:
if i in dc:
dc[i]+=1
else:
dc[i]=1
spare = dc.copy()
for i in arr:
check(i,1,[])
dc = spare.copy()
for i in dcans:
if dcans[i]:
ans.append(i)
break
print(*ans) | Title: Divide by three, multiply by two
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp likes to play with numbers. He takes some integer number $x$, writes it down on the board, and then performs with it $n - 1$ operations of the two kinds:
- divide the number $x$ by $3$ ($x$ must be divisible by $3$); - multiply the number $x$ by $2$.
After each operation, Polycarp writes down the result on the board and replaces $x$ by the result. So there will be $n$ numbers on the board after all.
You are given a sequence of length $n$ — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.
Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.
It is guaranteed that the answer exists.
Input Specification:
The first line of the input contatins an integer number $n$ ($2 \le n \le 100$) — the number of the elements in the sequence. The second line of the input contains $n$ integer numbers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 3 \cdot 10^{18}$) — rearranged (reordered) sequence that Polycarp can wrote down on the board.
Output Specification:
Print $n$ integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.
It is guaranteed that the answer exists.
Demo Input:
['6\n4 8 6 3 12 9\n', '4\n42 28 84 126\n', '2\n1000000000000000000 3000000000000000000\n']
Demo Output:
['9 3 6 12 4 8 \n', '126 42 84 28 \n', '3000000000000000000 1000000000000000000 \n']
Note:
In the first example the given sequence can be rearranged in the following way: $[9, 3, 6, 12, 4, 8]$. It can match possible Polycarp's game which started with $x = 9$. | ```python
ln = int(input())
arr = list(map(int,input().split()))
dc = {}
ans = []
dcans = {}
def check(num,cnt,anss):
global dc,ln,ans,dcans
if num in dc and dc[num]:
if cnt==ln:
ans = anss
dcans = dc.copy()
return
dc[num]-=1
else:
return
if cnt>ln:
return
if num%3==0:
check(int(num/3),cnt+1,anss+[num])
check(num*2,cnt+1,anss+[num])
for i in arr:
if i in dc:
dc[i]+=1
else:
dc[i]=1
spare = dc.copy()
for i in arr:
check(i,1,[])
dc = spare.copy()
for i in dcans:
if dcans[i]:
ans.append(i)
break
print(*ans)
``` | 0 | |
189 | A | Cut Ribbon | PROGRAMMING | 1,300 | [
"brute force",
"dp"
] | null | null | Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting. | The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide. | Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists. | [
"5 5 3 2\n",
"7 5 5 2\n"
] | [
"2\n",
"2\n"
] | In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2. | 500 | [
{
"input": "5 5 3 2",
"output": "2"
},
{
"input": "7 5 5 2",
"output": "2"
},
{
"input": "4 4 4 4",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "4000 1 2 3",
"output": "4000"
},
{
"input": "4000 3 4 5",
"output": "1333"
},
... | 1,683,968,840 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 4 | 1,000 | 0 | m = list(map(int,input().split()))
n = m.pop(0)
k = 0
for i in range(0,n//m[0]+1):
for j in range(0,(n-m[0]*i)//m[1]+1):
z = n-i*m[0]-j*m[1]
if z%m[2]==0:
k = max(k,i+j+z//m[2])
print(k)
| Title: Cut Ribbon
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
Input Specification:
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
Output Specification:
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
Demo Input:
['5 5 3 2\n', '7 5 5 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2. | ```python
m = list(map(int,input().split()))
n = m.pop(0)
k = 0
for i in range(0,n//m[0]+1):
for j in range(0,(n-m[0]*i)//m[1]+1):
z = n-i*m[0]-j*m[1]
if z%m[2]==0:
k = max(k,i+j+z//m[2])
print(k)
``` | 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... | 1,696,141,923 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 62 | 0 | m, n = [int(e) for e in input().strip().split()]
cells_amount = m * n
print((m*n)//2)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
m, n = [int(e) for e in input().strip().split()]
cells_amount = m * n
print((m*n)//2)
``` | 3.9845 |
0 | none | none | none | 0 | [
"none"
] | null | null | Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular *n*<=×<=*m* field, split into *nm* square 1<=×<=1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (*x*1,<=*y*1) and (*x*2,<=*y*2) equals exactly (*x*1<=-<=*x*2)2<=+<=(*y*1<=-<=*y*2)2. Now not all *nm* squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2,<=2), then he cannot put soldiers in the squares (1,<=4), (3,<=4), (4,<=1) and (4,<=3) — each of them will conflict with the soldier in the square (2,<=2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse. | The single line contains space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) that represent the size of the drill exercise field. | Print the desired maximum number of warriors. | [
"2 4\n",
"3 4\n"
] | [
"4",
"6"
] | In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner: | 0 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "4 3",
"output": "6"
},
{
"input": "4 2",
"output": "4"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "3 5",
"output"... | 1,676,467,876 | 4,576 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 218 | 4,198,400 | n,m = map(int,input().split())
ans = 0
for i in range(2):
rs = (n+2)//3
rrss = n-rs
p = m//10
q = m%10
tmp = rs*p*5+rs*min(q, 5) + p*rrss
if q >= 8:
tmp += rrss
ans = max(ans,tmp)
dp = [[0]*m for i in range(n)]
dp[0][0] = 1
cnt = 1
for i in range(n):
for j in range(m):
if i == 0 and j == 0: continue
now = 0
if i >= 5:
now |= dp[i-5][j]
if j >= 5:
now |= dp[i][j-5]
if i >= 2 and j >= 1:
now |= dp[i-2][j-1]
if i >= 2 and j+1 <m:
now |= dp[i-2][j+1]
if i >= 1 and j >= 2:
now |= dp[i-1][j-2]
if i >= 1 and j+2 < m:
now |= dp[i-1][j+2]
dp[i][j] = 1-now
cnt += 1-now
ans = max(ans, cnt)
n,m = m,n
print(ans) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular *n*<=×<=*m* field, split into *nm* square 1<=×<=1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (*x*1,<=*y*1) and (*x*2,<=*y*2) equals exactly (*x*1<=-<=*x*2)2<=+<=(*y*1<=-<=*y*2)2. Now not all *nm* squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2,<=2), then he cannot put soldiers in the squares (1,<=4), (3,<=4), (4,<=1) and (4,<=3) — each of them will conflict with the soldier in the square (2,<=2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input Specification:
The single line contains space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) that represent the size of the drill exercise field.
Output Specification:
Print the desired maximum number of warriors.
Demo Input:
['2 4\n', '3 4\n']
Demo Output:
['4', '6']
Note:
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner: | ```python
n,m = map(int,input().split())
ans = 0
for i in range(2):
rs = (n+2)//3
rrss = n-rs
p = m//10
q = m%10
tmp = rs*p*5+rs*min(q, 5) + p*rrss
if q >= 8:
tmp += rrss
ans = max(ans,tmp)
dp = [[0]*m for i in range(n)]
dp[0][0] = 1
cnt = 1
for i in range(n):
for j in range(m):
if i == 0 and j == 0: continue
now = 0
if i >= 5:
now |= dp[i-5][j]
if j >= 5:
now |= dp[i][j-5]
if i >= 2 and j >= 1:
now |= dp[i-2][j-1]
if i >= 2 and j+1 <m:
now |= dp[i-2][j+1]
if i >= 1 and j >= 2:
now |= dp[i-1][j-2]
if i >= 1 and j+2 < m:
now |= dp[i-1][j+2]
dp[i][j] = 1-now
cnt += 1-now
ans = max(ans, cnt)
n,m = m,n
print(ans)
``` | 0 | |
567 | A | Lineland Mail | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city | The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order. | Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city. | [
"4\n-5 -2 2 7\n",
"2\n-1 1\n"
] | [
"3 12\n3 9\n4 7\n5 12\n",
"2 2\n2 2\n"
] | none | 500 | [
{
"input": "4\n-5 -2 2 7",
"output": "3 12\n3 9\n4 7\n5 12"
},
{
"input": "2\n-1 1",
"output": "2 2\n2 2"
},
{
"input": "3\n-1 0 1",
"output": "1 2\n1 1\n1 2"
},
{
"input": "4\n-1 0 1 3",
"output": "1 4\n1 3\n1 2\n2 4"
},
{
"input": "3\n-1000000000 0 1000000000",
... | 1,675,081,715 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 54 | 187 | 16,179,200 | n=int(input())
City=[int( x ) for x in input().split()]
for i in range(n):
if i==0:
minNode=abs(City[i] - City[i+1])
maxNode = abs(City[i] - City[-1])
elif i==n-1:
minNode = abs(City[i] - City[i-1])
maxNode = abs(City[i] - City[0])
else:
minNode = min(abs(City[i]-City[i-1]), abs(City[i]-City[i+1]))
maxNode = max(abs(City[i]-City[0]), abs(City[i]-City[n-1]))
print(minNode,maxNode) | Title: Lineland Mail
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order.
Output Specification:
Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city.
Demo Input:
['4\n-5 -2 2 7\n', '2\n-1 1\n']
Demo Output:
['3 12\n3 9\n4 7\n5 12\n', '2 2\n2 2\n']
Note:
none | ```python
n=int(input())
City=[int( x ) for x in input().split()]
for i in range(n):
if i==0:
minNode=abs(City[i] - City[i+1])
maxNode = abs(City[i] - City[-1])
elif i==n-1:
minNode = abs(City[i] - City[i-1])
maxNode = abs(City[i] - City[0])
else:
minNode = min(abs(City[i]-City[i-1]), abs(City[i]-City[i+1]))
maxNode = max(abs(City[i]-City[0]), abs(City[i]-City[n-1]))
print(minNode,maxNode)
``` | 3 | |
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": "... | 1,509,712,828 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 530 | 16,486,400 | import heapq
n = int(input())
colors_lens = list(map(int, input().split()))
if len(colors_lens) % 2 == 0:
colors_lens.append(0)
heapq.heapify(colors_lens)
ans = 0
l = len(colors_lens)
while l > 1:
su = 0
su += heapq.heappop(colors_lens)
su += heapq.heappop(colors_lens)
su += heapq.heappop(colors_lens)
ans += su
heapq.heappush(colors_lens, su)
l -= 2
print(ans)
| 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
import heapq
n = int(input())
colors_lens = list(map(int, input().split()))
if len(colors_lens) % 2 == 0:
colors_lens.append(0)
heapq.heapify(colors_lens)
ans = 0
l = len(colors_lens)
while l > 1:
su = 0
su += heapq.heappop(colors_lens)
su += heapq.heappop(colors_lens)
su += heapq.heappop(colors_lens)
ans += su
heapq.heappush(colors_lens, su)
l -= 2
print(ans)
``` | 3 | |
514 | A | Chewbaсca and Number | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero. | The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca. | Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes. | [
"27\n",
"4545\n"
] | [
"22\n",
"4444\n"
] | none | 500 | [
{
"input": "27",
"output": "22"
},
{
"input": "4545",
"output": "4444"
},
{
"input": "1",
"output": "1"
},
{
"input": "9",
"output": "9"
},
{
"input": "8772",
"output": "1222"
},
{
"input": "81",
"output": "11"
},
{
"input": "71723447",
... | 1,677,128,912 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 46 | 0 | x = int(input())
y = str(x)
z = ""
for i in range(len(y)):
if (y[i]=='9'):
z+='9'
else:
if (y[i]>='5'):
z+=(str(9-int(y[i])))
else:
z+=y[i]
print(z)
| Title: Chewbaсca and Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
Output Specification:
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
Demo Input:
['27\n', '4545\n']
Demo Output:
['22\n', '4444\n']
Note:
none | ```python
x = int(input())
y = str(x)
z = ""
for i in range(len(y)):
if (y[i]=='9'):
z+='9'
else:
if (y[i]>='5'):
z+=(str(9-int(y[i])))
else:
z+=y[i]
print(z)
``` | 0 | |
788 | A | Functions again | PROGRAMMING | 1,600 | [
"dp",
"two pointers"
] | null | null | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function *f*, which is defined as follows:
In the above formula, 1<=≤<=*l*<=<<=*r*<=≤<=*n* must hold, where *n* is the size of the Main Uzhlyandian Array *a*, and |*x*| means absolute value of *x*. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of *f* among all possible values of *l* and *r* for the given array *a*. | The first line contains single integer *n* (2<=≤<=*n*<=≤<=105) — the size of the array *a*.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (-109<=≤<=*a**i*<=≤<=109) — the array elements. | Print the only integer — the maximum value of *f*. | [
"5\n1 4 2 3 1\n",
"4\n1 5 4 7\n"
] | [
"3",
"6"
] | In the first sample case, the optimal value of *f* is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of *f* is reachable only on the whole array. | 500 | [
{
"input": "5\n1 4 2 3 1",
"output": "3"
},
{
"input": "4\n1 5 4 7",
"output": "6"
},
{
"input": "8\n16 14 12 10 8 100 50 0",
"output": "92"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "50\n-5 -9 0 44 -10 37 34 -49 11 -22 -26 44 8 -13 23 -46 34 12 -24 2 -4... | 1,624,880,786 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 93 | 0 | def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().split())
def msi(): return map(str,input().split())
def li(): return list(mi())
n=ii()
a=li()
dp=[[0,0] for i in range(n)]
for i in range(1,n):
curr=abs(a[i]-a[i-1])
if dp[i-1][0]>0:
dp[i][1]=curr
dp[i][0]=curr-dp[i-1][1]
else:
dp[i][1]=curr-dp[i-1][0]
if dp[i-1][1]<0:
dp[i][0]=curr
else:
dp[i][0]=curr-dp[i-1][1]
print(dp[n-1][1])
| Title: Functions again
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function *f*, which is defined as follows:
In the above formula, 1<=≤<=*l*<=<<=*r*<=≤<=*n* must hold, where *n* is the size of the Main Uzhlyandian Array *a*, and |*x*| means absolute value of *x*. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of *f* among all possible values of *l* and *r* for the given array *a*.
Input Specification:
The first line contains single integer *n* (2<=≤<=*n*<=≤<=105) — the size of the array *a*.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (-109<=≤<=*a**i*<=≤<=109) — the array elements.
Output Specification:
Print the only integer — the maximum value of *f*.
Demo Input:
['5\n1 4 2 3 1\n', '4\n1 5 4 7\n']
Demo Output:
['3', '6']
Note:
In the first sample case, the optimal value of *f* is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of *f* is reachable only on the whole array. | ```python
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().split())
def msi(): return map(str,input().split())
def li(): return list(mi())
n=ii()
a=li()
dp=[[0,0] for i in range(n)]
for i in range(1,n):
curr=abs(a[i]-a[i-1])
if dp[i-1][0]>0:
dp[i][1]=curr
dp[i][0]=curr-dp[i-1][1]
else:
dp[i][1]=curr-dp[i-1][0]
if dp[i-1][1]<0:
dp[i][0]=curr
else:
dp[i][0]=curr-dp[i-1][1]
print(dp[n-1][1])
``` | 0 | |
137 | B | Permutation | PROGRAMMING | 1,000 | [
"greedy"
] | null | null | "Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer). | The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*). | Print the only number — the minimum number of changes needed to get the permutation. | [
"3\n3 1 2\n",
"2\n2 2\n",
"5\n5 3 3 3 1\n"
] | [
"0\n",
"1\n",
"2\n"
] | The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2. | 1,000 | [
{
"input": "3\n3 1 2",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "5\n5 3 3 3 1",
"output": "2"
},
{
"input": "5\n6 6 6 6 6",
"output": "5"
},
{
"input": "10\n1 1 2 2 8 8 7 7 9 9",
"output": "5"
},
{
"input": "8\n9 8 7 6 5 4 3 2"... | 1,610,569,594 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 154 | 819,200 | from collections import Counter
n = int(input())
s = [int(i) for i in input().split()]
c = Counter(s)
ans = 0
for k, v in c.items():
if k <= n:
ans += v - 1
else:
ans += v
print(ans)
| Title: Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer).
Input Specification:
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*).
Output Specification:
Print the only number — the minimum number of changes needed to get the permutation.
Demo Input:
['3\n3 1 2\n', '2\n2 2\n', '5\n5 3 3 3 1\n']
Demo Output:
['0\n', '1\n', '2\n']
Note:
The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2. | ```python
from collections import Counter
n = int(input())
s = [int(i) for i in input().split()]
c = Counter(s)
ans = 0
for k, v in c.items():
if k <= n:
ans += v - 1
else:
ans += v
print(ans)
``` | 3 | |
272 | A | Dima and Friends | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space. | In a single line print the answer to the problem. | [
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] | [
"3\n",
"2\n",
"3\n"
] | In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | 500 | [
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "... | 1,588,274,581 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 216 | 0 | n = int(input())+1
f = sum(map(int,input().split()))
l = [x for x in range(1,6) if (f+x)%n+1!=1]
print(len(l)) | Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | ```python
n = int(input())+1
f = sum(map(int,input().split()))
l = [x for x in range(1,6) if (f+x)%n+1!=1]
print(len(l))
``` | 0 | |
1,011 | A | Stages | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once. | The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once. | Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all. | [
"5 3\nxyabd\n",
"7 4\nproblem\n",
"2 2\nab\n",
"12 1\nabaabbaaabbb\n"
] | [
"29",
"34",
"-1",
"1"
] | In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1. | 500 | [
{
"input": "5 3\nxyabd",
"output": "29"
},
{
"input": "7 4\nproblem",
"output": "34"
},
{
"input": "2 2\nab",
"output": "-1"
},
{
"input": "12 1\nabaabbaaabbb",
"output": "1"
},
{
"input": "50 13\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": ... | 1,602,521,710 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 109 | 204,800 | inp = input().split(' ')
n,k = int(inp[0]),int(inp[1])
s = input()
arr = []
for i in s:
arr.append(ord(i)-96)
arr.sort()
sum,count,i = 0,0,0
while i < n:
sum+=arr[i]
if arr[i]+1 == arr[i+1]:
i+=2
else:
i+=1
count +=1
if count == k:
break
if count < k:
print(-1)
else:
print(sum) | Title: Stages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
Input Specification:
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
Output Specification:
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
Demo Input:
['5 3\nxyabd\n', '7 4\nproblem\n', '2 2\nab\n', '12 1\nabaabbaaabbb\n']
Demo Output:
['29', '34', '-1', '1']
Note:
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1. | ```python
inp = input().split(' ')
n,k = int(inp[0]),int(inp[1])
s = input()
arr = []
for i in s:
arr.append(ord(i)-96)
arr.sort()
sum,count,i = 0,0,0
while i < n:
sum+=arr[i]
if arr[i]+1 == arr[i+1]:
i+=2
else:
i+=1
count +=1
if count == k:
break
if count < k:
print(-1)
else:
print(sum)
``` | 0 | |
257 | C | View Angle | PROGRAMMING | 1,800 | [
"brute force",
"geometry",
"math"
] | null | null | Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of mannequins.
Next *n* lines contain two space-separated integers each: *x**i*,<=*y**i* (|*x**i*|,<=|*y**i*|<=≤<=1000) — the coordinates of the *i*-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane. | Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10<=-<=6. | [
"2\n2 0\n0 2\n",
"3\n2 0\n0 2\n-2 2\n",
"4\n2 0\n0 2\n-2 0\n0 -2\n",
"2\n2 1\n1 2\n"
] | [
"90.0000000000\n",
"135.0000000000\n",
"270.0000000000\n",
"36.8698976458\n"
] | Solution for the first sample test is shown below:
Solution for the second sample test is shown below:
Solution for the third sample test is shown below:
Solution for the fourth sample test is shown below: | 1,500 | [
{
"input": "2\n2 0\n0 2",
"output": "90.0000000000"
},
{
"input": "3\n2 0\n0 2\n-2 2",
"output": "135.0000000000"
},
{
"input": "4\n2 0\n0 2\n-2 0\n0 -2",
"output": "270.0000000000"
},
{
"input": "2\n2 1\n1 2",
"output": "36.8698976458"
},
{
"input": "1\n1 1",
... | 1,670,851,629 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | import math as m
n = int(input())
for i in range(n):
x,y = map(int,input().split())
if x==0 and y==0:
theta = 0
elif x==0 or y==0:
if x==0:
if y>0:
theta = 90
else:
theta = 270
if y==0:
if x>0:
theta = 0
else:
theta = 180
elif x>=0 and y>=0:
theta = 180*m.atan(y/x)/m.pi
elif x>=0 and y<0:
theta = 360+180*m.atan(y/x)/m.pi
elif x<0 and y>=0:
theta = 180+180*m.atan(y/x)/m.pi
else:
theta = 180+180*m.atan(y/x)/m.pi
a.append(theta)
a = sorted(a)
m = len(a)
mx = 0
for i in range(1,m):
mx = max(mx,a[i]-a[i-1])
#print(a)
y = 360-a[-1]
x = a[0]
mx = max(mx,x+y)
print(360-mx) | Title: View Angle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of mannequins.
Next *n* lines contain two space-separated integers each: *x**i*,<=*y**i* (|*x**i*|,<=|*y**i*|<=≤<=1000) — the coordinates of the *i*-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output Specification:
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10<=-<=6.
Demo Input:
['2\n2 0\n0 2\n', '3\n2 0\n0 2\n-2 2\n', '4\n2 0\n0 2\n-2 0\n0 -2\n', '2\n2 1\n1 2\n']
Demo Output:
['90.0000000000\n', '135.0000000000\n', '270.0000000000\n', '36.8698976458\n']
Note:
Solution for the first sample test is shown below:
Solution for the second sample test is shown below:
Solution for the third sample test is shown below:
Solution for the fourth sample test is shown below: | ```python
import math as m
n = int(input())
for i in range(n):
x,y = map(int,input().split())
if x==0 and y==0:
theta = 0
elif x==0 or y==0:
if x==0:
if y>0:
theta = 90
else:
theta = 270
if y==0:
if x>0:
theta = 0
else:
theta = 180
elif x>=0 and y>=0:
theta = 180*m.atan(y/x)/m.pi
elif x>=0 and y<0:
theta = 360+180*m.atan(y/x)/m.pi
elif x<0 and y>=0:
theta = 180+180*m.atan(y/x)/m.pi
else:
theta = 180+180*m.atan(y/x)/m.pi
a.append(theta)
a = sorted(a)
m = len(a)
mx = 0
for i in range(1,m):
mx = max(mx,a[i]-a[i-1])
#print(a)
y = 360-a[-1]
x = a[0]
mx = max(mx,x+y)
print(360-mx)
``` | -1 | |
621 | A | Wet Shark and Odd and Even | PROGRAMMING | 900 | [
"implementation"
] | null | null | Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0. | The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive. | Print the maximum possible even sum that can be obtained if we use some of the given integers. | [
"3\n1 2 3\n",
"5\n999999999 999999999 999999999 999999999 999999999\n"
] | [
"6",
"3999999996"
] | In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | 500 | [
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "5\n999999999 999999999 999999999 999999999 999999999",
"output": "3999999996"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98",
"output": "870"
},
{
"input": "15\... | 1,515,847,903 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 5,632,000 | l = int(input())
x = str(input())
Sum = 0
odd = []
temp = 0
for n in range(0,len(x)):
if x[n] == '\n':
if(temp%2 == 0):
Sum += temp
else:
odd += [temp]
temp = 0
else:
temp = temp*10 + int(x[n])
if len(odd)%2 == 0:
Sum+=add1(odd)
else:
Sum+=add2(odd)
print(sum)
def add1(x):
temp = 0
for n in range(0,len(x)):
temp += x[n]
def add2(x):
temp = 0
min = x[0]
for n in range(0,len(x)):
if x[n] < min:
temp += min
min = x[n]
else:
temp += x[n]
| Title: Wet Shark and Odd and Even
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
Input Specification:
The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Output Specification:
Print the maximum possible even sum that can be obtained if we use some of the given integers.
Demo Input:
['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n']
Demo Output:
['6', '3999999996']
Note:
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | ```python
l = int(input())
x = str(input())
Sum = 0
odd = []
temp = 0
for n in range(0,len(x)):
if x[n] == '\n':
if(temp%2 == 0):
Sum += temp
else:
odd += [temp]
temp = 0
else:
temp = temp*10 + int(x[n])
if len(odd)%2 == 0:
Sum+=add1(odd)
else:
Sum+=add2(odd)
print(sum)
def add1(x):
temp = 0
for n in range(0,len(x)):
temp += x[n]
def add2(x):
temp = 0
min = x[0]
for n in range(0,len(x)):
if x[n] < min:
temp += min
min = x[n]
else:
temp += x[n]
``` | -1 | |
228 | A | Is your horseshoe on the other hoof? | PROGRAMMING | 800 | [
"implementation"
] | null | null | Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. | The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers. | Print a single integer — the minimum number of horseshoes Valera needs to buy. | [
"1 7 3 3\n",
"7 7 7 7\n"
] | [
"1\n",
"3\n"
] | none | 500 | [
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 58697... | 1,699,084,210 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | n = str(input())
n = n.split()
moreuseful = set(n)
horseshoes = len(n) - len(moreuseful)
print(horseshoes) | Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
n = str(input())
n = n.split()
moreuseful = set(n)
horseshoes = len(n) - len(moreuseful)
print(horseshoes)
``` | 3 | |
805 | B | 3-palindrome | PROGRAMMING | 1,000 | [
"constructive algorithms"
] | null | null | In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.
He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the length of the string. | Print the string that satisfies all the constraints.
If there are multiple answers, print any of them. | [
"2\n",
"3\n"
] | [
"aa\n",
"bba\n"
] | A palindrome is a sequence of characters which reads the same backward and forward. | 1,000 | [
{
"input": "2",
"output": "aa"
},
{
"input": "3",
"output": "aab"
},
{
"input": "38",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaa"
},
{
"input": "47",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab"
},
{
"input": "59",
"output": "aabbaabbaabb... | 1,596,873,577 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 1,000 | 31,539,200 | import math
n = int(input())
ans = ""
last = None
if n % 2:
if n % 4 == 1:
last = "a"
else:
last = "b"
n -= 1
j = (n)//2
ans = ""
for i in range(j):
if i % 2 == 0:
ans += "aa"
else:
ans += "bb"
if last:
ans += last
print(ans)
| Title: 3-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.
He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the length of the string.
Output Specification:
Print the string that satisfies all the constraints.
If there are multiple answers, print any of them.
Demo Input:
['2\n', '3\n']
Demo Output:
['aa\n', 'bba\n']
Note:
A palindrome is a sequence of characters which reads the same backward and forward. | ```python
import math
n = int(input())
ans = ""
last = None
if n % 2:
if n % 4 == 1:
last = "a"
else:
last = "b"
n -= 1
j = (n)//2
ans = ""
for i in range(j):
if i % 2 == 0:
ans += "aa"
else:
ans += "bb"
if last:
ans += last
print(ans)
``` | 0 | |
6 | A | Triangle | PROGRAMMING | 900 | [
"brute force",
"geometry"
] | A. Triangle | 2 | 64 | Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him. | The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks. | Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length. | [
"4 2 1 3\n",
"7 2 2 4\n",
"3 5 9 1\n"
] | [
"TRIANGLE\n",
"SEGMENT\n",
"IMPOSSIBLE\n"
] | none | 0 | [
{
"input": "4 2 1 3",
"output": "TRIANGLE"
},
{
"input": "7 2 2 4",
"output": "SEGMENT"
},
{
"input": "3 5 9 1",
"output": "IMPOSSIBLE"
},
{
"input": "3 1 5 1",
"output": "IMPOSSIBLE"
},
{
"input": "10 10 10 10",
"output": "TRIANGLE"
},
{
"input": "11 ... | 1,619,247,909 | 2,147,483,647 | PyPy 3 | OK | TESTS | 83 | 186 | 0 | a=[]
a.extend(map(int,input().split()))
a.sort()
s1=a[0]+a[1]
s2=a[1]+a[2]
if s1>a[2] or s2>a[3]:
print("TRIANGLE")
elif s1==a[2] or s2==a[3]:
print("SEGMENT")
else:
print("IMPOSSIBLE") | Title: Triangle
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input Specification:
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output Specification:
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Demo Input:
['4 2 1 3\n', '7 2 2 4\n', '3 5 9 1\n']
Demo Output:
['TRIANGLE\n', 'SEGMENT\n', 'IMPOSSIBLE\n']
Note:
none | ```python
a=[]
a.extend(map(int,input().split()))
a.sort()
s1=a[0]+a[1]
s2=a[1]+a[2]
if s1>a[2] or s2>a[3]:
print("TRIANGLE")
elif s1==a[2] or s2==a[3]:
print("SEGMENT")
else:
print("IMPOSSIBLE")
``` | 3.9535 |
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",
... | 1,664,781,872 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | x=[]
k=input().split()
for i in k:
x.append(int(i))
m=7-max(x)
if 6%m==0:
print(f'{1}/{6//m}')
elif 6%m==3/2:
print('2/3')
else:
print(f'{m}/{6}') | 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
x=[]
k=input().split()
for i in k:
x.append(int(i))
m=7-max(x)
if 6%m==0:
print(f'{1}/{6//m}')
elif 6%m==3/2:
print('2/3')
else:
print(f'{m}/{6}')
``` | 0 |
996 | A | Hit the Lottery | PROGRAMMING | 800 | [
"dp",
"greedy"
] | null | null | Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance? | The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$). | Output the minimum number of bills that Allen could receive. | [
"125\n",
"43\n",
"1000000000\n"
] | [
"3\n",
"5\n",
"10000000\n"
] | In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills. | 500 | [
{
"input": "125",
"output": "3"
},
{
"input": "43",
"output": "5"
},
{
"input": "1000000000",
"output": "10000000"
},
{
"input": "4",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "1",
"output": "1"
},
{
"input": "74",
"o... | 1,683,467,528 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 49 | 61 | 0 | ct=0
a=int(input())
while a>0:
if a>=1000:
c=a//100
ct+=c
a=a-c*100
elif a>=100:
a-=100
ct+=1
elif a>=20:
a-=20
ct+=1
elif a>=10:
a-=10
ct+=1
elif a>=5:
a-=5
ct+=1
elif a>=1:
a-=1
ct+=1
print(ct) | Title: Hit the Lottery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input Specification:
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
Output Specification:
Output the minimum number of bills that Allen could receive.
Demo Input:
['125\n', '43\n', '1000000000\n']
Demo Output:
['3\n', '5\n', '10000000\n']
Note:
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills. | ```python
ct=0
a=int(input())
while a>0:
if a>=1000:
c=a//100
ct+=c
a=a-c*100
elif a>=100:
a-=100
ct+=1
elif a>=20:
a-=20
ct+=1
elif a>=10:
a-=10
ct+=1
elif a>=5:
a-=5
ct+=1
elif a>=1:
a-=1
ct+=1
print(ct)
``` | 3 | |
835 | A | Key races | PROGRAMMING | 800 | [
"math"
] | null | null | Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game. | The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant. | If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship". | [
"5 1 2 1 2\n",
"3 3 1 1 1\n",
"4 5 3 1 5\n"
] | [
"First\n",
"Second\n",
"Friendship\n"
] | In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | 500 | [
{
"input": "5 1 2 1 2",
"output": "First"
},
{
"input": "3 3 1 1 1",
"output": "Second"
},
{
"input": "4 5 3 1 5",
"output": "Friendship"
},
{
"input": "1000 1000 1000 1000 1000",
"output": "Friendship"
},
{
"input": "1 1 1 1 1",
"output": "Friendship"
},
... | 1,634,309,822 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | /******************************************************************************
Welcome to GDB Online.
GDB online is an online compiler and debugger tool for C, C++, Python, Java, PHP, Ruby, Perl,
C#, VB, Swift, Pascal, Fortran, Haskell, Objective-C, Assembly, HTML, CSS, JS, SQLite, Prolog.
Code, Compile, Run and Debug online from anywhere in world.
*******************************************************************************/
#include <iostream>
using namespace std;
int main()
{
int c,v1,v2,t1,t2;
cin>>c>>v1>>v2>>t1>>t2;
if (c*v1+2*t1<c*v2+t2*2){
cout<<"First";}
else if (c*v1+2*t1>c*v2+2*t2) {
cout<<"Second";}
else {
cout<<"Friendship";}
return 0;
}
| Title: Key races
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game.
Input Specification:
The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.
Output Specification:
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
Demo Input:
['5 1 2 1 2\n', '3 3 1 1 1\n', '4 5 3 1 5\n']
Demo Output:
['First\n', 'Second\n', 'Friendship\n']
Note:
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | ```python
/******************************************************************************
Welcome to GDB Online.
GDB online is an online compiler and debugger tool for C, C++, Python, Java, PHP, Ruby, Perl,
C#, VB, Swift, Pascal, Fortran, Haskell, Objective-C, Assembly, HTML, CSS, JS, SQLite, Prolog.
Code, Compile, Run and Debug online from anywhere in world.
*******************************************************************************/
#include <iostream>
using namespace std;
int main()
{
int c,v1,v2,t1,t2;
cin>>c>>v1>>v2>>t1>>t2;
if (c*v1+2*t1<c*v2+t2*2){
cout<<"First";}
else if (c*v1+2*t1>c*v2+2*t2) {
cout<<"Second";}
else {
cout<<"Friendship";}
return 0;
}
``` | -1 | |
780 | A | Andryusha and Socks | PROGRAMMING | 800 | [
"implementation"
] | null | null | Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time? | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair. | Print single integer — the maximum number of socks that were on the table at the same time. | [
"1\n1 1\n",
"3\n2 1 1 3 2 3\n"
] | [
"1\n",
"2\n"
] | In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe. | 500 | [
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n2 1 1 3 2 3",
"output": "2"
},
{
"input": "5\n5 1 3 2 4 3 1 2 4 5",
"output": "5"
},
{
"input": "10\n4 2 6 3 4 8 7 1 1 5 2 10 6 8 3 5 10 9 9 7",
"output": "6"
},
{
"input": "50\n30 47 31 38 37 50 36 43 9 23 2 2 ... | 1,645,810,852 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 2,000 | 13,107,200 | n= input()
box_socks = [int(x) for x in input().split()]
table = []
socks_on_table = 0
max_count = 0
for i in box_socks:
if i not in table:
socks_on_table +=1
table.append(i)
else:
if socks_on_table > max_count:
max_count = socks_on_table
socks_on_table -= 1
table.remove(i)
print(max_count)
| Title: Andryusha and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
Output Specification:
Print single integer — the maximum number of socks that were on the table at the same time.
Demo Input:
['1\n1 1\n', '3\n2 1 1 3 2 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe. | ```python
n= input()
box_socks = [int(x) for x in input().split()]
table = []
socks_on_table = 0
max_count = 0
for i in box_socks:
if i not in table:
socks_on_table +=1
table.append(i)
else:
if socks_on_table > max_count:
max_count = socks_on_table
socks_on_table -= 1
table.remove(i)
print(max_count)
``` | 0 | |
658 | A | Bear and Reverse Radewoosh | PROGRAMMING | 800 | [
"implementation"
] | null | null | Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1.
A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points.
Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems. | The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores.
The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem. | Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points. | [
"3 2\n50 85 250\n10 15 25\n",
"3 6\n50 85 250\n10 15 25\n",
"8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n"
] | [
"Limak\n",
"Radewoosh\n",
"Tie\n"
] | In the first sample, there are 3 problems. Limak solves them as follows:
1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4. | 500 | [
{
"input": "3 2\n50 85 250\n10 15 25",
"output": "Limak"
},
{
"input": "3 6\n50 85 250\n10 15 25",
"output": "Radewoosh"
},
{
"input": "8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76",
"output": "Tie"
},
{
"input": "4 1\n3 5 6 9\n1 2 4 8",
"output": "Limak"
},
{... | 1,550,129,984 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 109 | 0 | a, b = [int(x) for x in input().split()]
scores = [int(x) for x in input().split()]
times = [int(x) for x in input().split()]
l, r, cur_t_1, cur_t_2 = 0, 0, 0, 0
for i in range(0, a):
cur_t_1, cur_t_2 = cur_t_1+times[i], cur_t_2+times[a-i-1]
l, r = l+max(0, scores[i]-b*cur_t_1), r+max(0, scores[a-i-1]-b*cur_t_2)
if l == r: print('Tie')
elif l > r: print('Limak')
else: print('Radewoosh')
| Title: Bear and Reverse Radewoosh
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1.
A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points.
Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores.
The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem.
Output Specification:
Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.
Demo Input:
['3 2\n50 85 250\n10 15 25\n', '3 6\n50 85 250\n10 15 25\n', '8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n']
Demo Output:
['Limak\n', 'Radewoosh\n', 'Tie\n']
Note:
In the first sample, there are 3 problems. Limak solves them as follows:
1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4. | ```python
a, b = [int(x) for x in input().split()]
scores = [int(x) for x in input().split()]
times = [int(x) for x in input().split()]
l, r, cur_t_1, cur_t_2 = 0, 0, 0, 0
for i in range(0, a):
cur_t_1, cur_t_2 = cur_t_1+times[i], cur_t_2+times[a-i-1]
l, r = l+max(0, scores[i]-b*cur_t_1), r+max(0, scores[a-i-1]-b*cur_t_2)
if l == r: print('Tie')
elif l > r: print('Limak')
else: print('Radewoosh')
``` | 3 | |
550 | A | Two Substrings | PROGRAMMING | 1,500 | [
"brute force",
"dp",
"greedy",
"implementation",
"strings"
] | null | null | You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). | The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters. | Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. | [
"ABA\n",
"BACFAB\n",
"AXBYBXA\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | 1,000 | [
{
"input": "ABA",
"output": "NO"
},
{
"input": "BACFAB",
"output": "YES"
},
{
"input": "AXBYBXA",
"output": "NO"
},
{
"input": "ABABAB",
"output": "YES"
},
{
"input": "BBBBBBBBBB",
"output": "NO"
},
{
"input": "ABBA",
"output": "YES"
},
{
"... | 1,695,465,723 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 46 | 0 | s = input().strip()
found_ab = found_ba = False
for i in range(len(s) - 1):
if s[i:i+2] == "AB" and not found_ba:
found_ab = True
if s[i:i+2] == "BA" and not found_ab:
found_ba = True
if found_ab and found_ba:
print("YES")
else:
print("NO")
| Title: Two Substrings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).
Input Specification:
The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters.
Output Specification:
Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.
Demo Input:
['ABA\n', 'BACFAB\n', 'AXBYBXA\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | ```python
s = input().strip()
found_ab = found_ba = False
for i in range(len(s) - 1):
if s[i:i+2] == "AB" and not found_ba:
found_ab = True
if s[i:i+2] == "BA" and not found_ab:
found_ba = True
if found_ab and found_ba:
print("YES")
else:
print("NO")
``` | 0 | |
2 | A | Winner | PROGRAMMING | 1,500 | [
"hashing",
"implementation"
] | A. Winner | 1 | 64 | The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points. | The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive. | Print the name of the winner. | [
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] | [
"andrew\n",
"andrew\n"
] | none | 0 | [
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},... | 1,690,618,652 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 62 | 0 | #CodeForce Round 2a Winner
scorelist = {}
max = 0
winner = ""
instruction_list = []
ia = instruction_list.append
n = int(input())
name, score = input("Please give input for 2a : Winner\n").split()
scorelist[name] = int(score)
for i in range(n-1):
name, strscore = input().split()
score = int(strscore)
ia([name, score])
if name not in scorelist:
scorelist[name] = score
else:
scorelist[name] += score
if scorelist[name] > max:
max = scorelist[name]
winner = name
ls = list(scorelist[ppl] for ppl in scorelist)
realmax = max(ls)
if max != realmax:
if ls.count(realmax)> 1:
temp_ppl = []
for ppl in scorelist:
if scorelist[ppl] == realmax:
temp_ppl.append(ppl)
temp_score = {}
for ins in instruction_list:
if ins[0] in temp_ppl:
if ins[0] not in temp_score:
temp_score[ins[0]] = ins[1]
else:
temp_score[ins[0]] += ins[1]
if temp_score[ins[0]] == realmax:
winner = temp_ppl
break
else:
for ppl in scorelist:
if scorelist[ppl] == realmax:
winner = ppl
break
print(winner)
| Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none | ```python
#CodeForce Round 2a Winner
scorelist = {}
max = 0
winner = ""
instruction_list = []
ia = instruction_list.append
n = int(input())
name, score = input("Please give input for 2a : Winner\n").split()
scorelist[name] = int(score)
for i in range(n-1):
name, strscore = input().split()
score = int(strscore)
ia([name, score])
if name not in scorelist:
scorelist[name] = score
else:
scorelist[name] += score
if scorelist[name] > max:
max = scorelist[name]
winner = name
ls = list(scorelist[ppl] for ppl in scorelist)
realmax = max(ls)
if max != realmax:
if ls.count(realmax)> 1:
temp_ppl = []
for ppl in scorelist:
if scorelist[ppl] == realmax:
temp_ppl.append(ppl)
temp_score = {}
for ins in instruction_list:
if ins[0] in temp_ppl:
if ins[0] not in temp_score:
temp_score[ins[0]] = ins[1]
else:
temp_score[ins[0]] += ins[1]
if temp_score[ins[0]] == realmax:
winner = temp_ppl
break
else:
for ppl in scorelist:
if scorelist[ppl] == realmax:
winner = ppl
break
print(winner)
``` | -1 |
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... | 1,625,728,477 | 2,147,483,647 | PyPy 3 | OK | TESTS | 47 | 124 | 5,120,000 | n = int(input())
s = []
a = []
for _ in range(n):
t = input()
a.append(t)
s.extend(list(t))
s = set(s)
if len(s) != 2:
print("NO")
else:
d1 = ''
d2 = ''
for i in range(n):
for j in range(n):
if i == j:
d1 += a[i][j]
if i + j == n - 1:
d2 += a[i][j]
st = ''.join(a)
#print(d1, d2)
#dt = set(list(d1))
#print(dt)
if d1 == d2 and len(set(list(d1))) == 1:
if st.count(d1[0]) == n * 2 - 1:
if st.count(a[0][1]) == n * n - 2 * n + 1:
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("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
n = int(input())
s = []
a = []
for _ in range(n):
t = input()
a.append(t)
s.extend(list(t))
s = set(s)
if len(s) != 2:
print("NO")
else:
d1 = ''
d2 = ''
for i in range(n):
for j in range(n):
if i == j:
d1 += a[i][j]
if i + j == n - 1:
d2 += a[i][j]
st = ''.join(a)
#print(d1, d2)
#dt = set(list(d1))
#print(dt)
if d1 == d2 and len(set(list(d1))) == 1:
if st.count(d1[0]) == n * 2 - 1:
if st.count(a[0][1]) == n * n - 2 * n + 1:
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation. | The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits. | In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem. | [
"101\n",
"110010\n"
] | [
"11\n",
"11010\n"
] | In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>. | 0 | [
{
"input": "101",
"output": "11"
},
{
"input": "110010",
"output": "11010"
},
{
"input": "10000",
"output": "1000"
},
{
"input": "1111111110",
"output": "111111111"
},
{
"input": "10100101011110101",
"output": "1100101011110101"
},
{
"input": "11101001... | 1,377,327,522 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 9 | 92 | 0 | import re
print(re.sub('0','',input(),1)) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
Input Specification:
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
Output Specification:
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
Demo Input:
['101\n', '110010\n']
Demo Output:
['11\n', '11010\n']
Note:
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>. | ```python
import re
print(re.sub('0','',input(),1))
``` | 0 | |
998 | A | Balloons | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation"
] | null | null | There are quite a lot of ways to have fun with inflatable balloons. For example, you can fill them with water and see what happens.
Grigory and Andrew have the same opinion. So, once upon a time, they went to the shop and bought $n$ packets with inflatable balloons, where $i$-th of them has exactly $a_i$ balloons inside.
They want to divide the balloons among themselves. In addition, there are several conditions to hold:
- Do not rip the packets (both Grigory and Andrew should get unbroken packets); - Distribute all packets (every packet should be given to someone); - Give both Grigory and Andrew at least one packet; - To provide more fun, the total number of balloons in Grigory's packets should not be equal to the total number of balloons in Andrew's packets.
Help them to divide the balloons or determine that it's impossible under these conditions. | The first line of input contains a single integer $n$ ($1 \le n \le 10$) — the number of packets with balloons.
The second line contains $n$ integers: $a_1$, $a_2$, $\ldots$, $a_n$ ($1 \le a_i \le 1000$) — the number of balloons inside the corresponding packet. | If it's impossible to divide the balloons satisfying the conditions above, print $-1$.
Otherwise, print an integer $k$ — the number of packets to give to Grigory followed by $k$ distinct integers from $1$ to $n$ — the indices of those. The order of packets doesn't matter.
If there are multiple ways to divide balloons, output any of them. | [
"3\n1 2 1\n",
"2\n5 5\n",
"1\n10\n"
] | [
"2\n1 2\n",
"-1\n",
"-1\n"
] | In the first test Grigory gets $3$ balloons in total while Andrey gets $1$.
In the second test there's only one way to divide the packets which leads to equal numbers of balloons.
In the third test one of the boys won't get a packet at all. | 500 | [
{
"input": "3\n1 2 1",
"output": "1\n1"
},
{
"input": "2\n5 5",
"output": "-1"
},
{
"input": "1\n10",
"output": "-1"
},
{
"input": "1\n1",
"output": "-1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1\n1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 9",... | 1,587,409,178 | 2,147,483,647 | PyPy 3 | OK | TESTS | 49 | 140 | 0 | n=int(input())
s=list(map(int,input().split( )))
p=s[:]
s.sort()
if n==1:
print(-1)
elif n==2 and (s[0]==s[1]):
print(-1)
else:
print(1)
print(p.index(s[0])+1)
| Title: Balloons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are quite a lot of ways to have fun with inflatable balloons. For example, you can fill them with water and see what happens.
Grigory and Andrew have the same opinion. So, once upon a time, they went to the shop and bought $n$ packets with inflatable balloons, where $i$-th of them has exactly $a_i$ balloons inside.
They want to divide the balloons among themselves. In addition, there are several conditions to hold:
- Do not rip the packets (both Grigory and Andrew should get unbroken packets); - Distribute all packets (every packet should be given to someone); - Give both Grigory and Andrew at least one packet; - To provide more fun, the total number of balloons in Grigory's packets should not be equal to the total number of balloons in Andrew's packets.
Help them to divide the balloons or determine that it's impossible under these conditions.
Input Specification:
The first line of input contains a single integer $n$ ($1 \le n \le 10$) — the number of packets with balloons.
The second line contains $n$ integers: $a_1$, $a_2$, $\ldots$, $a_n$ ($1 \le a_i \le 1000$) — the number of balloons inside the corresponding packet.
Output Specification:
If it's impossible to divide the balloons satisfying the conditions above, print $-1$.
Otherwise, print an integer $k$ — the number of packets to give to Grigory followed by $k$ distinct integers from $1$ to $n$ — the indices of those. The order of packets doesn't matter.
If there are multiple ways to divide balloons, output any of them.
Demo Input:
['3\n1 2 1\n', '2\n5 5\n', '1\n10\n']
Demo Output:
['2\n1 2\n', '-1\n', '-1\n']
Note:
In the first test Grigory gets $3$ balloons in total while Andrey gets $1$.
In the second test there's only one way to divide the packets which leads to equal numbers of balloons.
In the third test one of the boys won't get a packet at all. | ```python
n=int(input())
s=list(map(int,input().split( )))
p=s[:]
s.sort()
if n==1:
print(-1)
elif n==2 and (s[0]==s[1]):
print(-1)
else:
print(1)
print(p.index(s[0])+1)
``` | 3 | |
353 | A | Domino | PROGRAMMING | 1,200 | [
"implementation",
"math"
] | null | null | Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half. | Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1. | [
"2\n4 2\n6 4\n",
"1\n2 3\n",
"3\n1 4\n2 3\n4 4\n"
] | [
"0\n",
"-1\n",
"1\n"
] | In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8. | 500 | [
{
"input": "2\n4 2\n6 4",
"output": "0"
},
{
"input": "1\n2 3",
"output": "-1"
},
{
"input": "3\n1 4\n2 3\n4 4",
"output": "1"
},
{
"input": "5\n5 4\n5 4\n1 5\n5 5\n3 3",
"output": "1"
},
{
"input": "20\n1 3\n5 2\n5 2\n2 6\n2 4\n1 1\n1 3\n1 4\n2 6\n4 2\n5 6\n2 2\n... | 1,611,426,076 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 1,000 | 1,740,800 | n = int(input())
upper, lower = [], []
for _ in range(n):
x, y = map(int, input().split())
upper += [x % 2]
lower += [y % 2]
c = 0
while sum(upper) % 2 == 1 and sum(lower) % 2 == 1:
for i in range(n):
if (upper[i] == 1 and lower[i] == 0) or (upper[i] == 0 and lower[i] == 1):
upper[i], lower[i] = lower[i], upper[i]
if sum(upper) % 2 == 0 and sum(lower) % 2 == 0:
break
c += 1
if c == 0 and (sum(upper) % 2 == 1 or sum(lower) % 2 == 1):
c = -1
print(c)
| Title: Domino
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half.
Output Specification:
Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1.
Demo Input:
['2\n4 2\n6 4\n', '1\n2 3\n', '3\n1 4\n2 3\n4 4\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8. | ```python
n = int(input())
upper, lower = [], []
for _ in range(n):
x, y = map(int, input().split())
upper += [x % 2]
lower += [y % 2]
c = 0
while sum(upper) % 2 == 1 and sum(lower) % 2 == 1:
for i in range(n):
if (upper[i] == 1 and lower[i] == 0) or (upper[i] == 0 and lower[i] == 1):
upper[i], lower[i] = lower[i], upper[i]
if sum(upper) % 2 == 0 and sum(lower) % 2 == 0:
break
c += 1
if c == 0 and (sum(upper) % 2 == 1 or sum(lower) % 2 == 1):
c = -1
print(c)
``` | 0 | |
182 | B | Vasya's Calendar | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Vasya lives in a strange world. The year has *n* months and the *i*-th month has *a**i* days. Vasya got a New Year present — the clock that shows not only the time, but also the date.
The clock's face can display any number from 1 to *d*. It is guaranteed that *a**i*<=≤<=*d* for all *i* from 1 to *n*. The clock does not keep information about the current month, so when a new day comes, it simply increases the current day number by one. The clock cannot display number *d*<=+<=1, so after day number *d* it shows day 1 (the current day counter resets). The mechanism of the clock allows you to increase the day number by one manually. When you execute this operation, day *d* is also followed by day 1.
Vasya begins each day checking the day number on the clock. If the day number on the clock does not match the actual day number in the current month, then Vasya manually increases it by one. Vasya is persistent and repeats this operation until the day number on the clock matches the actual number of the current day in the current month.
A year passed and Vasya wonders how many times he manually increased the day number by one, from the first day of the first month to the last day of the *n*-th month inclusive, considering that on the first day of the first month the clock display showed day 1. | The first line contains the single number *d* — the maximum number of the day that Vasya's clock can show (1<=≤<=*d*<=≤<=106).
The second line contains a single integer *n* — the number of months in the year (1<=≤<=*n*<=≤<=2000).
The third line contains *n* space-separated integers: *a**i* (1<=≤<=*a**i*<=≤<=*d*) — the number of days in each month in the order in which they follow, starting from the first one. | Print a single number — the number of times Vasya manually increased the day number by one throughout the last year. | [
"4\n2\n2 2\n",
"5\n3\n3 4 3\n",
"31\n12\n31 28 31 30 31 30 31 31 30 31 30 31\n"
] | [
"2\n",
"3\n",
"7\n"
] | In the first sample the situation is like this:
- Day 1. Month 1. The clock shows 1. Vasya changes nothing. - Day 2. Month 1. The clock shows 2. Vasya changes nothing. - Day 1. Month 2. The clock shows 3. Vasya manually increases the day number by 1. After that the clock shows 4. Vasya increases the day number by 1 manually. After that the clock shows 1. - Day 2. Month 2. The clock shows 2. Vasya changes nothing. | 500 | [
{
"input": "4\n2\n2 2",
"output": "2"
},
{
"input": "5\n3\n3 4 3",
"output": "3"
},
{
"input": "31\n12\n31 28 31 30 31 30 31 31 30 31 30 31",
"output": "7"
},
{
"input": "1\n1\n1",
"output": "0"
},
{
"input": "1\n2\n1 1",
"output": "0"
},
{
"input": "2... | 1,553,526,969 | 2,147,483,647 | PyPy 3 | OK | TESTS | 40 | 280 | 204,800 | d = int(input())
s = int(input())
a = list(map(int,input ().split ()))
print((s-1)*d-sum(a[:-1])) | Title: Vasya's Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya lives in a strange world. The year has *n* months and the *i*-th month has *a**i* days. Vasya got a New Year present — the clock that shows not only the time, but also the date.
The clock's face can display any number from 1 to *d*. It is guaranteed that *a**i*<=≤<=*d* for all *i* from 1 to *n*. The clock does not keep information about the current month, so when a new day comes, it simply increases the current day number by one. The clock cannot display number *d*<=+<=1, so after day number *d* it shows day 1 (the current day counter resets). The mechanism of the clock allows you to increase the day number by one manually. When you execute this operation, day *d* is also followed by day 1.
Vasya begins each day checking the day number on the clock. If the day number on the clock does not match the actual day number in the current month, then Vasya manually increases it by one. Vasya is persistent and repeats this operation until the day number on the clock matches the actual number of the current day in the current month.
A year passed and Vasya wonders how many times he manually increased the day number by one, from the first day of the first month to the last day of the *n*-th month inclusive, considering that on the first day of the first month the clock display showed day 1.
Input Specification:
The first line contains the single number *d* — the maximum number of the day that Vasya's clock can show (1<=≤<=*d*<=≤<=106).
The second line contains a single integer *n* — the number of months in the year (1<=≤<=*n*<=≤<=2000).
The third line contains *n* space-separated integers: *a**i* (1<=≤<=*a**i*<=≤<=*d*) — the number of days in each month in the order in which they follow, starting from the first one.
Output Specification:
Print a single number — the number of times Vasya manually increased the day number by one throughout the last year.
Demo Input:
['4\n2\n2 2\n', '5\n3\n3 4 3\n', '31\n12\n31 28 31 30 31 30 31 31 30 31 30 31\n']
Demo Output:
['2\n', '3\n', '7\n']
Note:
In the first sample the situation is like this:
- Day 1. Month 1. The clock shows 1. Vasya changes nothing. - Day 2. Month 1. The clock shows 2. Vasya changes nothing. - Day 1. Month 2. The clock shows 3. Vasya manually increases the day number by 1. After that the clock shows 4. Vasya increases the day number by 1 manually. After that the clock shows 1. - Day 2. Month 2. The clock shows 2. Vasya changes nothing. | ```python
d = int(input())
s = int(input())
a = list(map(int,input ().split ()))
print((s-1)*d-sum(a[:-1]))
``` | 3 | |
348 | A | Mafia | PROGRAMMING | 1,600 | [
"binary search",
"math",
"sortings"
] | null | null | One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want? | The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play. | In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"3\n3 2 2\n",
"4\n2 2 2 2\n"
] | [
"4\n",
"3\n"
] | You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game). | 500 | [
{
"input": "3\n3 2 2",
"output": "4"
},
{
"input": "4\n2 2 2 2",
"output": "3"
},
{
"input": "7\n9 7 7 8 8 7 8",
"output": "9"
},
{
"input": "10\n13 12 10 13 13 14 10 10 12 12",
"output": "14"
},
{
"input": "10\n94 96 91 95 99 94 96 92 95 99",
"output": "106"
... | 1,671,855,708 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n=input()
a=map(int,raw_input().split())
print max(max(a),-(-sum(a)/(n-1))) | Title: Mafia
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play.
Output Specification:
In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['3\n3 2 2\n', '4\n2 2 2 2\n']
Demo Output:
['4\n', '3\n']
Note:
You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game). | ```python
n=input()
a=map(int,raw_input().split())
print max(max(a),-(-sum(a)/(n-1)))
``` | -1 | |
821 | B | Okabe and Banana Trees | PROGRAMMING | 1,300 | [
"brute force",
"math"
] | null | null | Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees.
Consider the point (*x*,<=*y*) in the 2D plane such that *x* and *y* are integers and 0<=≤<=*x*,<=*y*. There is a tree in such a point, and it has *x*<=+<=*y* bananas. There are no trees nor bananas in other points. Now, Okabe draws a line with equation . Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it can be a line segment or even a point.
Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely.
Okabe is sure that the answer does not exceed 1018. You can trust him. | The first line of input contains two space-separated integers *m* and *b* (1<=≤<=*m*<=≤<=1000, 1<=≤<=*b*<=≤<=10000). | Print the maximum number of bananas Okabe can get from the trees he cuts. | [
"1 5\n",
"2 3\n"
] | [
"30\n",
"25\n"
] | The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has 30 bananas. | 1,000 | [
{
"input": "1 5",
"output": "30"
},
{
"input": "2 3",
"output": "25"
},
{
"input": "4 6",
"output": "459"
},
{
"input": "6 3",
"output": "171"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "10 1",
"output": "55"
},
{
"input": "20 10",
... | 1,498,405,294 | 3,994 | Python 3 | COMPILATION_ERROR | PRETESTS | 0 | 0 | 0 | #include <iostream>
using namespace std;
int main()
{
int m,b;
cin >> m >> b;
long long t = 0;
long long ans = 0;
for(int x=0;x<=m*b;x++)
{
int y = int(-float(x)/m+b);
t=((x+1)*(y+1)*(x+y))/2;
ans = max(ans,t);
}
cout << ans;
}
| Title: Okabe and Banana Trees
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees.
Consider the point (*x*,<=*y*) in the 2D plane such that *x* and *y* are integers and 0<=≤<=*x*,<=*y*. There is a tree in such a point, and it has *x*<=+<=*y* bananas. There are no trees nor bananas in other points. Now, Okabe draws a line with equation . Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it can be a line segment or even a point.
Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely.
Okabe is sure that the answer does not exceed 1018. You can trust him.
Input Specification:
The first line of input contains two space-separated integers *m* and *b* (1<=≤<=*m*<=≤<=1000, 1<=≤<=*b*<=≤<=10000).
Output Specification:
Print the maximum number of bananas Okabe can get from the trees he cuts.
Demo Input:
['1 5\n', '2 3\n']
Demo Output:
['30\n', '25\n']
Note:
The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has 30 bananas. | ```python
#include <iostream>
using namespace std;
int main()
{
int m,b;
cin >> m >> b;
long long t = 0;
long long ans = 0;
for(int x=0;x<=m*b;x++)
{
int y = int(-float(x)/m+b);
t=((x+1)*(y+1)*(x+y))/2;
ans = max(ans,t);
}
cout << ans;
}
``` | -1 | |
982 | A | Row | PROGRAMMING | 1,200 | [
"brute force",
"constructive algorithms"
] | null | null | You're given a row with $n$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated. 1. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($0$ means that the corresponding seat is empty, $1$ — occupied). The goal is to determine whether this seating is "maximal".
Note that the first and last seats are not adjacent (if $n \ne 2$). | The first line contains a single integer $n$ ($1 \leq n \leq 1000$) — the number of chairs.
The next line contains a string of $n$ characters, each of them is either zero or one, describing the seating. | Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase). | [
"3\n101\n",
"4\n1011\n",
"5\n10001\n"
] | [
"Yes\n",
"No\n",
"No\n"
] | In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | 500 | [
{
"input": "3\n101",
"output": "Yes"
},
{
"input": "4\n1011",
"output": "No"
},
{
"input": "5\n10001",
"output": "No"
},
{
"input": "1\n0",
"output": "No"
},
{
"input": "1\n1",
"output": "Yes"
},
{
"input": "100\n010100101010100101001001010100101010010... | 1,584,282,647 | 2,147,483,647 | PyPy 3 | OK | TESTS | 55 | 155 | 0 | n=int(input())
s=input()
s='0'+s+'0'
if('000' in s or '11' in s):
print('No')
else:
print('Yes')
| Title: Row
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You're given a row with $n$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated. 1. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($0$ means that the corresponding seat is empty, $1$ — occupied). The goal is to determine whether this seating is "maximal".
Note that the first and last seats are not adjacent (if $n \ne 2$).
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 1000$) — the number of chairs.
The next line contains a string of $n$ characters, each of them is either zero or one, describing the seating.
Output Specification:
Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase).
Demo Input:
['3\n101\n', '4\n1011\n', '5\n10001\n']
Demo Output:
['Yes\n', 'No\n', 'No\n']
Note:
In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | ```python
n=int(input())
s=input()
s='0'+s+'0'
if('000' in s or '11' in s):
print('No')
else:
print('Yes')
``` | 3 | |
577 | A | Multiplication Table | PROGRAMMING | 1,000 | [
"implementation",
"number theory"
] | null | null | Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*. | The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table. | Print a single number: the number of times *x* occurs in the table. | [
"10 5\n",
"6 12\n",
"5 13\n"
] | [
"2\n",
"4\n",
"0\n"
] | A table for the second sample test is given below. The occurrences of number 12 are marked bold. | 500 | [
{
"input": "10 5",
"output": "2"
},
{
"input": "6 12",
"output": "4"
},
{
"input": "5 13",
"output": "0"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "1 10000000... | 1,642,001,777 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | n, x = map(int, input().split())
count = 0
if n * n == x:
print('1')
else:
for i in range(1, n+1):
for j in range(i, n+1):
if i * j == x:
count += 1
print(count*2) | Title: Multiplication Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
Input Specification:
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
Output Specification:
Print a single number: the number of times *x* occurs in the table.
Demo Input:
['10 5\n', '6 12\n', '5 13\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
A table for the second sample test is given below. The occurrences of number 12 are marked bold. | ```python
n, x = map(int, input().split())
count = 0
if n * n == x:
print('1')
else:
for i in range(1, n+1):
for j in range(i, n+1):
if i * j == x:
count += 1
print(count*2)
``` | 0 | |
378 | A | Playing with Dice | PROGRAMMING | 800 | [
"brute force"
] | null | null | Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.
The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins? | The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly. | Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly. | [
"2 5\n",
"2 4\n"
] | [
"3 0 3\n",
"2 1 3\n"
] | The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.
You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|. | 500 | [
{
"input": "2 5",
"output": "3 0 3"
},
{
"input": "2 4",
"output": "2 1 3"
},
{
"input": "5 3",
"output": "2 1 3"
},
{
"input": "1 6",
"output": "3 0 3"
},
{
"input": "5 1",
"output": "3 1 2"
},
{
"input": "6 3",
"output": "2 0 4"
},
{
"inp... | 1,614,076,643 | 2,147,483,647 | Python 3 | OK | TESTS | 38 | 62 | 307,200 | a, b = list(map(int, input().split()))
first_player_wins = 0
second_player_wins = 0
draw = 0
for i in range(1, 7):
p = abs(a-i)
q = abs(b-i)
if p < q:
first_player_wins += 1
elif p > q:
second_player_wins += 1
elif p == q:
draw += 1
print(first_player_wins, end = " ")
print(draw, end = " ")
print(second_player_wins)
| Title: Playing with Dice
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.
The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?
Input Specification:
The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly.
Output Specification:
Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.
Demo Input:
['2 5\n', '2 4\n']
Demo Output:
['3 0 3\n', '2 1 3\n']
Note:
The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.
You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|. | ```python
a, b = list(map(int, input().split()))
first_player_wins = 0
second_player_wins = 0
draw = 0
for i in range(1, 7):
p = abs(a-i)
q = abs(b-i)
if p < q:
first_player_wins += 1
elif p > q:
second_player_wins += 1
elif p == q:
draw += 1
print(first_player_wins, end = " ")
print(draw, end = " ")
print(second_player_wins)
``` | 3 | |
451 | A | Game With Sticks | PROGRAMMING | 900 | [
"implementation"
] | null | null | After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of *n* horizontal and *m* vertical sticks.
An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.
In the grid shown below, *n*<==<=3 and *m*<==<=3. There are *n*<=+<=*m*<==<=6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are *n*·*m*<==<=9 intersection points, numbered from 1 to 9.
The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).
Assume that both players play optimally. Who will win the game? | The first line of input contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). | Print a single line containing "Akshat" or "Malvika" (without the quotes), depending on the winner of the game. | [
"2 2\n",
"2 3\n",
"3 3\n"
] | [
"Malvika\n",
"Malvika\n",
"Akshat\n"
] | Explanation of the first sample:
The grid has four intersection points, numbered from 1 to 4.
If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3). The resulting grid will look like this.
Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.
In the empty grid, Akshat cannot make any move, hence he will lose.
Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks. | 500 | [
{
"input": "2 2",
"output": "Malvika"
},
{
"input": "2 3",
"output": "Malvika"
},
{
"input": "3 3",
"output": "Akshat"
},
{
"input": "20 68",
"output": "Malvika"
},
{
"input": "1 1",
"output": "Akshat"
},
{
"input": "1 2",
"output": "Akshat"
},
... | 1,683,898,777 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 61 | 2,764,800 | n=int(input())
m=int(input())
turn=0
while(n*m!=0):
n=n-1
m=m-1
turn=turn+1
if turn%2!=0:
print("malvika")
else:
print("akshat") | Title: Game With Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of *n* horizontal and *m* vertical sticks.
An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.
In the grid shown below, *n*<==<=3 and *m*<==<=3. There are *n*<=+<=*m*<==<=6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are *n*·*m*<==<=9 intersection points, numbered from 1 to 9.
The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).
Assume that both players play optimally. Who will win the game?
Input Specification:
The first line of input contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Output Specification:
Print a single line containing "Akshat" or "Malvika" (without the quotes), depending on the winner of the game.
Demo Input:
['2 2\n', '2 3\n', '3 3\n']
Demo Output:
['Malvika\n', 'Malvika\n', 'Akshat\n']
Note:
Explanation of the first sample:
The grid has four intersection points, numbered from 1 to 4.
If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3). The resulting grid will look like this.
Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.
In the empty grid, Akshat cannot make any move, hence he will lose.
Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks. | ```python
n=int(input())
m=int(input())
turn=0
while(n*m!=0):
n=n-1
m=m-1
turn=turn+1
if turn%2!=0:
print("malvika")
else:
print("akshat")
``` | -1 | |
808 | B | Average Sleep Time | PROGRAMMING | 1,300 | [
"data structures",
"implementation",
"math"
] | null | null | It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts *k* days!
When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last *n* days. So now he has a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sleep time on the *i*-th day.
The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider *k* consecutive days as a week. So there will be *n*<=-<=*k*<=+<=1 weeks to take into consideration. For example, if *k*<==<=2, *n*<==<=3 and *a*<==<=[3,<=4,<=7], then the result is .
You should write a program which will calculate average sleep times of Polycarp over all weeks. | The first line contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105). | Output average sleeping time over all weeks.
The answer is considered to be correct if its absolute or relative error does not exceed 10<=-<=6. In particular, it is enough to output real number with at least 6 digits after the decimal point. | [
"3 2\n3 4 7\n",
"1 1\n10\n",
"8 2\n1 2 4 100000 123 456 789 1\n"
] | [
"9.0000000000\n",
"10.0000000000\n",
"28964.2857142857\n"
] | In the third example there are *n* - *k* + 1 = 7 weeks, so the answer is sums of all weeks divided by 7. | 0 | [
{
"input": "3 2\n3 4 7",
"output": "9.0000000000"
},
{
"input": "1 1\n10",
"output": "10.0000000000"
},
{
"input": "8 2\n1 2 4 100000 123 456 789 1",
"output": "28964.2857142857"
},
{
"input": "1 1\n1",
"output": "1.0000000000"
},
{
"input": "1 1\n100000",
"ou... | 1,497,223,771 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | n,k = map(int,input().split())
a = list(map(int,input().split()))
pref = [0]
for i in a:
pref.append(i+pref[-1])
cnt = 0
cur = 0
for i in range(k,n):
cnt += 1
cur += pref[i]-pref[i-k]
print(cur/cnt) | Title: Average Sleep Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts *k* days!
When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last *n* days. So now he has a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sleep time on the *i*-th day.
The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider *k* consecutive days as a week. So there will be *n*<=-<=*k*<=+<=1 weeks to take into consideration. For example, if *k*<==<=2, *n*<==<=3 and *a*<==<=[3,<=4,<=7], then the result is .
You should write a program which will calculate average sleep times of Polycarp over all weeks.
Input Specification:
The first line contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Output average sleeping time over all weeks.
The answer is considered to be correct if its absolute or relative error does not exceed 10<=-<=6. In particular, it is enough to output real number with at least 6 digits after the decimal point.
Demo Input:
['3 2\n3 4 7\n', '1 1\n10\n', '8 2\n1 2 4 100000 123 456 789 1\n']
Demo Output:
['9.0000000000\n', '10.0000000000\n', '28964.2857142857\n']
Note:
In the third example there are *n* - *k* + 1 = 7 weeks, so the answer is sums of all weeks divided by 7. | ```python
n,k = map(int,input().split())
a = list(map(int,input().split()))
pref = [0]
for i in a:
pref.append(i+pref[-1])
cnt = 0
cur = 0
for i in range(k,n):
cnt += 1
cur += pref[i]-pref[i-k]
print(cur/cnt)
``` | 0 | |
731 | A | Night at the Museum | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it. | The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters. | Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input. | [
"zeus\n",
"map\n",
"ares\n"
] | [
"18\n",
"35\n",
"34\n"
] | To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | 500 | [
{
"input": "zeus",
"output": "18"
},
{
"input": "map",
"output": "35"
},
{
"input": "ares",
"output": "34"
},
{
"input": "l",
"output": "11"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv",
"... | 1,630,336,963 | 2,147,483,647 | PyPy 3 | OK | TESTS | 44 | 155 | 20,172,800 | def rotating(a,b):
rotation=0
singleDir= abs(ord(a)-ord(b))
if singleDir>13:
rotation=26-singleDir
else:
rotation=singleDir
return rotation
name=input().lower()
firstRotation=abs(ord("a")-ord(name[0]))
if firstRotation > 13:
count = 26 - firstRotation
else:
count = firstRotation
for i in range(len(name)-1):
count=count+rotating(name[i],name[i+1])
print(count) | Title: Night at the Museum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
Input Specification:
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
Output Specification:
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
Demo Input:
['zeus\n', 'map\n', 'ares\n']
Demo Output:
['18\n', '35\n', '34\n']
Note:
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | ```python
def rotating(a,b):
rotation=0
singleDir= abs(ord(a)-ord(b))
if singleDir>13:
rotation=26-singleDir
else:
rotation=singleDir
return rotation
name=input().lower()
firstRotation=abs(ord("a")-ord(name[0]))
if firstRotation > 13:
count = 26 - firstRotation
else:
count = firstRotation
for i in range(len(name)-1):
count=count+rotating(name[i],name[i+1])
print(count)
``` | 3 | |
760 | B | Frodo and pillows | PROGRAMMING | 1,500 | [
"binary search",
"greedy"
] | null | null | *n* hobbits are planning to spend the night at Frodo's house. Frodo has *n* beds standing in a row and *m* pillows (*n*<=≤<=*m*). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it's not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have.
Frodo will sleep on the *k*-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt? | The only line contain three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=*m*<=≤<=109, 1<=≤<=*k*<=≤<=*n*) — the number of hobbits, the number of pillows and the number of Frodo's bed. | Print single integer — the maximum number of pillows Frodo can have so that no one is hurt. | [
"4 6 2\n",
"3 10 3\n",
"3 6 1\n"
] | [
"2\n",
"4\n",
"3\n"
] | In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds.
In the second example Frodo can take at most four pillows, giving three pillows to each of the others.
In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed. | 1,000 | [
{
"input": "4 6 2",
"output": "2"
},
{
"input": "3 10 3",
"output": "4"
},
{
"input": "3 6 1",
"output": "3"
},
{
"input": "3 3 3",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "1 1000000000 1",
"output": "1000000000"
},
{
... | 1,485,111,578 | 2,678 | Python 3 | WRONG_ANSWER | PRETESTS | 8 | 61 | 4,608,000 | def partSum(a, b):
if a < 0:a = 1
return b * (b + 1) // 2 - (a - 1) * (a - 1 + 1) // 2
def f(v, l, r):
return v + partSum(v - l, v - 1) + partSum(v - r, v - 1)
n, m, k = [int(item) for item in input().split()]
l = k - 1
r = n - k
bound = m - (n - 1)
a, b = 1, bound
while a <= b:
mi = (a + b) // 2
if f(mi, l, r) > m:
b = mi - 1
elif f(mi, l, r) < m:
a = mi + 1
else:
print(mi)
break
#print(a,b, m)
else:
if f(mi, l, r) < m : print(mi)
else:print(mi - 1) | Title: Frodo and pillows
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* hobbits are planning to spend the night at Frodo's house. Frodo has *n* beds standing in a row and *m* pillows (*n*<=≤<=*m*). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it's not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have.
Frodo will sleep on the *k*-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt?
Input Specification:
The only line contain three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=*m*<=≤<=109, 1<=≤<=*k*<=≤<=*n*) — the number of hobbits, the number of pillows and the number of Frodo's bed.
Output Specification:
Print single integer — the maximum number of pillows Frodo can have so that no one is hurt.
Demo Input:
['4 6 2\n', '3 10 3\n', '3 6 1\n']
Demo Output:
['2\n', '4\n', '3\n']
Note:
In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds.
In the second example Frodo can take at most four pillows, giving three pillows to each of the others.
In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed. | ```python
def partSum(a, b):
if a < 0:a = 1
return b * (b + 1) // 2 - (a - 1) * (a - 1 + 1) // 2
def f(v, l, r):
return v + partSum(v - l, v - 1) + partSum(v - r, v - 1)
n, m, k = [int(item) for item in input().split()]
l = k - 1
r = n - k
bound = m - (n - 1)
a, b = 1, bound
while a <= b:
mi = (a + b) // 2
if f(mi, l, r) > m:
b = mi - 1
elif f(mi, l, r) < m:
a = mi + 1
else:
print(mi)
break
#print(a,b, m)
else:
if f(mi, l, r) < m : print(mi)
else:print(mi - 1)
``` | 0 | |
47 | A | Triangular numbers | PROGRAMMING | 800 | [
"brute force",
"math"
] | A. Triangular numbers | 2 | 256 | A triangular number is the number of dots in an equilateral triangle uniformly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangular number. The *n*-th triangular number is the number of dots in a triangle with *n* dots on a side. . You can learn more about these numbers from Wikipedia (http://en.wikipedia.org/wiki/Triangular_number).
Your task is to find out if a given integer is a triangular number. | The first line contains the single number *n* (1<=≤<=*n*<=≤<=500) — the given integer. | If the given integer is a triangular number output YES, otherwise output NO. | [
"1\n",
"2\n",
"3\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "1",
"output": "YES"
},
{
"input": "2",
"output": "NO"
},
{
"input": "3",
"output": "YES"
},
{
"input": "4",
"output": "NO"
},
{
"input": "5",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "7",
"output": "NO... | 1,655,903,161 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 92 | 0 | n=int(input())
i=1
while n>0:
n-=i
i+=1
if n==0:
print("YES")
else:
print("NO") | Title: Triangular numbers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A triangular number is the number of dots in an equilateral triangle uniformly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangular number. The *n*-th triangular number is the number of dots in a triangle with *n* dots on a side. . You can learn more about these numbers from Wikipedia (http://en.wikipedia.org/wiki/Triangular_number).
Your task is to find out if a given integer is a triangular number.
Input Specification:
The first line contains the single number *n* (1<=≤<=*n*<=≤<=500) — the given integer.
Output Specification:
If the given integer is a triangular number output YES, otherwise output NO.
Demo Input:
['1\n', '2\n', '3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
n=int(input())
i=1
while n>0:
n-=i
i+=1
if n==0:
print("YES")
else:
print("NO")
``` | 3.977 |
727 | A | Transformation: from A to B | PROGRAMMING | 1,000 | [
"brute force",
"dfs and similar",
"math"
] | null | null | Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*. | The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have. | If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them. | [
"2 162\n",
"4 42\n",
"100 40021\n"
] | [
"YES\n5\n2 4 8 81 162 \n",
"NO\n",
"YES\n5\n100 200 2001 4002 40021 \n"
] | none | 1,000 | [
{
"input": "2 162",
"output": "YES\n5\n2 4 8 81 162 "
},
{
"input": "4 42",
"output": "NO"
},
{
"input": "100 40021",
"output": "YES\n5\n100 200 2001 4002 40021 "
},
{
"input": "1 111111111",
"output": "YES\n9\n1 11 111 1111 11111 111111 1111111 11111111 111111111 "
},
... | 1,630,518,396 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 61 | 20,172,800 | a,b=map(int,input().split())
can=1
arr=[]
while(b>=a):
arr.append(b)
if(a==b):
break
elif(b%2==0):
b=b//2
elif(b%10==1):
b=b//10
else:
can=0
break
print(len(arr))
for i in range(len(arr)):
print(arr[len(arr)-1-i],end=" ")
print("") | Title: Transformation: from A to B
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*.
Input Specification:
The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have.
Output Specification:
If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them.
Demo Input:
['2 162\n', '4 42\n', '100 40021\n']
Demo Output:
['YES\n5\n2 4 8 81 162 \n', 'NO\n', 'YES\n5\n100 200 2001 4002 40021 \n']
Note:
none | ```python
a,b=map(int,input().split())
can=1
arr=[]
while(b>=a):
arr.append(b)
if(a==b):
break
elif(b%2==0):
b=b//2
elif(b%10==1):
b=b//10
else:
can=0
break
print(len(arr))
for i in range(len(arr)):
print(arr[len(arr)-1-i],end=" ")
print("")
``` | 0 | |
26 | A | Almost Prime | PROGRAMMING | 900 | [
"number theory"
] | A. Almost Prime | 2 | 256 | A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive. | Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000). | Output the amount of almost prime numbers between 1 and *n*, inclusive. | [
"10\n",
"21\n"
] | [
"2\n",
"8\n"
] | none | 500 | [
{
"input": "10",
"output": "2"
},
{
"input": "21",
"output": "8"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "8",
"output": "1"
},
... | 1,595,235,769 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 560 | 6,963,200 | def multiple(i):
j = i
while j <= n:
yield j
j += i
def find_factor(n):
divisor = []
for i in range(1,int(n**(1/2))+1):
if n%i == 0:
divisor.append(i)
divisor.append(n//i)
return set(divisor)
count = 0
n = int(input())
prime = [i for i in range(2,n+1)]
true_prime = []
while len(prime) > 0:
# print(prime[0])
true_prime.append(prime[0])
cross = [a for a in multiple(prime[0])]
prime = [i for i in prime if i not in cross]
prime = [i for i in range(2,n+1)]
for n in prime:
factor = find_factor(n)
count = count + 1 if len([i for i in factor if i in true_prime]) == 2 else count
print(count)
| Title: Almost Prime
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
Input Specification:
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
Output Specification:
Output the amount of almost prime numbers between 1 and *n*, inclusive.
Demo Input:
['10\n', '21\n']
Demo Output:
['2\n', '8\n']
Note:
none | ```python
def multiple(i):
j = i
while j <= n:
yield j
j += i
def find_factor(n):
divisor = []
for i in range(1,int(n**(1/2))+1):
if n%i == 0:
divisor.append(i)
divisor.append(n//i)
return set(divisor)
count = 0
n = int(input())
prime = [i for i in range(2,n+1)]
true_prime = []
while len(prime) > 0:
# print(prime[0])
true_prime.append(prime[0])
cross = [a for a in multiple(prime[0])]
prime = [i for i in prime if i not in cross]
prime = [i for i in range(2,n+1)]
for n in prime:
factor = find_factor(n)
count = count + 1 if len([i for i in factor if i in true_prime]) == 2 else count
print(count)
``` | 3.84703 |
31 | A | Worms Evolution | PROGRAMMING | 1,200 | [
"implementation"
] | A. Worms Evolution | 2 | 256 | Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form. | Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*. | [
"5\n1 2 3 5 7\n",
"5\n1 8 1 5 1\n"
] | [
"3 2 1\n",
"-1\n"
] | none | 500 | [
{
"input": "5\n1 2 3 5 7",
"output": "3 2 1"
},
{
"input": "5\n1 8 1 5 1",
"output": "-1"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "6\n86 402 133 524 405 610",
"output": "6 4 1"
},
{
"input": "8\n217 779 418 895 996 473 3 22",
"output":... | 1,467,877,132 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 154 | 0 | n = int(input())
a = list(map(int, input().split()))
for i in range(n):
for j in range(i + 1 , n):
for k in range(j + 1 , n):
x = -1;y = -1;z = -1
if a[i] + a[j] == a[k]:
x, y, z = k, i, j
elif a[k] + a[j] == a[i]:
x, y, z = i, k, j
elif a[i] + a[k] == a[j]:
x, y, z = j, i, k
if x != -1 and y != -1 and z != -1:
print(str(x + 1) + " " + str(y + 1) + " " + str(z + 1))
exit(0)
print(-1)
| Title: Worms Evolution
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form.
Output Specification:
Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*.
Demo Input:
['5\n1 2 3 5 7\n', '5\n1 8 1 5 1\n']
Demo Output:
['3 2 1\n', '-1\n']
Note:
none | ```python
n = int(input())
a = list(map(int, input().split()))
for i in range(n):
for j in range(i + 1 , n):
for k in range(j + 1 , n):
x = -1;y = -1;z = -1
if a[i] + a[j] == a[k]:
x, y, z = k, i, j
elif a[k] + a[j] == a[i]:
x, y, z = i, k, j
elif a[i] + a[k] == a[j]:
x, y, z = j, i, k
if x != -1 and y != -1 and z != -1:
print(str(x + 1) + " " + str(y + 1) + " " + str(z + 1))
exit(0)
print(-1)
``` | 3.9615 |
460 | A | Vasya and Socks | PROGRAMMING | 900 | [
"brute force",
"implementation",
"math"
] | null | null | Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks? | The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space. | Print a single integer — the answer to the problem. | [
"2 2\n",
"9 3\n"
] | [
"3\n",
"13\n"
] | In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day. | 500 | [
{
"input": "2 2",
"output": "3"
},
{
"input": "9 3",
"output": "13"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "1 99",
"output": "1"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "10 2",
"outp... | 1,681,804,288 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 12 | 46 | 0 | import math
n, m = map(int, input().split())
original_n = n
arr1 = []
count = 0
temp = 0
flag = True
if n == m:
print(n + 1)
elif n < m:
print(n)
else:
if m == 2:
print((original_n*2)-1)
flag = False
while flag:
if n > 0:
n = math.floor(n/m)
original_n+=n
else:
break
if flag:
print(original_n)
| Title: Vasya and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
Input Specification:
The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['2 2\n', '9 3\n']
Demo Output:
['3\n', '13\n']
Note:
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day. | ```python
import math
n, m = map(int, input().split())
original_n = n
arr1 = []
count = 0
temp = 0
flag = True
if n == m:
print(n + 1)
elif n < m:
print(n)
else:
if m == 2:
print((original_n*2)-1)
flag = False
while flag:
if n > 0:
n = math.floor(n/m)
original_n+=n
else:
break
if flag:
print(original_n)
``` | 0 | |
19 | A | World Football Cup | PROGRAMMING | 1,400 | [
"implementation"
] | A. World Football Cup | 2 | 64 | Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
- the final tournament features *n* teams (*n* is always even) - the first *n*<=/<=2 teams (according to the standings) come through to the knockout stage - the standings are made on the following principle: for a victory a team gets 3 points, for a draw — 1 point, for a defeat — 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place — in decreasing order of the difference between scored and missed goals; in the third place — in the decreasing order of scored goals - it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage. | The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=50) — amount of the teams, taking part in the final tournament of World Cup. The following *n* lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following *n*·(*n*<=-<=1)<=/<=2 lines describe the held matches in the format name1-name2 num1:num2, where *name*1, *name*2 — names of the teams; *num*1, *num*2 (0<=≤<=*num*1,<=*num*2<=≤<=100) — amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once. | Output *n*<=/<=2 lines — names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity. | [
"4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3\n",
"2\na\nA\na-A 2:1\n"
] | [
"A\nD\n",
"a\n"
] | none | 0 | [
{
"input": "4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3",
"output": "A\nD"
},
{
"input": "2\na\nA\na-A 2:1",
"output": "a"
},
{
"input": "2\nEULEUbCmfrmqxtzvg\nuHGRmKUhDcxcfqyruwzen\nuHGRmKUhDcxcfqyruwzen-EULEUbCmfrmqxtzvg 13:92",
"output": "EULEUbCmfrmqxtzvg"... | 1,587,291,564 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 2 | 186 | 307,200 | def completeTeamInfo(team,scored,missed):
team.scoredGoals += scored
team.missedGoals += missed
team.differenceScoredAndMissed += scored - missed
if scored > missed:
team.points += 3
elif scored == missed:
team.points += 1
def arrangeTeams(totals):
totals.sort(key=lambda team: team.points, reverse=True)
for teamIndex in range(len(totals)-1):
if totals[teamIndex].points == totals[teamIndex + 1].points:
if totals[teamIndex].differenceScoredAndMissed < totals[teamIndex + 1].differenceScoredAndMissed:
totals[teamIndex], totals[teamIndex + 1] = totals[teamIndex + 1], totals[teamIndex]
elif totals[teamIndex].differenceScoredAndMissed == totals[teamIndex + 1].differenceScoredAndMissed:
if totals[teamIndex].scoredGoals < totals[teamIndex + 1].scoredGoals:
totals[teamIndex], totals[teamIndex + 1] = totals[teamIndex + 1], totals[teamIndex]
def main():
class Team:
def __init__(self, name, scoredGoals, missedGoals, differenceScoredAndMissed, points):
self.name = name
self.scoredGoals = scoredGoals
self.missedGoals = missedGoals
self.differenceScoredAndMissed = differenceScoredAndMissed
self.points = points
teamsNr = int(input())
totals = []
for _ in range(teamsNr):
name = input()
totals.append(Team(name,0,0,0,0))
for _ in range(teamsNr*(teamsNr-1)//2):
playingTeams,score = input().split()
#we want only 0 and 2 position because the input is in this format
#A-B 1:1 and we are not interested in - or :
teamOne = playingTeams[0]
teamTwo = playingTeams[2]
scoreTeamOne = int(score[0])
scoreTeamTwo = int(score[2])
for team in totals:
if team.name == teamOne:
completeTeamInfo(team,scoreTeamOne,scoreTeamTwo)
if team.name == teamTwo:
completeTeamInfo(team,scoreTeamTwo,scoreTeamOne)
arrangeTeams(totals)
#print results
i = teamsNr//2 - 1
while i >= 0:
print(totals[i].name)
i -= 1
main() | Title: World Football Cup
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
- the final tournament features *n* teams (*n* is always even) - the first *n*<=/<=2 teams (according to the standings) come through to the knockout stage - the standings are made on the following principle: for a victory a team gets 3 points, for a draw — 1 point, for a defeat — 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place — in decreasing order of the difference between scored and missed goals; in the third place — in the decreasing order of scored goals - it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=50) — amount of the teams, taking part in the final tournament of World Cup. The following *n* lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following *n*·(*n*<=-<=1)<=/<=2 lines describe the held matches in the format name1-name2 num1:num2, where *name*1, *name*2 — names of the teams; *num*1, *num*2 (0<=≤<=*num*1,<=*num*2<=≤<=100) — amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output Specification:
Output *n*<=/<=2 lines — names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Demo Input:
['4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3\n', '2\na\nA\na-A 2:1\n']
Demo Output:
['A\nD\n', 'a\n']
Note:
none | ```python
def completeTeamInfo(team,scored,missed):
team.scoredGoals += scored
team.missedGoals += missed
team.differenceScoredAndMissed += scored - missed
if scored > missed:
team.points += 3
elif scored == missed:
team.points += 1
def arrangeTeams(totals):
totals.sort(key=lambda team: team.points, reverse=True)
for teamIndex in range(len(totals)-1):
if totals[teamIndex].points == totals[teamIndex + 1].points:
if totals[teamIndex].differenceScoredAndMissed < totals[teamIndex + 1].differenceScoredAndMissed:
totals[teamIndex], totals[teamIndex + 1] = totals[teamIndex + 1], totals[teamIndex]
elif totals[teamIndex].differenceScoredAndMissed == totals[teamIndex + 1].differenceScoredAndMissed:
if totals[teamIndex].scoredGoals < totals[teamIndex + 1].scoredGoals:
totals[teamIndex], totals[teamIndex + 1] = totals[teamIndex + 1], totals[teamIndex]
def main():
class Team:
def __init__(self, name, scoredGoals, missedGoals, differenceScoredAndMissed, points):
self.name = name
self.scoredGoals = scoredGoals
self.missedGoals = missedGoals
self.differenceScoredAndMissed = differenceScoredAndMissed
self.points = points
teamsNr = int(input())
totals = []
for _ in range(teamsNr):
name = input()
totals.append(Team(name,0,0,0,0))
for _ in range(teamsNr*(teamsNr-1)//2):
playingTeams,score = input().split()
#we want only 0 and 2 position because the input is in this format
#A-B 1:1 and we are not interested in - or :
teamOne = playingTeams[0]
teamTwo = playingTeams[2]
scoreTeamOne = int(score[0])
scoreTeamTwo = int(score[2])
for team in totals:
if team.name == teamOne:
completeTeamInfo(team,scoreTeamOne,scoreTeamTwo)
if team.name == teamTwo:
completeTeamInfo(team,scoreTeamTwo,scoreTeamOne)
arrangeTeams(totals)
#print results
i = teamsNr//2 - 1
while i >= 0:
print(totals[i].name)
i -= 1
main()
``` | -1 |
721 | A | One-dimensional Japanese Crossword | PROGRAMMING | 800 | [
"implementation"
] | null | null | Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). | The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. | [
"3\nBBW\n",
"5\nBWBWB\n",
"4\nWWWW\n",
"4\nBBBB\n",
"13\nWBBBBWWBWBBBW\n"
] | [
"1\n2 ",
"3\n1 1 1 ",
"0\n",
"1\n4 ",
"3\n4 1 3 "
] | The last sample case correspond to the picture in the statement. | 500 | [
{
"input": "3\nBBW",
"output": "1\n2 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "4\nWWWW",
"output": "0"
},
{
"input": "4\nBBBB",
"output": "1\n4 "
},
{
"input": "13\nWBBBBWWBWBBBW",
"output": "3\n4 1 3 "
},
{
"input": "1\nB",
... | 1,667,333,460 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 30 | 0 | n = int(input())
crossword = input()
b = 0
co = []
for i in crossword:
if i == 'B':
b += 1
if i == 'W':
if b != 0:
co.append(b)
b = 0
print(len(co))
for i in co:
print(f"{i} ", end="")
print() | Title: One-dimensional Japanese Crossword
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output Specification:
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Demo Input:
['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n']
Demo Output:
['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 ']
Note:
The last sample case correspond to the picture in the statement. | ```python
n = int(input())
crossword = input()
b = 0
co = []
for i in crossword:
if i == 'B':
b += 1
if i == 'W':
if b != 0:
co.append(b)
b = 0
print(len(co))
for i in co:
print(f"{i} ", end="")
print()
``` | 0 | |
993 | A | Two Squares | PROGRAMMING | 1,600 | [
"geometry",
"implementation"
] | null | null | You are given two squares, one with sides parallel to the coordinate axes, and another one with sides at 45 degrees to the coordinate axes. Find whether the two squares intersect.
The interior of the square is considered to be part of the square, i.e. if one square is completely inside another, they intersect. If the two squares only share one common point, they are also considered to intersect. | The input data consists of two lines, one for each square, both containing 4 pairs of integers. Each pair represents coordinates of one vertex of the square. Coordinates within each line are either in clockwise or counterclockwise order.
The first line contains the coordinates of the square with sides parallel to the coordinate axes, the second line contains the coordinates of the square at 45 degrees.
All the values are integer and between $-100$ and $100$. | Print "Yes" if squares intersect, otherwise print "No".
You can print each letter in any case (upper or lower). | [
"0 0 6 0 6 6 0 6\n1 3 3 5 5 3 3 1\n",
"0 0 6 0 6 6 0 6\n7 3 9 5 11 3 9 1\n",
"6 0 6 6 0 6 0 0\n7 4 4 7 7 10 10 7\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first example the second square lies entirely within the first square, so they do intersect.
In the second sample squares do not have any points in common.
Here are images corresponding to the samples: | 500 | [
{
"input": "0 0 6 0 6 6 0 6\n1 3 3 5 5 3 3 1",
"output": "YES"
},
{
"input": "0 0 6 0 6 6 0 6\n7 3 9 5 11 3 9 1",
"output": "NO"
},
{
"input": "6 0 6 6 0 6 0 0\n7 4 4 7 7 10 10 7",
"output": "YES"
},
{
"input": "0 0 6 0 6 6 0 6\n8 4 4 8 8 12 12 8",
"output": "YES"
},
... | 1,672,806,969 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | import math
ax, ay, bx, by, cx, cy, dx, dy = map(int, input().split(' '))
ls = list(map(int, input().split(' ')))
left, right = min(ax, bx, cx, dx), max(ax, bx, cx, dx)
down, up = min(ay, by, cy, dy), max(ay, by, cy, dy)
yes = False
for i in range(0, len(ls), 2):
if left <= ls[i] <= right and down <= ls[i + 1] <= up:
yes = True
break
if yes:
print('YES')
else:
centerX, centerY = (ls[2] + ls[6]) / 2, (ls[1] + ls[5]) / 2
length = abs(ls[1] - ls[5]) / (2 * math.sqrt(2))
ls = [ax - centerX, ay - centerY, bx - centerX, by - centerY, cx - centerX, cy - centerY, dx - centerX, dy - centerY]
for i in range(0, len(ls), 2):
x, y = ls[i], ls[i + 1]
r = math.sqrt(x * x + y * y)
theta = math.acos(x / r) + math.pi / 4
x, y = r * math.cos(theta), r * math.sin(theta)
ls[i], ls[i + 1] = x, y
if -length <= ls[i] <= length and -length <= ls[i + 1] <= length:
yes = True
break
print('YES' if yes else 'NO')
| Title: Two Squares
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two squares, one with sides parallel to the coordinate axes, and another one with sides at 45 degrees to the coordinate axes. Find whether the two squares intersect.
The interior of the square is considered to be part of the square, i.e. if one square is completely inside another, they intersect. If the two squares only share one common point, they are also considered to intersect.
Input Specification:
The input data consists of two lines, one for each square, both containing 4 pairs of integers. Each pair represents coordinates of one vertex of the square. Coordinates within each line are either in clockwise or counterclockwise order.
The first line contains the coordinates of the square with sides parallel to the coordinate axes, the second line contains the coordinates of the square at 45 degrees.
All the values are integer and between $-100$ and $100$.
Output Specification:
Print "Yes" if squares intersect, otherwise print "No".
You can print each letter in any case (upper or lower).
Demo Input:
['0 0 6 0 6 6 0 6\n1 3 3 5 5 3 3 1\n', '0 0 6 0 6 6 0 6\n7 3 9 5 11 3 9 1\n', '6 0 6 6 0 6 0 0\n7 4 4 7 7 10 10 7\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example the second square lies entirely within the first square, so they do intersect.
In the second sample squares do not have any points in common.
Here are images corresponding to the samples: | ```python
import math
ax, ay, bx, by, cx, cy, dx, dy = map(int, input().split(' '))
ls = list(map(int, input().split(' ')))
left, right = min(ax, bx, cx, dx), max(ax, bx, cx, dx)
down, up = min(ay, by, cy, dy), max(ay, by, cy, dy)
yes = False
for i in range(0, len(ls), 2):
if left <= ls[i] <= right and down <= ls[i + 1] <= up:
yes = True
break
if yes:
print('YES')
else:
centerX, centerY = (ls[2] + ls[6]) / 2, (ls[1] + ls[5]) / 2
length = abs(ls[1] - ls[5]) / (2 * math.sqrt(2))
ls = [ax - centerX, ay - centerY, bx - centerX, by - centerY, cx - centerX, cy - centerY, dx - centerX, dy - centerY]
for i in range(0, len(ls), 2):
x, y = ls[i], ls[i + 1]
r = math.sqrt(x * x + y * y)
theta = math.acos(x / r) + math.pi / 4
x, y = r * math.cos(theta), r * math.sin(theta)
ls[i], ls[i + 1] = x, y
if -length <= ls[i] <= length and -length <= ls[i + 1] <= length:
yes = True
break
print('YES' if yes else 'NO')
``` | 0 | |
271 | A | Beautiful Year | PROGRAMMING | 800 | [
"brute force"
] | null | null | It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits. | The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number. | Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists. | [
"1987\n",
"2013\n"
] | [
"2013\n",
"2014\n"
] | none | 500 | [
{
"input": "1987",
"output": "2013"
},
{
"input": "2013",
"output": "2014"
},
{
"input": "1000",
"output": "1023"
},
{
"input": "1001",
"output": "1023"
},
{
"input": "1234",
"output": "1235"
},
{
"input": "5555",
"output": "5601"
},
{
"inp... | 1,693,013,628 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 92 | 0 | num = int(input())
num += 1
while True:
if (len(set(str(num))) == len(str(num))):
break
else:
num += 1
print(num) | Title: Beautiful Year
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
Input Specification:
The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.
Output Specification:
Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.
Demo Input:
['1987\n', '2013\n']
Demo Output:
['2013\n', '2014\n']
Note:
none | ```python
num = int(input())
num += 1
while True:
if (len(set(str(num))) == len(str(num))):
break
else:
num += 1
print(num)
``` | 3 | |
202 | A | LLPS | PROGRAMMING | 800 | [
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null | This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z". | The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10. | Print the lexicographically largest palindromic subsequence of string *s*. | [
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] | [
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] | Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | 500 | [
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
... | 1,592,124,870 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 216 | 0 | a=input()
m=""
ma=max(a)
for i in range(len(a)):
if a[i]==ma:
m+=a[i]
print(m) | Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | ```python
a=input()
m=""
ma=max(a)
for i in range(len(a)):
if a[i]==ma:
m+=a[i]
print(m)
``` | 3 | |
588 | B | Duff in Love | PROGRAMMING | 1,300 | [
"math"
] | null | null | Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store. | The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012). | Print the answer in one line. | [
"10\n",
"12\n"
] | [
"10\n",
"6\n"
] | In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely. | 1,000 | [
{
"input": "10",
"output": "10"
},
{
"input": "12",
"output": "6"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "4",
"output": "2"
},
{
"input": "8",
"output": "2"
},
{
"input": "3",
"output": "3"
},
... | 1,694,935,551 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 0 | n=int(input())
sq=int(n**0.5)
i=1
maxx=0
while(i<=sq):
if n%i==0:
l=n//i
s=i
if maxx>=l and maxx>=s:
print(maxx)
break
flag1=False
flag2=False
j=2
k=2
while(j<=l**0.5):
# print(j<=l**0.5)
if l%(j**2)==0:
flag1=True
break
else:
j+=1
if flag1:
# print('flag1 not worked')
while(k<=s**0.5):
if s%(k**2)==0:
flag=True
break
else:
k+=1
if flag1 and flag2:
continue
elif flag1== False:
print(l)
break
elif flag2==False:
maxx=max(maxx,s)
i+=1
| Title: Duff in Love
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.
Input Specification:
The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012).
Output Specification:
Print the answer in one line.
Demo Input:
['10\n', '12\n']
Demo Output:
['10\n', '6\n']
Note:
In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely. | ```python
n=int(input())
sq=int(n**0.5)
i=1
maxx=0
while(i<=sq):
if n%i==0:
l=n//i
s=i
if maxx>=l and maxx>=s:
print(maxx)
break
flag1=False
flag2=False
j=2
k=2
while(j<=l**0.5):
# print(j<=l**0.5)
if l%(j**2)==0:
flag1=True
break
else:
j+=1
if flag1:
# print('flag1 not worked')
while(k<=s**0.5):
if s%(k**2)==0:
flag=True
break
else:
k+=1
if flag1 and flag2:
continue
elif flag1== False:
print(l)
break
elif flag2==False:
maxx=max(maxx,s)
i+=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... | 1,628,904,439 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 124 | 6,656,000 | #loser707
month=['January','February','March','April','May','June','July','August','September','October','November','December']
current_month=input()
months_left=int(input())
a=month.index(current_month)+1+months_left%12-1
print(month[a])
| 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
#loser707
month=['January','February','March','April','May','June','July','August','September','October','November','December']
current_month=input()
months_left=int(input())
a=month.index(current_month)+1+months_left%12-1
print(month[a])
``` | -1 |
862 | A | Mahmoud and Ehab and the MEX | PROGRAMMING | 1,000 | [
"greedy",
"implementation"
] | null | null | Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil? | The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set. | The only line should contain one integer — the minimal number of operations Dr. Evil should perform. | [
"5 3\n0 4 5 6 7\n",
"1 0\n0\n",
"5 0\n1 2 3 4 5\n"
] | [
"2\n",
"1\n",
"0\n"
] | For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil. | 500 | [
{
"input": "5 3\n0 4 5 6 7",
"output": "2"
},
{
"input": "1 0\n0",
"output": "1"
},
{
"input": "5 0\n1 2 3 4 5",
"output": "0"
},
{
"input": "10 5\n57 1 47 9 93 37 76 70 78 15",
"output": "4"
},
{
"input": "10 5\n99 98 93 97 95 100 92 94 91 96",
"output": "5"
... | 1,517,735,466 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 62 | 5,632,000 | def u_b(li,x,low,high):
while low<=high:
mid=(high+low)//2
if li[mid]<x:
low=mid+1
else:
high=mid-1
return low
n,x=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
index=u_b(l,x,0,n-1)
print(x-index if x not in l else x-index+1)
| Title: Mahmoud and Ehab and the MEX
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
Input Specification:
The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
Output Specification:
The only line should contain one integer — the minimal number of operations Dr. Evil should perform.
Demo Input:
['5 3\n0 4 5 6 7\n', '1 0\n0\n', '5 0\n1 2 3 4 5\n']
Demo Output:
['2\n', '1\n', '0\n']
Note:
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil. | ```python
def u_b(li,x,low,high):
while low<=high:
mid=(high+low)//2
if li[mid]<x:
low=mid+1
else:
high=mid-1
return low
n,x=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
index=u_b(l,x,0,n-1)
print(x-index if x not in l else x-index+1)
``` | 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... | 1,637,168,323 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 |
s=input()
l=len(S)
v=['a','e','i','o','u']
result="YES"
for i in range(1-l):
if s[i]!='n' and s[i] not in v:
if s[i+1] not in v:
result="NO"
if s[-1] not in v and s[-1]!='n':
result="NO"
print(result) | 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()
l=len(S)
v=['a','e','i','o','u']
result="YES"
for i in range(1-l):
if s[i]!='n' and s[i] not in v:
if s[i+1] not in v:
result="NO"
if s[-1] not in v and s[-1]!='n':
result="NO"
print(result)
``` | -1 | |
525 | B | Pasha and String | PROGRAMMING | 1,400 | [
"constructive algorithms",
"greedy",
"math",
"strings"
] | null | null | Pasha got a very beautiful string *s* for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |*s*| from left to right, where |*s*| is the length of the given string.
Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent *m* days performing the following transformations on his string — each day he chose integer *a**i* and reversed a piece of string (a segment) from position *a**i* to position |*s*|<=-<=*a**i*<=+<=1. It is guaranteed that 2·*a**i*<=≤<=|*s*|.
You face the following task: determine what Pasha's string will look like after *m* days. | The first line of the input contains Pasha's string *s* of length from 2 to 2·105 characters, consisting of lowercase Latin letters.
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=105) — the number of days when Pasha changed his string.
The third line contains *m* space-separated elements *a**i* (1<=≤<=*a**i*; 2·*a**i*<=≤<=|*s*|) — the position from which Pasha started transforming the string on the *i*-th day. | In the first line of the output print what Pasha's string *s* will look like after *m* days. | [
"abcdef\n1\n2\n",
"vwxyz\n2\n2 2\n",
"abcdef\n3\n1 2 3\n"
] | [
"aedcbf\n",
"vwxyz\n",
"fbdcea\n"
] | none | 750 | [
{
"input": "abcdef\n1\n2",
"output": "aedcbf"
},
{
"input": "vwxyz\n2\n2 2",
"output": "vwxyz"
},
{
"input": "abcdef\n3\n1 2 3",
"output": "fbdcea"
},
{
"input": "jc\n5\n1 1 1 1 1",
"output": "cj"
},
{
"input": "wljqgdlxyc\n13\n3 4 3 3 5 4 4 2 4 4 5 3 3",
"out... | 1,485,206,357 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 233 | 12,697,600 | a =list(input())
n = int(input())
t = [int(i) for i in input().split()]
t = sorted(t)
s = len(a)
k=0
for i in range(s//2):
while k<n and t[k]-1 <= i:
k+=1
if k%2==1:
a[i],a[s-i-1] = a[s-i-1],a[i]
print(''.join(a)) | Title: Pasha and String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha got a very beautiful string *s* for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |*s*| from left to right, where |*s*| is the length of the given string.
Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent *m* days performing the following transformations on his string — each day he chose integer *a**i* and reversed a piece of string (a segment) from position *a**i* to position |*s*|<=-<=*a**i*<=+<=1. It is guaranteed that 2·*a**i*<=≤<=|*s*|.
You face the following task: determine what Pasha's string will look like after *m* days.
Input Specification:
The first line of the input contains Pasha's string *s* of length from 2 to 2·105 characters, consisting of lowercase Latin letters.
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=105) — the number of days when Pasha changed his string.
The third line contains *m* space-separated elements *a**i* (1<=≤<=*a**i*; 2·*a**i*<=≤<=|*s*|) — the position from which Pasha started transforming the string on the *i*-th day.
Output Specification:
In the first line of the output print what Pasha's string *s* will look like after *m* days.
Demo Input:
['abcdef\n1\n2\n', 'vwxyz\n2\n2 2\n', 'abcdef\n3\n1 2 3\n']
Demo Output:
['aedcbf\n', 'vwxyz\n', 'fbdcea\n']
Note:
none | ```python
a =list(input())
n = int(input())
t = [int(i) for i in input().split()]
t = sorted(t)
s = len(a)
k=0
for i in range(s//2):
while k<n and t[k]-1 <= i:
k+=1
if k%2==1:
a[i],a[s-i-1] = a[s-i-1],a[i]
print(''.join(a))
``` | 3 | |
248 | A | Cupboards | PROGRAMMING | 800 | [
"implementation"
] | null | null | One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*. | The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces. | In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs. | [
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 ... | 1,620,111,409 | 2,147,483,647 | PyPy 3 | OK | TESTS | 57 | 404 | 2,969,600 | # cases = int(input())
cases = 1
def solve():
n = int(input())
ones_in_left = 0
ones_in_right = 0
for i in range(n):
left, right = map(int, input().split())
if left == 1:
ones_in_left += 1
if right == 1:
ones_in_right += 1
ans = min(ones_in_left, n - ones_in_left) + min(ones_in_right, n - ones_in_right)
return ans
for case in range(1, cases+1):
ans = solve()
print(ans)
# print(f"Case #{case}: {ans}") | Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none | ```python
# cases = int(input())
cases = 1
def solve():
n = int(input())
ones_in_left = 0
ones_in_right = 0
for i in range(n):
left, right = map(int, input().split())
if left == 1:
ones_in_left += 1
if right == 1:
ones_in_right += 1
ans = min(ones_in_left, n - ones_in_left) + min(ones_in_right, n - ones_in_right)
return ans
for case in range(1, cases+1):
ans = solve()
print(ans)
# print(f"Case #{case}: {ans}")
``` | 3 | |
16 | B | Burglar and Matches | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | B. Burglar and Matches | 0 | 64 | A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer. | Output the only number — answer to the problem. | [
"7 3\n5 10\n2 5\n3 6\n",
"3 3\n1 3\n2 2\n3 1\n"
] | [
"62\n",
"7\n"
] | none | 0 | [
{
"input": "7 3\n5 10\n2 5\n3 6",
"output": "62"
},
{
"input": "3 3\n1 3\n2 2\n3 1",
"output": "7"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "1 2\n1 9\n1 6",
"output": "9"
},
{
"input": "1 10\n1 1\n1 9\n1 3\n1 9\n1 7\n1 10\n1 4\n1 7\n1 3\n1 1",
"ou... | 1,634,126,598 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 8 | 92 | 6,963,200 | n, m = map(int, input().split())
v = 0
l = sorted((list(map(int, input().split()))[::-1] for i in range(m)), reverse=True)
i = 0
while n > 0:
x = min(l[i][1], n)
v += l[i][0] * x
n -= x
i += 1
print(v) | Title: Burglar and Matches
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
Output Specification:
Output the only number — answer to the problem.
Demo Input:
['7 3\n5 10\n2 5\n3 6\n', '3 3\n1 3\n2 2\n3 1\n']
Demo Output:
['62\n', '7\n']
Note:
none | ```python
n, m = map(int, input().split())
v = 0
l = sorted((list(map(int, input().split()))[::-1] for i in range(m)), reverse=True)
i = 0
while n > 0:
x = min(l[i][1], n)
v += l[i][0] * x
n -= x
i += 1
print(v)
``` | -1 |
262 | A | Roma and Lucky Numbers | PROGRAMMING | 800 | [
"implementation"
] | null | null | Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers.
Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem. | The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has.
The numbers in the lines are separated by single spaces. | In a single line print a single integer — the answer to the problem. | [
"3 4\n1 2 4\n",
"3 2\n447 44 77\n"
] | [
"3\n",
"2\n"
] | In the first sample all numbers contain at most four lucky digits, so the answer is 3.
In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2. | 500 | [
{
"input": "3 4\n1 2 4",
"output": "3"
},
{
"input": "3 2\n447 44 77",
"output": "2"
},
{
"input": "2 2\n507978501 180480073",
"output": "2"
},
{
"input": "9 6\n655243746 167613748 1470546 57644035 176077477 56984809 44677 215706823 369042089",
"output": "9"
},
{
... | 1,679,244,748 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 62 | 0 | x=input().split(" ")
n,k=x
count=0
y=input().split(" ")
for i in y:
counter=0
for j in i:
if "4"==j:
counter+=1
elif "7"==j:
counter+=1
if counter<=int(k):
count+=1
print(count)
| Title: Roma and Lucky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers.
Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem.
Input Specification:
The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has.
The numbers in the lines are separated by single spaces.
Output Specification:
In a single line print a single integer — the answer to the problem.
Demo Input:
['3 4\n1 2 4\n', '3 2\n447 44 77\n']
Demo Output:
['3\n', '2\n']
Note:
In the first sample all numbers contain at most four lucky digits, so the answer is 3.
In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2. | ```python
x=input().split(" ")
n,k=x
count=0
y=input().split(" ")
for i in y:
counter=0
for j in i:
if "4"==j:
counter+=1
elif "7"==j:
counter+=1
if counter<=int(k):
count+=1
print(count)
``` | 3 | |
682 | B | Alyona and Mex | PROGRAMMING | 1,200 | [
"sortings"
] | null | null | Someone gave Alyona an array containing *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*. In one operation, Alyona can choose any element of the array and decrease it, i.e. replace with any positive integer that is smaller than the current one. Alyona can repeat this operation as many times as she wants. In particular, she may not apply any operation to the array at all.
Formally, after applying some operations Alyona will get an array of *n* positive integers *b*1,<=*b*2,<=...,<=*b**n* such that 1<=≤<=*b**i*<=≤<=*a**i* for every 1<=≤<=*i*<=≤<=*n*. Your task is to determine the maximum possible value of mex of this array.
Mex of an array in this problem is the minimum positive integer that doesn't appear in this array. For example, mex of the array containing 1, 3 and 4 is equal to 2, while mex of the array containing 2, 3 and 2 is equal to 1. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of elements in the Alyona's array.
The second line of the input contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array. | Print one positive integer — the maximum possible value of mex of the array after Alyona applies some (possibly none) operations. | [
"5\n1 3 3 3 6\n",
"2\n2 1\n"
] | [
"5\n",
"3\n"
] | In the first sample case if one will decrease the second element value to 2 and the fifth element value to 4 then the mex value of resulting array 1 2 3 3 4 will be equal to 5.
To reach the answer to the second sample case one must not decrease any of the array elements. | 1,000 | [
{
"input": "5\n1 3 3 3 6",
"output": "5"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "1\n1",
"output": "2"
},
{
"input": "1\n1000000000",
"output": "2"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"... | 1,617,203,616 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 62 | 0 |
n = int(input())
l = list(map(int, input().rstrip().split()))
l.sort()
ans = [0] * (n + 1)
cu = n
ex = 0
j = 1
for i in range(n):
if l[i] >= j:
j += 1
else:
print(j+1)
exit()
if i != 0 and l[i] > j:
l[i] = j
print(n + 1)
| Title: Alyona and Mex
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Someone gave Alyona an array containing *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*. In one operation, Alyona can choose any element of the array and decrease it, i.e. replace with any positive integer that is smaller than the current one. Alyona can repeat this operation as many times as she wants. In particular, she may not apply any operation to the array at all.
Formally, after applying some operations Alyona will get an array of *n* positive integers *b*1,<=*b*2,<=...,<=*b**n* such that 1<=≤<=*b**i*<=≤<=*a**i* for every 1<=≤<=*i*<=≤<=*n*. Your task is to determine the maximum possible value of mex of this array.
Mex of an array in this problem is the minimum positive integer that doesn't appear in this array. For example, mex of the array containing 1, 3 and 4 is equal to 2, while mex of the array containing 2, 3 and 2 is equal to 1.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of elements in the Alyona's array.
The second line of the input contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
Output Specification:
Print one positive integer — the maximum possible value of mex of the array after Alyona applies some (possibly none) operations.
Demo Input:
['5\n1 3 3 3 6\n', '2\n2 1\n']
Demo Output:
['5\n', '3\n']
Note:
In the first sample case if one will decrease the second element value to 2 and the fifth element value to 4 then the mex value of resulting array 1 2 3 3 4 will be equal to 5.
To reach the answer to the second sample case one must not decrease any of the array elements. | ```python
n = int(input())
l = list(map(int, input().rstrip().split()))
l.sort()
ans = [0] * (n + 1)
cu = n
ex = 0
j = 1
for i in range(n):
if l[i] >= j:
j += 1
else:
print(j+1)
exit()
if i != 0 and l[i] > j:
l[i] = j
print(n + 1)
``` | 0 | |
231 | A | Team | PROGRAMMING | 800 | [
"brute force",
"greedy"
] | null | null | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. | Print a single integer — the number of problems the friends will implement on the contest. | [
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] | [
"2\n",
"1\n"
] | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | 500 | [
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
... | 1,696,265,499 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 |
n=int(input())
def conter(ch):
n1=0
n=0
for i in range (len(ch)):
if ch[i]!=" ":
if ch[i]=='1':
n1+=1
else:
n0+=1
return str(n0)+'.'+str(n1)
x=0
for i in range (n):
ch=input()
c=conter(ch)
n0=int(c[:c.find('.')])
n1=int(c[c.find('.')+1:])
if n1>n0:
x+=1
print(x) | Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | ```python
n=int(input())
def conter(ch):
n1=0
n=0
for i in range (len(ch)):
if ch[i]!=" ":
if ch[i]=='1':
n1+=1
else:
n0+=1
return str(n0)+'.'+str(n1)
x=0
for i in range (n):
ch=input()
c=conter(ch)
n0=int(c[:c.find('.')])
n1=int(c[c.find('.')+1:])
if n1>n0:
x+=1
print(x)
``` | -1 | |
2 | B | The least round way | PROGRAMMING | 2,000 | [
"dp",
"math"
] | B. The least round way | 2 | 64 | There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that
- starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros. | The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109). | In the first line print the least number of trailing zeros. In the second line print the correspondent way itself. | [
"3\n1 2 3\n4 5 6\n7 8 9\n"
] | [
"0\nDDRR\n"
] | none | 0 | [
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "0\nDDRR"
},
{
"input": "2\n7 6\n3 8",
"output": "0\nDR"
},
{
"input": "3\n4 10 5\n10 9 4\n6 5 3",
"output": "1\nDRRD"
},
{
"input": "4\n1 1 9 9\n3 4 7 3\n7 9 1 7\n1 7 1 5",
"output": "0\nDDDRRR"
},
{
"input": "5\n8 3... | 1,439,996,322 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 122 | 0 | def circle(x, i, j):
x = str(x)
point = -1
answer = 0
while x[point] == '0':
answer += 1
point -= 1
return [answer if i >= 0 and j >= 0 else 10 ** 18, int(x), [i, j]]
n = int(input())
matrix = [list(map(int, input().split())) for i in range(n)]
dp = [[0 for i in range(n)] for j in range(n)]
prev = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j in range(n):
dp[i][j], matrix[i][j], prev[i][j] = min(circle(matrix[i][j - 1] * matrix[i][j], i, j - 1), circle(matrix[i - 1][j] * matrix[i][j], i - 1, j)) if i > 0 or j > 0 else circle(matrix[i][j], i, j)
print(dp[-1][-1])
i, j = n - 1, n - 1
ans = list()
while (i != 0 or j != 0):
li, lj = i, j
i, j = prev[i][j]
ans += ['D' if j < lj else 'R']
print(''.join(reversed(ans))) | Title: The least round way
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that
- starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
Input Specification:
The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109).
Output Specification:
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n']
Demo Output:
['0\nDDRR\n']
Note:
none | ```python
def circle(x, i, j):
x = str(x)
point = -1
answer = 0
while x[point] == '0':
answer += 1
point -= 1
return [answer if i >= 0 and j >= 0 else 10 ** 18, int(x), [i, j]]
n = int(input())
matrix = [list(map(int, input().split())) for i in range(n)]
dp = [[0 for i in range(n)] for j in range(n)]
prev = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j in range(n):
dp[i][j], matrix[i][j], prev[i][j] = min(circle(matrix[i][j - 1] * matrix[i][j], i, j - 1), circle(matrix[i - 1][j] * matrix[i][j], i - 1, j)) if i > 0 or j > 0 else circle(matrix[i][j], i, j)
print(dp[-1][-1])
i, j = n - 1, n - 1
ans = list()
while (i != 0 or j != 0):
li, lj = i, j
i, j = prev[i][j]
ans += ['D' if j < lj else 'R']
print(''.join(reversed(ans)))
``` | 0 |
447 | A | DZY Loves Hash | PROGRAMMING | 800 | [
"implementation"
] | null | null | DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x* *mod* *p*. Operation *a* *mod* *b* denotes taking a remainder after division *a* by *b*.
However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1. | The first line contains two integers, *p* and *n* (2<=≤<=*p*,<=*n*<=≤<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=≤<=*x**i*<=≤<=109). | Output a single integer — the answer to the problem. | [
"10 5\n0\n21\n53\n41\n53\n",
"5 5\n0\n1\n2\n3\n4\n"
] | [
"4\n",
"-1\n"
] | none | 500 | [
{
"input": "10 5\n0\n21\n53\n41\n53",
"output": "4"
},
{
"input": "5 5\n0\n1\n2\n3\n4",
"output": "-1"
},
{
"input": "10 6\n811966798\n734823552\n790326404\n929189974\n414343256\n560346537",
"output": "4"
},
{
"input": "2 2\n788371161\n801743052",
"output": "-1"
},
{
... | 1,406,888,526 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 77 | 0 | l=[]
a=-1
p,n=map(int,input().split())
for i in range(n):
temp=int(input())
t=temp%p
if (t in l) and a==-1:
a=i+1
l.append(t)
print(a)
| Title: DZY Loves Hash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x* *mod* *p*. Operation *a* *mod* *b* denotes taking a remainder after division *a* by *b*.
However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1.
Input Specification:
The first line contains two integers, *p* and *n* (2<=≤<=*p*,<=*n*<=≤<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=≤<=*x**i*<=≤<=109).
Output Specification:
Output a single integer — the answer to the problem.
Demo Input:
['10 5\n0\n21\n53\n41\n53\n', '5 5\n0\n1\n2\n3\n4\n']
Demo Output:
['4\n', '-1\n']
Note:
none | ```python
l=[]
a=-1
p,n=map(int,input().split())
for i in range(n):
temp=int(input())
t=temp%p
if (t in l) and a==-1:
a=i+1
l.append(t)
print(a)
``` | 3 | |
929 | B | Места в самолёте | PROGRAMMING | 1,300 | [
"*special",
"implementation"
] | null | null | В самолёте есть *n* рядов мест. Если смотреть на ряды сверху, то в каждом ряду есть 3 места слева, затем проход между рядами, затем 4 центральных места, затем ещё один проход между рядами, а затем ещё 3 места справа.
Известно, что некоторые места уже заняты пассажирами. Всего есть два вида пассажиров — статусные (те, которые часто летают) и обычные.
Перед вами стоит задача рассадить ещё *k* обычных пассажиров так, чтобы суммарное число соседей у статусных пассажиров было минимально возможным. Два пассажира считаются соседями, если они сидят в одном ряду и между ними нет других мест и прохода между рядами. Если пассажир является соседним пассажиром для двух статусных пассажиров, то его следует учитывать в сумме соседей дважды. | В первой строке следуют два целых числа *n* и *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10·*n*) — количество рядов мест в самолёте и количество пассажиров, которых нужно рассадить.
Далее следует описание рядов мест самолёта по одному ряду в строке. Если очередной символ равен '-', то это проход между рядами. Если очередной символ равен '.', то это свободное место. Если очередной символ равен 'S', то на текущем месте будет сидеть статусный пассажир. Если очередной символ равен 'P', то на текущем месте будет сидеть обычный пассажир.
Гарантируется, что количество свободных мест не меньше *k*. Гарантируется, что все ряды удовлетворяют описанному в условии формату. | В первую строку выведите минимальное суммарное число соседей у статусных пассажиров.
Далее выведите план рассадки пассажиров, который минимизирует суммарное количество соседей у статусных пассажиров, в том же формате, что и во входных данных. Если в свободное место нужно посадить одного из *k* пассажиров, выведите строчную букву 'x' вместо символа '.'. | [
"1 2\nSP.-SS.S-S.S\n",
"4 9\nPP.-PPPS-S.S\nPSP-PPSP-.S.\n.S.-S..P-SS.\nP.S-P.PP-PSP\n"
] | [
"5\nSPx-SSxS-S.S\n",
"15\nPPx-PPPS-S.S\nPSP-PPSP-xSx\nxSx-SxxP-SSx\nP.S-PxPP-PSP\n"
] | В первом примере нужно посадить ещё двух обычных пассажиров. Для минимизации соседей у статусных пассажиров, нужно посадить первого из них на третье слева место, а второго на любое из оставшихся двух мест, так как независимо от выбора места он станет соседом двух статусных пассажиров.
Изначально, у статусного пассажира, который сидит на самом левом месте уже есть сосед. Также на четвёртом и пятом местах слева сидят статусные пассажиры, являющиеся соседями друг для друга (что добавляет к сумме 2).
Таким образом, после посадки ещё двух обычных пассажиров, итоговое суммарное количество соседей у статусных пассажиров станет равно пяти. | 1,000 | [
{
"input": "1 2\nSP.-SS.S-S.S",
"output": "5\nSPx-SSxS-S.S"
},
{
"input": "4 9\nPP.-PPPS-S.S\nPSP-PPSP-.S.\n.S.-S..P-SS.\nP.S-P.PP-PSP",
"output": "15\nPPx-PPPS-S.S\nPSP-PPSP-xSx\nxSx-SxxP-SSx\nP.S-PxPP-PSP"
},
{
"input": "3 7\n.S.-SSSP-..S\nS..-.SPP-S.P\n.S.-PPPP-PSP",
"output": "13... | 1,520,089,747 | 84,847 | Python 3 | OK | TESTS | 47 | 77 | 6,348,800 | n, k = map(int, input().split())
a = b = sits = ""
for _ in range(n):
sits += "+"+input()+"+"
sits = list(sits)
length = len(sits)
perfect = []
ok = []
bad = []
neighbors = 0
for i in range(length):
if sits[i] == "S":
a, b = sits[i - 1], sits[i + 1]
a, b = sits[i - 1], sits[i + 1]
st = "".join([a, b])
neighbors += st.count("S")
neighbors += st.count("P")
continue
if sits[i] != ".":
continue
a, b = sits[i - 1], sits[i + 1]
s = "".join([a, b]).count("S")
if s == 0:
perfect.append(i)
elif s == 1:
ok.append(i)
else:
bad.append(i)
for index in perfect:
if k <= 0:
break
sits[index] = "x"
k -= 1
for index in ok:
if k <= 0:
break
sits[index] = "x"
k -= 1
neighbors += 1
for index in bad:
if k <= 0:
break
sits[index] = "x"
k -= 1
neighbors += 2
answer = "".join(sits).split("+")
print(neighbors)
for item in answer:
if item != "":
print(item)
| Title: Места в самолёте
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
В самолёте есть *n* рядов мест. Если смотреть на ряды сверху, то в каждом ряду есть 3 места слева, затем проход между рядами, затем 4 центральных места, затем ещё один проход между рядами, а затем ещё 3 места справа.
Известно, что некоторые места уже заняты пассажирами. Всего есть два вида пассажиров — статусные (те, которые часто летают) и обычные.
Перед вами стоит задача рассадить ещё *k* обычных пассажиров так, чтобы суммарное число соседей у статусных пассажиров было минимально возможным. Два пассажира считаются соседями, если они сидят в одном ряду и между ними нет других мест и прохода между рядами. Если пассажир является соседним пассажиром для двух статусных пассажиров, то его следует учитывать в сумме соседей дважды.
Input Specification:
В первой строке следуют два целых числа *n* и *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10·*n*) — количество рядов мест в самолёте и количество пассажиров, которых нужно рассадить.
Далее следует описание рядов мест самолёта по одному ряду в строке. Если очередной символ равен '-', то это проход между рядами. Если очередной символ равен '.', то это свободное место. Если очередной символ равен 'S', то на текущем месте будет сидеть статусный пассажир. Если очередной символ равен 'P', то на текущем месте будет сидеть обычный пассажир.
Гарантируется, что количество свободных мест не меньше *k*. Гарантируется, что все ряды удовлетворяют описанному в условии формату.
Output Specification:
В первую строку выведите минимальное суммарное число соседей у статусных пассажиров.
Далее выведите план рассадки пассажиров, который минимизирует суммарное количество соседей у статусных пассажиров, в том же формате, что и во входных данных. Если в свободное место нужно посадить одного из *k* пассажиров, выведите строчную букву 'x' вместо символа '.'.
Demo Input:
['1 2\nSP.-SS.S-S.S\n', '4 9\nPP.-PPPS-S.S\nPSP-PPSP-.S.\n.S.-S..P-SS.\nP.S-P.PP-PSP\n']
Demo Output:
['5\nSPx-SSxS-S.S\n', '15\nPPx-PPPS-S.S\nPSP-PPSP-xSx\nxSx-SxxP-SSx\nP.S-PxPP-PSP\n']
Note:
В первом примере нужно посадить ещё двух обычных пассажиров. Для минимизации соседей у статусных пассажиров, нужно посадить первого из них на третье слева место, а второго на любое из оставшихся двух мест, так как независимо от выбора места он станет соседом двух статусных пассажиров.
Изначально, у статусного пассажира, который сидит на самом левом месте уже есть сосед. Также на четвёртом и пятом местах слева сидят статусные пассажиры, являющиеся соседями друг для друга (что добавляет к сумме 2).
Таким образом, после посадки ещё двух обычных пассажиров, итоговое суммарное количество соседей у статусных пассажиров станет равно пяти. | ```python
n, k = map(int, input().split())
a = b = sits = ""
for _ in range(n):
sits += "+"+input()+"+"
sits = list(sits)
length = len(sits)
perfect = []
ok = []
bad = []
neighbors = 0
for i in range(length):
if sits[i] == "S":
a, b = sits[i - 1], sits[i + 1]
a, b = sits[i - 1], sits[i + 1]
st = "".join([a, b])
neighbors += st.count("S")
neighbors += st.count("P")
continue
if sits[i] != ".":
continue
a, b = sits[i - 1], sits[i + 1]
s = "".join([a, b]).count("S")
if s == 0:
perfect.append(i)
elif s == 1:
ok.append(i)
else:
bad.append(i)
for index in perfect:
if k <= 0:
break
sits[index] = "x"
k -= 1
for index in ok:
if k <= 0:
break
sits[index] = "x"
k -= 1
neighbors += 1
for index in bad:
if k <= 0:
break
sits[index] = "x"
k -= 1
neighbors += 2
answer = "".join(sits).split("+")
print(neighbors)
for item in answer:
if item != "":
print(item)
``` | 3 | |
242 | C | King's Path | PROGRAMMING | 1,800 | [
"dfs and similar",
"graphs",
"hashing",
"shortest paths"
] | null | null | The black king is standing on a chess field consisting of 109 rows and 109 columns. We will consider the rows of the field numbered with integers from 1 to 109 from top to bottom. The columns are similarly numbered with integers from 1 to 109 from left to right. We will denote a cell of the field that is located in the *i*-th row and *j*-th column as (*i*,<=*j*).
You know that some squares of the given chess field are allowed. All allowed cells of the chess field are given as *n* segments. Each segment is described by three integers *r**i*,<=*a**i*,<=*b**i* (*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed.
Your task is to find the minimum number of moves the king needs to get from square (*x*0,<=*y*0) to square (*x*1,<=*y*1), provided that he only moves along the allowed cells. In other words, the king can be located only on allowed cells on his way.
Let us remind you that a chess king can move to any of the neighboring cells in one move. Two cells of a chess field are considered neighboring if they share at least one point. | The first line contains four space-separated integers *x*0,<=*y*0,<=*x*1,<=*y*1 (1<=≤<=*x*0,<=*y*0,<=*x*1,<=*y*1<=≤<=109), denoting the initial and the final positions of the king.
The second line contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of segments of allowed cells. Next *n* lines contain the descriptions of these segments. The *i*-th line contains three space-separated integers *r**i*,<=*a**i*,<=*b**i* (1<=≤<=*r**i*,<=*a**i*,<=*b**i*<=≤<=109,<=*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Note that the segments of the allowed cells can intersect and embed arbitrarily.
It is guaranteed that the king's initial and final position are allowed cells. It is guaranteed that the king's initial and the final positions do not coincide. It is guaranteed that the total length of all given segments doesn't exceed 105. | If there is no path between the initial and final position along allowed cells, print -1.
Otherwise print a single integer — the minimum number of moves the king needs to get from the initial position to the final one. | [
"5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5\n",
"3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10\n",
"1 1 2 10\n2\n1 1 3\n2 6 10\n"
] | [
"4\n",
"6\n",
"-1\n"
] | none | 1,500 | [
{
"input": "5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5",
"output": "4"
},
{
"input": "3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10",
"output": "6"
},
{
"input": "1 1 2 10\n2\n1 1 3\n2 6 10",
"output": "-1"
},
{
"input": "9 8 7 8\n9\n10 6 6\n10 6 6\n7 7 8\n9 5 6\n8 9 9\n9 5 5\n9 8 8\n8 5 6\n9 10... | 1,658,395,390 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 43 | 1,902 | 32,358,400 | import sys
import math
import collections
import heapq
input=sys.stdin.readline
x0,y0,x1,y1=(int(i) for i in input().split())
n=int(input())
d1={}
for i in range(n):
r,a,b=(int(i) for i in input().split())
for i in range(a,b+1):
d1[(r,i)]=1
vis={}
c=0
c1=-1
queue=[[x0,y0,0]]
vis[(x0,y0)]=1
while(queue!=[]):
s=queue.pop(0)
if(s[0]==x1 and s[1]==y1):
c1=1
c=s[2]
break
else:
if(s[0]-1>0 and (s[0]-1,s[1]) not in vis and (s[0]-1,s[1]) in d1):
vis[(s[0]-1,s[1])]=1
queue.append([s[0]-1,s[1],s[2]+1])
if(s[1]-1>0 and (s[0],s[1]-1) not in vis and (s[0],s[1]-1) in d1):
vis[(s[0],s[1]-1)]=1
queue.append([s[0],s[1]-1,s[2]+1])
if(s[0]+1<1000000001 and (s[0]+1,s[1]) not in vis and (s[0]+1,s[1]) in d1):
vis[(s[0]+1,s[1])]=1
queue.append([s[0]+1,s[1],s[2]+1])
if(s[1]+1<1000000001 and (s[0],s[1]+1) not in vis and (s[0],s[1]+1) in d1):
vis[(s[0],s[1]+1)]=1
queue.append([s[0],s[1]+1,s[2]+1])
if(s[0]-1>0 and s[1]-1>0 and (s[0]-1,s[1]-1) not in vis and (s[0]-1,s[1]-1) in d1):
vis[(s[0]-1,s[1]-1)]=1
queue.append([s[0]-1,s[1]-1,s[2]+1])
if(s[0]-1>0 and s[1]+1<1000000001 and (s[0]-1,s[1]+1) not in vis and (s[0]-1,s[1]+1) in d1):
vis[(s[0]-1,s[1]+1)]=1
queue.append([s[0]-1,s[1]+1,s[2]+1])
if(s[0]+1<1000000001 and s[1]-1>0 and (s[0]+1,s[1]-1) not in vis and (s[0]+1,s[1]-1) in d1):
vis[(s[0]+1,s[1]-1)]=1
queue.append([s[0]+1,s[1]-1,s[2]+1])
if(s[0]+1<1000000001 and s[1]+1<1000000001 and (s[0]+1,s[1]+1) not in vis and (s[0]+1,s[1]+1) in d1):
vis[(s[0]+1,s[1]+1)]=1
queue.append([s[0]+1,s[1]+1,s[2]+1])
if(c1==-1):
print(c1)
else:
print(c) | Title: King's Path
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The black king is standing on a chess field consisting of 109 rows and 109 columns. We will consider the rows of the field numbered with integers from 1 to 109 from top to bottom. The columns are similarly numbered with integers from 1 to 109 from left to right. We will denote a cell of the field that is located in the *i*-th row and *j*-th column as (*i*,<=*j*).
You know that some squares of the given chess field are allowed. All allowed cells of the chess field are given as *n* segments. Each segment is described by three integers *r**i*,<=*a**i*,<=*b**i* (*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed.
Your task is to find the minimum number of moves the king needs to get from square (*x*0,<=*y*0) to square (*x*1,<=*y*1), provided that he only moves along the allowed cells. In other words, the king can be located only on allowed cells on his way.
Let us remind you that a chess king can move to any of the neighboring cells in one move. Two cells of a chess field are considered neighboring if they share at least one point.
Input Specification:
The first line contains four space-separated integers *x*0,<=*y*0,<=*x*1,<=*y*1 (1<=≤<=*x*0,<=*y*0,<=*x*1,<=*y*1<=≤<=109), denoting the initial and the final positions of the king.
The second line contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of segments of allowed cells. Next *n* lines contain the descriptions of these segments. The *i*-th line contains three space-separated integers *r**i*,<=*a**i*,<=*b**i* (1<=≤<=*r**i*,<=*a**i*,<=*b**i*<=≤<=109,<=*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Note that the segments of the allowed cells can intersect and embed arbitrarily.
It is guaranteed that the king's initial and final position are allowed cells. It is guaranteed that the king's initial and the final positions do not coincide. It is guaranteed that the total length of all given segments doesn't exceed 105.
Output Specification:
If there is no path between the initial and final position along allowed cells, print -1.
Otherwise print a single integer — the minimum number of moves the king needs to get from the initial position to the final one.
Demo Input:
['5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5\n', '3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10\n', '1 1 2 10\n2\n1 1 3\n2 6 10\n']
Demo Output:
['4\n', '6\n', '-1\n']
Note:
none | ```python
import sys
import math
import collections
import heapq
input=sys.stdin.readline
x0,y0,x1,y1=(int(i) for i in input().split())
n=int(input())
d1={}
for i in range(n):
r,a,b=(int(i) for i in input().split())
for i in range(a,b+1):
d1[(r,i)]=1
vis={}
c=0
c1=-1
queue=[[x0,y0,0]]
vis[(x0,y0)]=1
while(queue!=[]):
s=queue.pop(0)
if(s[0]==x1 and s[1]==y1):
c1=1
c=s[2]
break
else:
if(s[0]-1>0 and (s[0]-1,s[1]) not in vis and (s[0]-1,s[1]) in d1):
vis[(s[0]-1,s[1])]=1
queue.append([s[0]-1,s[1],s[2]+1])
if(s[1]-1>0 and (s[0],s[1]-1) not in vis and (s[0],s[1]-1) in d1):
vis[(s[0],s[1]-1)]=1
queue.append([s[0],s[1]-1,s[2]+1])
if(s[0]+1<1000000001 and (s[0]+1,s[1]) not in vis and (s[0]+1,s[1]) in d1):
vis[(s[0]+1,s[1])]=1
queue.append([s[0]+1,s[1],s[2]+1])
if(s[1]+1<1000000001 and (s[0],s[1]+1) not in vis and (s[0],s[1]+1) in d1):
vis[(s[0],s[1]+1)]=1
queue.append([s[0],s[1]+1,s[2]+1])
if(s[0]-1>0 and s[1]-1>0 and (s[0]-1,s[1]-1) not in vis and (s[0]-1,s[1]-1) in d1):
vis[(s[0]-1,s[1]-1)]=1
queue.append([s[0]-1,s[1]-1,s[2]+1])
if(s[0]-1>0 and s[1]+1<1000000001 and (s[0]-1,s[1]+1) not in vis and (s[0]-1,s[1]+1) in d1):
vis[(s[0]-1,s[1]+1)]=1
queue.append([s[0]-1,s[1]+1,s[2]+1])
if(s[0]+1<1000000001 and s[1]-1>0 and (s[0]+1,s[1]-1) not in vis and (s[0]+1,s[1]-1) in d1):
vis[(s[0]+1,s[1]-1)]=1
queue.append([s[0]+1,s[1]-1,s[2]+1])
if(s[0]+1<1000000001 and s[1]+1<1000000001 and (s[0]+1,s[1]+1) not in vis and (s[0]+1,s[1]+1) in d1):
vis[(s[0]+1,s[1]+1)]=1
queue.append([s[0]+1,s[1]+1,s[2]+1])
if(c1==-1):
print(c1)
else:
print(c)
``` | 3 | |
691 | B | s-palindrome | PROGRAMMING | 1,600 | [
"implementation",
"strings"
] | null | null | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
You are given a string *s*. Check if the string is "s-palindrome". | The only line contains the string *s* (1<=≤<=|*s*|<=≤<=1000) which consists of only English letters. | Print "TAK" if the string *s* is "s-palindrome" and "NIE" otherwise. | [
"oXoxoXo\n",
"bod\n",
"ER\n"
] | [
"TAK\n",
"TAK\n",
"NIE\n"
] | none | 0 | [
{
"input": "oXoxoXo",
"output": "TAK"
},
{
"input": "bod",
"output": "TAK"
},
{
"input": "ER",
"output": "NIE"
},
{
"input": "o",
"output": "TAK"
},
{
"input": "a",
"output": "NIE"
},
{
"input": "opo",
"output": "NIE"
},
{
"input": "HCMoxkg... | 1,551,448,788 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 93 | 0 | s=input().split();s=s[0];k=[]
if len(s)==1 or len(s)==2:
print('NIE')
elif len(s)==3:
print('TAK')
else:
for i in range(1,len(s)-2):
k.append(s[i])
if k==list(reversed(k)):
print('TAK')
else:
print('NIE') | Title: s-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
You are given a string *s*. Check if the string is "s-palindrome".
Input Specification:
The only line contains the string *s* (1<=≤<=|*s*|<=≤<=1000) which consists of only English letters.
Output Specification:
Print "TAK" if the string *s* is "s-palindrome" and "NIE" otherwise.
Demo Input:
['oXoxoXo\n', 'bod\n', 'ER\n']
Demo Output:
['TAK\n', 'TAK\n', 'NIE\n']
Note:
none | ```python
s=input().split();s=s[0];k=[]
if len(s)==1 or len(s)==2:
print('NIE')
elif len(s)==3:
print('TAK')
else:
for i in range(1,len(s)-2):
k.append(s[i])
if k==list(reversed(k)):
print('TAK')
else:
print('NIE')
``` | 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",
... | 1,644,249,388 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 92 | 512,000 | import fractions as fr
a = input()
b = a.split(' ')
y = int(b[0])
w = int(b[1])
c = 7
asa = 0
if y > w:
for i in range(y,7):
asa += 1
else:
for aga in range(w,7):
asa +=1
probabi = fr.Fraction(asa / 6)
print (probabi)
| 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
import fractions as fr
a = input()
b = a.split(' ')
y = int(b[0])
w = int(b[1])
c = 7
asa = 0
if y > w:
for i in range(y,7):
asa += 1
else:
for aga in range(w,7):
asa +=1
probabi = fr.Fraction(asa / 6)
print (probabi)
``` | 0 |
584 | A | Olesya and Rodion | PROGRAMMING | 1,000 | [
"math"
] | null | null | Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1. | The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by. | Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them. | [
"3 2\n"
] | [
"712"
] | none | 500 | [
{
"input": "3 2",
"output": "222"
},
{
"input": "2 2",
"output": "22"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 3",
"output": "33333"
},
{
"input": "10 7",
"output": "7777777777"
},
{
"input": "2 9",
"output": "99"
},
{
"input"... | 1,699,095,037 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 31 | 0 | from random import randint
n,m = map(int,input().split())
print(int(str(m) *n)) | Title: Olesya and Rodion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
Input Specification:
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
Output Specification:
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Demo Input:
['3 2\n']
Demo Output:
['712']
Note:
none | ```python
from random import randint
n,m = map(int,input().split())
print(int(str(m) *n))
``` | 0 | |
416 | A | Guess a number! | PROGRAMMING | 1,400 | [
"greedy",
"implementation",
"two pointers"
] | null | null | A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.
The rules are simple. The host thinks of an integer *y* and the participants guess it by asking questions to the host. There are four types of acceptable questions:
- Is it true that *y* is strictly larger than number *x*? - Is it true that *y* is strictly smaller than number *x*? - Is it true that *y* is larger than or equal to number *x*? - Is it true that *y* is smaller than or equal to number *x*?
On each question the host answers truthfully, "yes" or "no".
Given the sequence of questions and answers, find any integer value of *y* that meets the criteria of all answers. If there isn't such value, print "Impossible". | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10000) — the number of questions (and answers). Next *n* lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:
- ">" (for the first type queries), - "<" (for the second type queries), - ">=" (for the third type queries), - "<=" (for the fourth type queries).
All values of *x* are integer and meet the inequation <=-<=109<=≤<=*x*<=≤<=109. The answer is an English letter "Y" (for "yes") or "N" (for "no").
Consequtive elements in lines are separated by a single space. | Print any of such integers *y*, that the answers to all the queries are correct. The printed number *y* must meet the inequation <=-<=2·109<=≤<=*y*<=≤<=2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). | [
"4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N\n",
"2\n> 100 Y\n< -100 Y\n"
] | [
"17\n",
"Impossible\n"
] | none | 500 | [
{
"input": "4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N",
"output": "17"
},
{
"input": "2\n> 100 Y\n< -100 Y",
"output": "Impossible"
},
{
"input": "4\n< 1 N\n> 1 N\n> 1 N\n> 1 N",
"output": "1"
},
{
"input": "4\n<= 1 Y\n>= 1 Y\n>= 1 Y\n<= 1 Y",
"output": "1"
},
{
"input": ... | 1,669,989,146 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | def ExtractNum(A):
a=A.find(' ')
B=A[::-1]
b=len(A)-1-B.find(' ')
#print(a,b)
return int(A[a+1:b])
n=int(input())
l1=[]
l2=[]
for i in range(n):
a=input()
if(a[0]== '>' and a[1]=='='):
if(a[-1]=='Y'):
l1.append(ExtractNum(a))
else:
l2.append(ExtractNum(a)-1)
elif(a[0]=='>'):
if(a[-1]=='Y'):
l1.append(ExtractNum(a)-1)
else:
l2.append(ExtractNum(a))
elif(a[0]== '<' and a[1]=='='):
if(a[-1]=='Y'):
l2.append(ExtractNum(a))
else:
l1.append(ExtractNum(a)-1)
elif(a[0]=='<'):
if(a[-1]=='Y'):
l2.append(ExtractNum(a)-1)
else:
l1.append(ExtractNum(a))
#print(l1)
#print(l2)
if(max(l1)<=min(l2)):
print(max(l1))
else:
print("Impossible")
| Title: Guess a number!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.
The rules are simple. The host thinks of an integer *y* and the participants guess it by asking questions to the host. There are four types of acceptable questions:
- Is it true that *y* is strictly larger than number *x*? - Is it true that *y* is strictly smaller than number *x*? - Is it true that *y* is larger than or equal to number *x*? - Is it true that *y* is smaller than or equal to number *x*?
On each question the host answers truthfully, "yes" or "no".
Given the sequence of questions and answers, find any integer value of *y* that meets the criteria of all answers. If there isn't such value, print "Impossible".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10000) — the number of questions (and answers). Next *n* lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:
- ">" (for the first type queries), - "<" (for the second type queries), - ">=" (for the third type queries), - "<=" (for the fourth type queries).
All values of *x* are integer and meet the inequation <=-<=109<=≤<=*x*<=≤<=109. The answer is an English letter "Y" (for "yes") or "N" (for "no").
Consequtive elements in lines are separated by a single space.
Output Specification:
Print any of such integers *y*, that the answers to all the queries are correct. The printed number *y* must meet the inequation <=-<=2·109<=≤<=*y*<=≤<=2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes).
Demo Input:
['4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N\n', '2\n> 100 Y\n< -100 Y\n']
Demo Output:
['17\n', 'Impossible\n']
Note:
none | ```python
def ExtractNum(A):
a=A.find(' ')
B=A[::-1]
b=len(A)-1-B.find(' ')
#print(a,b)
return int(A[a+1:b])
n=int(input())
l1=[]
l2=[]
for i in range(n):
a=input()
if(a[0]== '>' and a[1]=='='):
if(a[-1]=='Y'):
l1.append(ExtractNum(a))
else:
l2.append(ExtractNum(a)-1)
elif(a[0]=='>'):
if(a[-1]=='Y'):
l1.append(ExtractNum(a)-1)
else:
l2.append(ExtractNum(a))
elif(a[0]== '<' and a[1]=='='):
if(a[-1]=='Y'):
l2.append(ExtractNum(a))
else:
l1.append(ExtractNum(a)-1)
elif(a[0]=='<'):
if(a[-1]=='Y'):
l2.append(ExtractNum(a)-1)
else:
l1.append(ExtractNum(a))
#print(l1)
#print(l2)
if(max(l1)<=min(l2)):
print(max(l1))
else:
print("Impossible")
``` | 0 | |
377 | A | Maze | PROGRAMMING | 1,600 | [
"dfs and similar"
] | null | null | Pavel loves grid mazes. A grid maze is an *n*<=×<=*m* rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly *k* empty cells into walls so that all the remaining cells still formed a connected area. Help him. | The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=<<=*s*), where *n* and *m* are the maze's height and width, correspondingly, *k* is the number of walls Pavel wants to add and letter *s* represents the number of empty cells in the original maze.
Each of the next *n* lines contains *m* characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall. | Print *n* lines containing *m* characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them. | [
"3 4 2\n#..#\n..#.\n#...\n",
"5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#\n"
] | [
"#.X#\nX.#.\n#...\n",
"#XXX\n#X#.\nX#..\n...#\n.#.#\n"
] | none | 500 | [
{
"input": "5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#",
"output": "#XXX\n#X#.\nX#..\n...#\n.#.#"
},
{
"input": "3 3 2\n#.#\n...\n#.#",
"output": "#X#\nX..\n#.#"
},
{
"input": "7 7 18\n#.....#\n..#.#..\n.#...#.\n...#...\n.#...#.\n..#.#..\n#.....#",
"output": "#XXXXX#\nXX#X#X.\nX#XXX#.\nXXX#... | 1,638,632,638 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 7 | 311 | 19,763,200 | n, m, k = map(int, input().split())
grid = [list(input()) for _ in range(n)]
visited = {}
def is_valid(i, j):
return 0 <= i < n and 0 <= j < m
def dfs(i, j):
global k
if k <= 0:
return False
visited[(i, j)] = True
children_blocked = True
for ni, nj in [(i - 1, j), (i, j - 1), (i, j + 1), (i + 1, j)]:
if not is_valid(ni, nj):
continue
if grid[ni][nj] == '.' and (ni, nj) not in visited:
cb = dfs(ni, nj)
children_blocked = cb and children_blocked
if children_blocked and k > 0:
grid[i][j] = 'X'
k -= 1
return children_blocked
return False
for i in range(n):
f = False
for j in range(m):
if grid[i][j] == '.':
dfs(i, j)
break
if f:
break
for i in range(n):
print(''.join(grid[i]))
| Title: Maze
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pavel loves grid mazes. A grid maze is an *n*<=×<=*m* rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly *k* empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input Specification:
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=<<=*s*), where *n* and *m* are the maze's height and width, correspondingly, *k* is the number of walls Pavel wants to add and letter *s* represents the number of empty cells in the original maze.
Each of the next *n* lines contains *m* characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output Specification:
Print *n* lines containing *m* characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Demo Input:
['3 4 2\n#..#\n..#.\n#...\n', '5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#\n']
Demo Output:
['#.X#\nX.#.\n#...\n', '#XXX\n#X#.\nX#..\n...#\n.#.#\n']
Note:
none | ```python
n, m, k = map(int, input().split())
grid = [list(input()) for _ in range(n)]
visited = {}
def is_valid(i, j):
return 0 <= i < n and 0 <= j < m
def dfs(i, j):
global k
if k <= 0:
return False
visited[(i, j)] = True
children_blocked = True
for ni, nj in [(i - 1, j), (i, j - 1), (i, j + 1), (i + 1, j)]:
if not is_valid(ni, nj):
continue
if grid[ni][nj] == '.' and (ni, nj) not in visited:
cb = dfs(ni, nj)
children_blocked = cb and children_blocked
if children_blocked and k > 0:
grid[i][j] = 'X'
k -= 1
return children_blocked
return False
for i in range(n):
f = False
for j in range(m):
if grid[i][j] == '.':
dfs(i, j)
break
if f:
break
for i in range(n):
print(''.join(grid[i]))
``` | -1 |
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