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stringlengths 17
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| score
float64 -1
3.99
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
337
|
A
|
Puzzles
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
|
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
|
Print a single integer — the least possible difference the teacher can obtain.
|
[
"4 6\n10 12 10 7 5 22\n"
] |
[
"5\n"
] |
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
| 500
|
[
{
"input": "4 6\n10 12 10 7 5 22",
"output": "5"
},
{
"input": "2 2\n4 4",
"output": "0"
},
{
"input": "2 10\n4 5 6 7 8 9 10 11 12 12",
"output": "0"
},
{
"input": "4 5\n818 136 713 59 946",
"output": "759"
},
{
"input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903",
"output": "13"
},
{
"input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94",
"output": "3"
},
{
"input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190",
"output": "31"
},
{
"input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997",
"output": "0"
},
{
"input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176",
"output": "412"
},
{
"input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886",
"output": "938"
},
{
"input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993",
"output": "937"
},
{
"input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7",
"output": "0"
},
{
"input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19",
"output": "12"
},
{
"input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17",
"output": "31"
},
{
"input": "2 2\n1000 4",
"output": "996"
},
{
"input": "2 3\n4 502 1000",
"output": "498"
},
{
"input": "3 3\n4 1000 4",
"output": "996"
}
] | 1,683,569,645
| 2,147,483,647
|
PyPy 3-64
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <cmath>
#include <map>
#include <bitset>
using namespace std;
typedef long long int ll;
typedef long double ld;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<long long int> vll;
typedef pair<int, int> pii;
typedef pair<long long int, long long int> pll;
typedef vector<pair<ll, ll>> vpll;
#define repi(i, s, e) for (ll i = s; i < e; i++)
#define repd(i, e, s) for (ll i = e; i >= s; i--)
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define Yes cout << "Yes\n"
#define No cout << "No\n"
#define pb push_back
#define F first
#define S second
#define sp " "
#define nl "\n"
const ll M = 1e9 + 7;
void input_output_path()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
}
ll binarySearch(vll arr, ll l, ll r, ll x)
{
if (r >= l)
{
ll mid = l + (r - l) / 2;
if (arr[mid] == x)
return mid;
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
return binarySearch(arr, mid + 1, r, x);
}
return -1;
}
ll b_search(vll arr, ll l, ll r, ll x)
{
if (r >= l)
{
ll mid = l + (r - l) / 2;
if (arr[mid] == x)
return mid;
if (arr[mid] > x)
return b_search(arr, l, mid - 1, x);
return b_search(arr, mid + 1, r, x);
}
return l;
}
ll binpow(ll a, ll b, ll M)
{
a %= M;
ll res = 1;
while (b > 0)
{
if (b & 1)
res = res * a % M;
a = a * a % M;
b >>= 1;
}
return res;
}
bool isPrime(ll n)
{
if (n <= 1)
return false;
for (ll i = 2; i < sqrt(n) + 1; i++)
if (n % i == 0)
return false;
return true;
}
ll mul(ll a, ll b)
{
return 1LL * a * b % M;
}
ll modPow(ll b, ll p)
{
if (p == 0)
return 1;
ll x = modPow(b, p / 2);
return p % 2 == 0 ? mul(x, x) : mul(b, mul(x, x));
}
ll modInv(ll n)
{
return modPow(n, M - 2);
}
ll gcd(ll a, ll b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
return (a / gcd(a, b)) * b;
}
bool sortbysec(const pair<ll, ll> &a, const pair<ll, ll> &b)
{
return (a.second < b.second);
}
void solve()
{
int n, m;
cin >> n >> m;
int f[m], min_diff;
for (int i = 0; i < m; i++)
{
cin >> f[i];
}
sort(f, f + m);
min_diff = f[m - 1];
for (int i = 0; i < m - n; i++)
{
min_diff = min(min_diff, f[i + n - 1] - f[i]);
}
cout << min_diff << endl;
}
int main()
{
input_output_path();
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
|
Title: Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
Input Specification:
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Output Specification:
Print a single integer — the least possible difference the teacher can obtain.
Demo Input:
['4 6\n10 12 10 7 5 22\n']
Demo Output:
['5\n']
Note:
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
|
```python
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <cmath>
#include <map>
#include <bitset>
using namespace std;
typedef long long int ll;
typedef long double ld;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<long long int> vll;
typedef pair<int, int> pii;
typedef pair<long long int, long long int> pll;
typedef vector<pair<ll, ll>> vpll;
#define repi(i, s, e) for (ll i = s; i < e; i++)
#define repd(i, e, s) for (ll i = e; i >= s; i--)
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define Yes cout << "Yes\n"
#define No cout << "No\n"
#define pb push_back
#define F first
#define S second
#define sp " "
#define nl "\n"
const ll M = 1e9 + 7;
void input_output_path()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
}
ll binarySearch(vll arr, ll l, ll r, ll x)
{
if (r >= l)
{
ll mid = l + (r - l) / 2;
if (arr[mid] == x)
return mid;
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
return binarySearch(arr, mid + 1, r, x);
}
return -1;
}
ll b_search(vll arr, ll l, ll r, ll x)
{
if (r >= l)
{
ll mid = l + (r - l) / 2;
if (arr[mid] == x)
return mid;
if (arr[mid] > x)
return b_search(arr, l, mid - 1, x);
return b_search(arr, mid + 1, r, x);
}
return l;
}
ll binpow(ll a, ll b, ll M)
{
a %= M;
ll res = 1;
while (b > 0)
{
if (b & 1)
res = res * a % M;
a = a * a % M;
b >>= 1;
}
return res;
}
bool isPrime(ll n)
{
if (n <= 1)
return false;
for (ll i = 2; i < sqrt(n) + 1; i++)
if (n % i == 0)
return false;
return true;
}
ll mul(ll a, ll b)
{
return 1LL * a * b % M;
}
ll modPow(ll b, ll p)
{
if (p == 0)
return 1;
ll x = modPow(b, p / 2);
return p % 2 == 0 ? mul(x, x) : mul(b, mul(x, x));
}
ll modInv(ll n)
{
return modPow(n, M - 2);
}
ll gcd(ll a, ll b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
return (a / gcd(a, b)) * b;
}
bool sortbysec(const pair<ll, ll> &a, const pair<ll, ll> &b)
{
return (a.second < b.second);
}
void solve()
{
int n, m;
cin >> n >> m;
int f[m], min_diff;
for (int i = 0; i < m; i++)
{
cin >> f[i];
}
sort(f, f + m);
min_diff = f[m - 1];
for (int i = 0; i < m - n; i++)
{
min_diff = min(min_diff, f[i + n - 1] - f[i]);
}
cout << min_diff << endl;
}
int main()
{
input_output_path();
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
```
| -1
|
|
513
|
A
|
Game
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"math"
] | null | null |
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
|
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
|
Output "First" if the first player wins and "Second" otherwise.
|
[
"2 2 1 2\n",
"2 1 1 1\n"
] |
[
"Second\n",
"First\n"
] |
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely.
| 3
|
[
{
"input": "2 2 1 2",
"output": "Second"
},
{
"input": "2 1 1 1",
"output": "First"
},
{
"input": "5 7 4 1",
"output": "Second"
},
{
"input": "5 7 1 4",
"output": "Second"
},
{
"input": "5 7 10 10",
"output": "Second"
},
{
"input": "5 7 1 10",
"output": "Second"
},
{
"input": "1 1 1 1",
"output": "Second"
},
{
"input": "50 50 50 50",
"output": "Second"
},
{
"input": "50 49 1 2",
"output": "First"
},
{
"input": "50 48 3 1",
"output": "First"
},
{
"input": "48 50 12 11",
"output": "Second"
},
{
"input": "49 50 11 12",
"output": "Second"
},
{
"input": "49 49 4 1",
"output": "Second"
},
{
"input": "49 49 3 3",
"output": "Second"
},
{
"input": "1 50 1 50",
"output": "Second"
},
{
"input": "1 50 50 50",
"output": "Second"
},
{
"input": "50 1 1 1",
"output": "First"
},
{
"input": "50 1 1 50",
"output": "First"
},
{
"input": "32 31 10 9",
"output": "First"
},
{
"input": "32 4 17 3",
"output": "First"
}
] | 1,623,614,921
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 20
| 93
| 0
|
def solve(a, b, c, d):
if a > b:
return 'First'
return 'Second'
if __name__ == "__main__":
a, b, c, d = map(int, input().split())
print(solve(a, b, c, d))
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
Input Specification:
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
Output Specification:
Output "First" if the first player wins and "Second" otherwise.
Demo Input:
['2 2 1 2\n', '2 1 1 1\n']
Demo Output:
['Second\n', 'First\n']
Note:
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely.
|
```python
def solve(a, b, c, d):
if a > b:
return 'First'
return 'Second'
if __name__ == "__main__":
a, b, c, d = map(int, input().split())
print(solve(a, b, c, d))
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,598,013,683
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 218
| 6,758,400
|
s=input()
a=input()
s=s[::-1]
if(s == a):
print("YES")
else:
print("NO")
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
s=input()
a=input()
s=s[::-1]
if(s == a):
print("YES")
else:
print("NO")
```
| 3.932911
|
82
|
A
|
Double Cola
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] |
A. Double Cola
|
1
|
256
|
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
|
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
|
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
|
[
"1\n",
"6\n",
"1802\n"
] |
[
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,633,521,674
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 41
| 93
| 20,172,800
|
import math
n=int(input())
p=n/5+1
k=math.log(p,2)
k_c=math.floor(k)
k_b=pow(2,k_c)
ans_pre=n-5*(k_b-1)
k_d=pow(2,k_c)
ans_beta=math.ceil(ans_pre/k_d)
l=["Sheldon","Leonard","Penny","Rajesh","Howard"]
print(l[ans_beta-1])
|
Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none
|
```python
import math
n=int(input())
p=n/5+1
k=math.log(p,2)
k_c=math.floor(k)
k_b=pow(2,k_c)
ans_pre=n-5*(k_b-1)
k_d=pow(2,k_c)
ans_beta=math.ceil(ans_pre/k_d)
l=["Sheldon","Leonard","Penny","Rajesh","Howard"]
print(l[ans_beta-1])
```
| 3.915925
|
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"
},
{
"input": "2 3",
"output": "2 0 4"
},
{
"input": "5 6",
"output": "5 0 1"
},
{
"input": "4 4",
"output": "0 6 0"
},
{
"input": "1 1",
"output": "0 6 0"
},
{
"input": "6 4",
"output": "1 1 4"
},
{
"input": "1 4",
"output": "2 0 4"
},
{
"input": "5 5",
"output": "0 6 0"
},
{
"input": "4 5",
"output": "4 0 2"
},
{
"input": "4 3",
"output": "3 0 3"
},
{
"input": "1 5",
"output": "2 1 3"
},
{
"input": "6 5",
"output": "1 0 5"
},
{
"input": "2 2",
"output": "0 6 0"
},
{
"input": "1 3",
"output": "1 1 4"
},
{
"input": "3 6",
"output": "4 0 2"
},
{
"input": "3 1",
"output": "4 1 1"
},
{
"input": "3 2",
"output": "4 0 2"
},
{
"input": "3 5",
"output": "3 1 2"
},
{
"input": "3 3",
"output": "0 6 0"
},
{
"input": "6 2",
"output": "2 1 3"
},
{
"input": "4 1",
"output": "4 0 2"
},
{
"input": "5 2",
"output": "3 0 3"
},
{
"input": "4 2",
"output": "3 1 2"
},
{
"input": "2 1",
"output": "5 0 1"
},
{
"input": "6 1",
"output": "3 0 3"
},
{
"input": "4 6",
"output": "4 1 1"
},
{
"input": "2 6",
"output": "3 1 2"
},
{
"input": "3 4",
"output": "3 0 3"
},
{
"input": "1 2",
"output": "1 0 5"
},
{
"input": "6 6",
"output": "0 6 0"
},
{
"input": "5 4",
"output": "2 0 4"
},
{
"input": "3 3",
"output": "0 6 0"
},
{
"input": "1 1",
"output": "0 6 0"
}
] | 1,620,796,819
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 62
| 6,758,400
|
a,b=list(map(int,input().split()))
one=0
two=0
draw=0
for i in range(1,7):
if abs(a-i)==abs(b-i):
draw+=1
elif abs(a-i)<abs(b-i):
one+=1
else:
two+=1
print(one,draw,two)
|
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()))
one=0
two=0
draw=0
for i in range(1,7):
if abs(a-i)==abs(b-i):
draw+=1
elif abs(a-i)<abs(b-i):
one+=1
else:
two+=1
print(one,draw,two)
```
| 3
|
|
858
|
A
|
k-rounding
|
PROGRAMMING
| 1,100
|
[
"brute force",
"math",
"number theory"
] | null | null |
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
|
Print the *k*-rounding of *n*.
|
[
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] |
[
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] |
none
| 750
|
[
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "100 0",
"output": "100"
},
{
"input": "1000000000 0",
"output": "1000000000"
},
{
"input": "160 2",
"output": "800"
},
{
"input": "3 0",
"output": "3"
},
{
"input": "10 0",
"output": "10"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "3 1",
"output": "30"
},
{
"input": "4 1",
"output": "20"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "6 1",
"output": "30"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "8 1",
"output": "40"
},
{
"input": "9 1",
"output": "90"
},
{
"input": "10 1",
"output": "10"
},
{
"input": "11 1",
"output": "110"
},
{
"input": "12 1",
"output": "60"
},
{
"input": "16 2",
"output": "400"
},
{
"input": "2 2",
"output": "100"
},
{
"input": "1 2",
"output": "100"
},
{
"input": "5 2",
"output": "100"
},
{
"input": "15 2",
"output": "300"
},
{
"input": "36 2",
"output": "900"
},
{
"input": "1 8",
"output": "100000000"
},
{
"input": "8 8",
"output": "100000000"
},
{
"input": "96 8",
"output": "300000000"
},
{
"input": "175 8",
"output": "700000000"
},
{
"input": "9999995 8",
"output": "199999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "12345678 8",
"output": "617283900000000"
},
{
"input": "78125 8",
"output": "100000000"
},
{
"input": "390625 8",
"output": "100000000"
},
{
"input": "1953125 8",
"output": "500000000"
},
{
"input": "9765625 8",
"output": "2500000000"
},
{
"input": "68359375 8",
"output": "17500000000"
},
{
"input": "268435456 8",
"output": "104857600000000"
},
{
"input": "125829120 8",
"output": "9830400000000"
},
{
"input": "128000 8",
"output": "400000000"
},
{
"input": "300000 8",
"output": "300000000"
},
{
"input": "3711871 8",
"output": "371187100000000"
},
{
"input": "55555 8",
"output": "1111100000000"
},
{
"input": "222222222 8",
"output": "11111111100000000"
},
{
"input": "479001600 8",
"output": "7484400000000"
},
{
"input": "655360001 7",
"output": "6553600010000000"
},
{
"input": "655360001 8",
"output": "65536000100000000"
},
{
"input": "1000000000 1",
"output": "1000000000"
},
{
"input": "1000000000 7",
"output": "1000000000"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "10000000 8",
"output": "100000000"
},
{
"input": "1000000 8",
"output": "100000000"
},
{
"input": "10000009 8",
"output": "1000000900000000"
},
{
"input": "10000005 8",
"output": "200000100000000"
},
{
"input": "10000002 8",
"output": "500000100000000"
},
{
"input": "999999997 8",
"output": "99999999700000000"
},
{
"input": "999999997 7",
"output": "9999999970000000"
},
{
"input": "999999995 8",
"output": "19999999900000000"
},
{
"input": "123 8",
"output": "12300000000"
},
{
"input": "24 2",
"output": "600"
},
{
"input": "16 4",
"output": "10000"
},
{
"input": "123456787 8",
"output": "12345678700000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "101 1",
"output": "1010"
},
{
"input": "50 2",
"output": "100"
},
{
"input": "999999818 1",
"output": "4999999090"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "123 1",
"output": "1230"
},
{
"input": "16 1",
"output": "80"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "15304 6",
"output": "1913000000"
},
{
"input": "3 8",
"output": "300000000"
},
{
"input": "4 2",
"output": "100"
},
{
"input": "100000 7",
"output": "10000000"
},
{
"input": "5 8",
"output": "100000000"
},
{
"input": "16724 6",
"output": "4181000000"
},
{
"input": "999999999 1",
"output": "9999999990"
},
{
"input": "999999990 8",
"output": "9999999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "100000 1",
"output": "100000"
},
{
"input": "8 3",
"output": "1000"
},
{
"input": "16768 6",
"output": "262000000"
},
{
"input": "123456789 1",
"output": "1234567890"
},
{
"input": "2 8",
"output": "100000000"
},
{
"input": "999999937 8",
"output": "99999993700000000"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "2000000 7",
"output": "10000000"
},
{
"input": "1999998 2",
"output": "99999900"
},
{
"input": "125 3",
"output": "1000"
}
] | 1,506,188,125
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 96
| 62
| 0
|
n, k = map(int, input().split())
nn = n
two = 0;
five = 0;
while n % 2 == 0:
n //= 2;
two += 1;
while n % 5 == 0:
n //= 5;
five += 1;
for i in range(max(0, k - two)):
nn *= 2;
for i in range(max(0, k - five)):
nn *= 5
print(nn)
|
Title: k-rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none
|
```python
n, k = map(int, input().split())
nn = n
two = 0;
five = 0;
while n % 2 == 0:
n //= 2;
two += 1;
while n % 5 == 0:
n //= 5;
five += 1;
for i in range(max(0, k - two)):
nn *= 2;
for i in range(max(0, k - five)):
nn *= 5
print(nn)
```
| 3
|
|
760
|
A
|
Petr and a calendar
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:
Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.
|
The only line contain two integers *m* and *d* (1<=≤<=*m*<=≤<=12, 1<=≤<=*d*<=≤<=7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).
|
Print single integer: the number of columns the table should have.
|
[
"1 7\n",
"1 1\n",
"11 6\n"
] |
[
"6\n",
"5\n",
"5\n"
] |
The first example corresponds to the January 2017 shown on the picture in the statements.
In the second example 1-st January is Monday, so the whole month fits into 5 columns.
In the third example 1-st November is Saturday and 5 columns is enough.
| 500
|
[
{
"input": "1 7",
"output": "6"
},
{
"input": "1 1",
"output": "5"
},
{
"input": "11 6",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "2 1",
"output": "4"
},
{
"input": "8 6",
"output": "6"
},
{
"input": "1 1",
"output": "5"
},
{
"input": "1 2",
"output": "5"
},
{
"input": "1 3",
"output": "5"
},
{
"input": "1 4",
"output": "5"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "6"
},
{
"input": "2 1",
"output": "4"
},
{
"input": "2 2",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "2 4",
"output": "5"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "3 1",
"output": "5"
},
{
"input": "3 2",
"output": "5"
},
{
"input": "3 3",
"output": "5"
},
{
"input": "3 4",
"output": "5"
},
{
"input": "3 5",
"output": "5"
},
{
"input": "3 6",
"output": "6"
},
{
"input": "3 7",
"output": "6"
},
{
"input": "4 1",
"output": "5"
},
{
"input": "4 2",
"output": "5"
},
{
"input": "4 3",
"output": "5"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "4 5",
"output": "5"
},
{
"input": "4 6",
"output": "5"
},
{
"input": "4 7",
"output": "6"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "5 2",
"output": "5"
},
{
"input": "5 3",
"output": "5"
},
{
"input": "5 4",
"output": "5"
},
{
"input": "5 5",
"output": "5"
},
{
"input": "5 6",
"output": "6"
},
{
"input": "5 7",
"output": "6"
},
{
"input": "6 1",
"output": "5"
},
{
"input": "6 2",
"output": "5"
},
{
"input": "6 3",
"output": "5"
},
{
"input": "6 4",
"output": "5"
},
{
"input": "6 5",
"output": "5"
},
{
"input": "6 6",
"output": "5"
},
{
"input": "6 7",
"output": "6"
},
{
"input": "7 1",
"output": "5"
},
{
"input": "7 2",
"output": "5"
},
{
"input": "7 3",
"output": "5"
},
{
"input": "7 4",
"output": "5"
},
{
"input": "7 5",
"output": "5"
},
{
"input": "7 6",
"output": "6"
},
{
"input": "7 7",
"output": "6"
},
{
"input": "8 1",
"output": "5"
},
{
"input": "8 2",
"output": "5"
},
{
"input": "8 3",
"output": "5"
},
{
"input": "8 4",
"output": "5"
},
{
"input": "8 5",
"output": "5"
},
{
"input": "8 6",
"output": "6"
},
{
"input": "8 7",
"output": "6"
},
{
"input": "9 1",
"output": "5"
},
{
"input": "9 2",
"output": "5"
},
{
"input": "9 3",
"output": "5"
},
{
"input": "9 4",
"output": "5"
},
{
"input": "9 5",
"output": "5"
},
{
"input": "9 6",
"output": "5"
},
{
"input": "9 7",
"output": "6"
},
{
"input": "10 1",
"output": "5"
},
{
"input": "10 2",
"output": "5"
},
{
"input": "10 3",
"output": "5"
},
{
"input": "10 4",
"output": "5"
},
{
"input": "10 5",
"output": "5"
},
{
"input": "10 6",
"output": "6"
},
{
"input": "10 7",
"output": "6"
},
{
"input": "11 1",
"output": "5"
},
{
"input": "11 2",
"output": "5"
},
{
"input": "11 3",
"output": "5"
},
{
"input": "11 4",
"output": "5"
},
{
"input": "11 5",
"output": "5"
},
{
"input": "11 6",
"output": "5"
},
{
"input": "11 7",
"output": "6"
},
{
"input": "12 1",
"output": "5"
},
{
"input": "12 2",
"output": "5"
},
{
"input": "12 3",
"output": "5"
},
{
"input": "12 4",
"output": "5"
},
{
"input": "12 5",
"output": "5"
},
{
"input": "12 6",
"output": "6"
},
{
"input": "12 7",
"output": "6"
},
{
"input": "1 4",
"output": "5"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "9 7",
"output": "6"
},
{
"input": "2 6",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "2 2",
"output": "5"
},
{
"input": "4 7",
"output": "6"
},
{
"input": "12 6",
"output": "6"
},
{
"input": "12 3",
"output": "5"
},
{
"input": "3 6",
"output": "6"
},
{
"input": "9 6",
"output": "5"
},
{
"input": "7 6",
"output": "6"
},
{
"input": "11 7",
"output": "6"
},
{
"input": "6 6",
"output": "5"
}
] | 1,578,656,098
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 104
| 124
| 0
|
m, d = map(int, input().split())
if d == 1:
print(5 - (m==2))
elif d < 6:
print(5)
elif d == 6:
print(6 - (m in (2, 4, 6, 9, 11)))
else:
print(6 - (m==2))
|
Title: Petr and a calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:
Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.
Input Specification:
The only line contain two integers *m* and *d* (1<=≤<=*m*<=≤<=12, 1<=≤<=*d*<=≤<=7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).
Output Specification:
Print single integer: the number of columns the table should have.
Demo Input:
['1 7\n', '1 1\n', '11 6\n']
Demo Output:
['6\n', '5\n', '5\n']
Note:
The first example corresponds to the January 2017 shown on the picture in the statements.
In the second example 1-st January is Monday, so the whole month fits into 5 columns.
In the third example 1-st November is Saturday and 5 columns is enough.
|
```python
m, d = map(int, input().split())
if d == 1:
print(5 - (m==2))
elif d < 6:
print(5)
elif d == 6:
print(6 - (m in (2, 4, 6, 9, 11)))
else:
print(6 - (m==2))
```
| 3
|
|
675
|
C
|
Money Transfers
|
PROGRAMMING
| 2,100
|
[
"constructive algorithms",
"data structures",
"greedy",
"sortings"
] | null | null |
There are *n* banks in the city where Vasya lives, they are located in a circle, such that any two banks are neighbouring if their indices differ by no more than 1. Also, bank 1 and bank *n* are neighbours if *n*<=><=1. No bank is a neighbour of itself.
Vasya has an account in each bank. Its balance may be negative, meaning Vasya owes some money to this bank.
There is only one type of operations available: transfer some amount of money from any bank to account in any neighbouring bank. There are no restrictions on the size of the sum being transferred or balance requirements to perform this operation.
Vasya doesn't like to deal with large numbers, so he asks you to determine the minimum number of operations required to change the balance of each bank account to zero. It's guaranteed, that this is possible to achieve, that is, the total balance of Vasya in all banks is equal to zero.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of banks.
The second line contains *n* integers *a**i* (<=-<=109<=≤<=*a**i*<=≤<=109), the *i*-th of them is equal to the initial balance of the account in the *i*-th bank. It's guaranteed that the sum of all *a**i* is equal to 0.
|
Print the minimum number of operations required to change balance in each bank to zero.
|
[
"3\n5 0 -5\n",
"4\n-1 0 1 0\n",
"4\n1 2 3 -6\n"
] |
[
"1\n",
"2\n",
"3\n"
] |
In the first sample, Vasya may transfer 5 from the first bank to the third.
In the second sample, Vasya may first transfer 1 from the third bank to the second, and then 1 from the second to the first.
In the third sample, the following sequence provides the optimal answer:
1. transfer 1 from the first bank to the second bank; 1. transfer 3 from the second bank to the third; 1. transfer 6 from the third bank to the fourth.
| 1,500
|
[
{
"input": "3\n5 0 -5",
"output": "1"
},
{
"input": "4\n-1 0 1 0",
"output": "2"
},
{
"input": "4\n1 2 3 -6",
"output": "3"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "50\n108431864 128274949 -554057370 -384620666 -202862975 -803855410 -482167063 -55139054 -215901009 0 0 0 0 0 94325701 730397219 358214459 -673647271 -131397668 -377892440 0 0 0 0 0 -487994257 -360271553 639988328 489338210 -281060728 250208758 0 993242346 -213071841 -59752620 -864351041 -114363541 506279952 999648597 -173503559 -144629749 -559693009 0 -46793577 511999017 -343503822 -741715911 647437511 821346413 993112810",
"output": "36"
},
{
"input": "6\n1 -1 1 -1 1 -1",
"output": "3"
}
] | 1,464,268,088
| 2,707
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 4,608,000
|
n = int(input())
a = list(map(int,input().split()))
if a == [0]*n:
print(0)
exit(0)
f = 0
for i in range(n):
if a[i] != 0:
f = i
break
res = 0
tmp = 0
s = 0
for j in range(f, f+n):
i = j % n
s += a[i]
if s == 0:
res += tmp
tmp = 0
else:
tmp += 1
a = a[::-1]
f = n-f-1
res1 = 0
tmp = 0
s = 0
for j in range(f, f+n):
i = j % n
s += a[i]
if s == 0:
res1 += tmp
tmp = 0
else:
tmp += 1
print(min(res,res1))
|
Title: Money Transfers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* banks in the city where Vasya lives, they are located in a circle, such that any two banks are neighbouring if their indices differ by no more than 1. Also, bank 1 and bank *n* are neighbours if *n*<=><=1. No bank is a neighbour of itself.
Vasya has an account in each bank. Its balance may be negative, meaning Vasya owes some money to this bank.
There is only one type of operations available: transfer some amount of money from any bank to account in any neighbouring bank. There are no restrictions on the size of the sum being transferred or balance requirements to perform this operation.
Vasya doesn't like to deal with large numbers, so he asks you to determine the minimum number of operations required to change the balance of each bank account to zero. It's guaranteed, that this is possible to achieve, that is, the total balance of Vasya in all banks is equal to zero.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of banks.
The second line contains *n* integers *a**i* (<=-<=109<=≤<=*a**i*<=≤<=109), the *i*-th of them is equal to the initial balance of the account in the *i*-th bank. It's guaranteed that the sum of all *a**i* is equal to 0.
Output Specification:
Print the minimum number of operations required to change balance in each bank to zero.
Demo Input:
['3\n5 0 -5\n', '4\n-1 0 1 0\n', '4\n1 2 3 -6\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In the first sample, Vasya may transfer 5 from the first bank to the third.
In the second sample, Vasya may first transfer 1 from the third bank to the second, and then 1 from the second to the first.
In the third sample, the following sequence provides the optimal answer:
1. transfer 1 from the first bank to the second bank; 1. transfer 3 from the second bank to the third; 1. transfer 6 from the third bank to the fourth.
|
```python
n = int(input())
a = list(map(int,input().split()))
if a == [0]*n:
print(0)
exit(0)
f = 0
for i in range(n):
if a[i] != 0:
f = i
break
res = 0
tmp = 0
s = 0
for j in range(f, f+n):
i = j % n
s += a[i]
if s == 0:
res += tmp
tmp = 0
else:
tmp += 1
a = a[::-1]
f = n-f-1
res1 = 0
tmp = 0
s = 0
for j in range(f, f+n):
i = j % n
s += a[i]
if s == 0:
res1 += tmp
tmp = 0
else:
tmp += 1
print(min(res,res1))
```
| 0
|
|
447
|
B
|
DZY Loves Strings
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where
Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?
|
The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103).
The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103).
The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.
|
Print a single integer — the largest possible value of the resulting string DZY could get.
|
[
"abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n"
] |
[
"41\n"
] |
In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.
| 1,000
|
[
{
"input": "abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "41"
},
{
"input": "mmzhr\n3\n443 497 867 471 195 670 453 413 579 466 553 881 847 642 269 996 666 702 487 209 257 741 974 133 519 453",
"output": "29978"
},
{
"input": "ajeeseerqnpaujubmajpibxrccazaawetywxmifzehojf\n23\n359 813 772 413 733 654 33 87 890 433 395 311 801 852 376 148 914 420 636 695 583 733 664 394 407 314",
"output": "1762894"
},
{
"input": "uahngxejpomhbsebcxvelfsojbaouynnlsogjyvktpwwtcyddkcdqcqs\n34\n530 709 150 660 947 830 487 142 208 276 885 542 138 214 76 184 273 753 30 195 722 236 82 691 572 585",
"output": "2960349"
},
{
"input": "xnzeqmouqyzvblcidmhbkqmtusszuczadpooslqxegldanwopilmdwzbczvrwgnwaireykwpugvpnpafbxlyggkgawghysufuegvmzvpgcqyjkoadcreaguzepbendwnowsuekxxivkziibxvxfoilofxcgnxvfefyezfhevfvtetsuhwtyxdlkccdkvqjl\n282\n170 117 627 886 751 147 414 187 150 960 410 70 576 681 641 729 798 877 611 108 772 643 683 166 305 933",
"output": "99140444"
},
{
"input": "pplkqmluhfympkjfjnfdkwrkpumgdmbkfbbldpepicbbmdgafttpopzdxsevlqbtywzkoxyviglbbxsohycbdqksrhlumsldiwzjmednbkcjishkiekfrchzuztkcxnvuykhuenqojrmzaxlaoxnljnvqgnabtmcftisaazzgbmubmpsorygyusmeonrhrgphnfhlaxrvyhuxsnnezjxmdoklpquzpvjbxgbywppmegzxknhfzyygrmejleesoqfwheulmqhonqaukyuejtwxskjldplripyihbfpookxkuehiwqthbfafyrgmykuxglpplozycgydyecqkgfjljfqvigqhuxssqqtfanwszduwbsoytnrtgc\n464\n838 95 473 955 690 84 436 19 179 437 674 626 377 365 781 4 733 776 462 203 119 256 381 668 855 686",
"output": "301124161"
},
{
"input": "qkautnuilwlhjsldfcuwhiqtgtoihifszlyvfaygrnivzgvwthkrzzdtfjcirrjjlrmjtbjlzmjeqmuffsjorjyggzefwgvmblvotvzffnwjhqxorpowzdcnfksdibezdtfjjxfozaghieksbmowrbeehuxlesmvqjsphlvauxiijm\n98\n121 622 0 691 616 959 838 161 581 862 876 830 267 812 598 106 337 73 588 323 999 17 522 399 657 495",
"output": "30125295"
},
{
"input": "tghyxqfmhz\n8\n191 893 426 203 780 326 148 259 182 140 847 636 778 97 167 773 219 891 758 993 695 603 223 779 368 165",
"output": "136422"
},
{
"input": "nyawbfjxnxjiyhwkydaruozobpphgjqdpfdqzezcsoyvurnapu\n30\n65 682 543 533 990 148 815 821 315 916 632 771 332 513 472 864 12 73 548 687 660 572 507 192 226 348",
"output": "2578628"
},
{
"input": "pylrnkrbcjgoytvdnhmlvnkknijkdgdhworlvtwuonrkhrilkewcnofodaumgvnsisxooswgrgtvdeauyxhkipfoxrrtysuepjcf\n60\n894 206 704 179 272 337 413 828 119 182 330 46 440 102 250 191 242 539 678 783 843 431 612 567 33 338",
"output": "9168707"
},
{
"input": "vhjnkrxbyhjhnjrxvwxmhxwoxttbtqosfxtcuvhfjlkyfspeypthsdkkwnqdpxdlnxsgtzvkrgqosgfjrwetqbxgoarkjhrjbspzgblsapifltkfxbfdbxqwoohlgyzijmiwnpmveybyzvasoctxsmgjehpyysmqblwnmkappbecklqjfmxhlyceordroflnposohfplrvijxbwvqdtvzhobtrumiujnyrfbwthvciinuveoizkccelxtaveiiagryqnyvsgfnipnavrtmdqlcnldepocbpzmqnarkdvykds\n276\n364 244 798 82 582 9 309 950 286 547 892 371 569 159 705 975 740 845 655 179 130 993 255 552 882 657",
"output": "144901921"
},
{
"input": "gsaddmezrnttfalbwlqbnedumvikplfosw\n12\n290 850 872 361 483 895 152 118 974 619 701 154 899 285 328 712 669 984 407 340 851 775 324 892 554 860",
"output": "809931"
},
{
"input": "a\n0\n5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "5"
},
{
"input": "lol\n3\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "21"
}
] | 1,536,408,604
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 124
| 0
|
s = input()
k = int(input())
w = [int(el) for el in input().split()]
ans = 0
zp = ord('a')
i = 1
for el in s:
ans += i * w[ord(el) - zp]
i += 1
maxw = max(w)
while k:
ans += i * maxw
k -= 1
i += 1
print(ans)
|
Title: DZY Loves Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where
Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?
Input Specification:
The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103).
The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103).
The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.
Output Specification:
Print a single integer — the largest possible value of the resulting string DZY could get.
Demo Input:
['abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n']
Demo Output:
['41\n']
Note:
In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.
|
```python
s = input()
k = int(input())
w = [int(el) for el in input().split()]
ans = 0
zp = ord('a')
i = 1
for el in s:
ans += i * w[ord(el) - zp]
i += 1
maxw = max(w)
while k:
ans += i * maxw
k -= 1
i += 1
print(ans)
```
| 3
|
|
124
|
A
|
The number of positions
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
|
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
|
Print the single number — the number of the sought positions.
|
[
"3 1 1\n",
"5 2 3\n"
] |
[
"2\n",
"3\n"
] |
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
| 500
|
[
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 7",
"output": "5"
},
{
"input": "14 5 5",
"output": "6"
},
{
"input": "16 6 9",
"output": "10"
},
{
"input": "20 13 17",
"output": "7"
},
{
"input": "22 4 8",
"output": "9"
},
{
"input": "23 8 14",
"output": "15"
},
{
"input": "26 18 22",
"output": "8"
},
{
"input": "28 6 1",
"output": "2"
},
{
"input": "29 5 23",
"output": "24"
},
{
"input": "32 27 15",
"output": "5"
},
{
"input": "33 11 5",
"output": "6"
},
{
"input": "37 21 15",
"output": "16"
},
{
"input": "39 34 33",
"output": "5"
},
{
"input": "41 27 11",
"output": "12"
},
{
"input": "42 25 16",
"output": "17"
},
{
"input": "45 7 43",
"output": "38"
},
{
"input": "47 16 17",
"output": "18"
},
{
"input": "49 11 37",
"output": "38"
},
{
"input": "51 38 39",
"output": "13"
},
{
"input": "52 29 7",
"output": "8"
},
{
"input": "56 43 12",
"output": "13"
},
{
"input": "58 57 28",
"output": "1"
},
{
"input": "59 12 39",
"output": "40"
},
{
"input": "62 9 52",
"output": "53"
},
{
"input": "63 29 44",
"output": "34"
},
{
"input": "65 30 22",
"output": "23"
},
{
"input": "66 27 38",
"output": "39"
},
{
"input": "71 33 53",
"output": "38"
},
{
"input": "73 14 12",
"output": "13"
},
{
"input": "73 37 35",
"output": "36"
},
{
"input": "76 69 44",
"output": "7"
},
{
"input": "79 25 20",
"output": "21"
},
{
"input": "81 60 20",
"output": "21"
},
{
"input": "81 79 14",
"output": "2"
},
{
"input": "84 0 42",
"output": "43"
},
{
"input": "88 79 8",
"output": "9"
},
{
"input": "90 76 59",
"output": "14"
},
{
"input": "92 2 22",
"output": "23"
},
{
"input": "94 5 88",
"output": "89"
},
{
"input": "94 62 48",
"output": "32"
},
{
"input": "96 22 72",
"output": "73"
},
{
"input": "100 11 88",
"output": "89"
},
{
"input": "100 81 91",
"output": "19"
},
{
"input": "1 0 0",
"output": "1"
}
] | 1,619,360,490
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 122
| 0
|
def soughts(n, a, b):
if a == 0:
return 0
elif a == n:
return 1
else:
return n-a
n, a, b = map(int, input().split())
print(soughts(n, a, b))
|
Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
|
```python
def soughts(n, a, b):
if a == 0:
return 0
elif a == n:
return 1
else:
return n-a
n, a, b = map(int, input().split())
print(soughts(n, a, b))
```
| 0
|
|
349
|
A
|
Cinema Line
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
|
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
|
[
"4\n25 25 50 50\n",
"2\n25 100\n",
"4\n50 50 25 25\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "4\n25 25 50 50",
"output": "YES"
},
{
"input": "2\n25 100",
"output": "NO"
},
{
"input": "4\n50 50 25 25",
"output": "NO"
},
{
"input": "3\n25 50 100",
"output": "NO"
},
{
"input": "10\n25 25 25 25 25 25 25 25 25 25",
"output": "YES"
},
{
"input": "10\n50 50 50 50 50 50 50 50 50 50",
"output": "NO"
},
{
"input": "10\n100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "7\n25 25 25 25 50 100 50",
"output": "YES"
},
{
"input": "2\n25 25",
"output": "YES"
},
{
"input": "3\n50 100 100",
"output": "NO"
},
{
"input": "3\n25 50 50",
"output": "NO"
},
{
"input": "2\n50 25",
"output": "NO"
},
{
"input": "14\n25 25 25 25 25 25 25 50 50 50 100 100 100 100",
"output": "NO"
},
{
"input": "5\n25 25 25 100 100",
"output": "NO"
},
{
"input": "7\n25 50 25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 25 50 50 50",
"output": "NO"
},
{
"input": "5\n25 25 50 100 50",
"output": "NO"
},
{
"input": "5\n25 25 50 50 100",
"output": "NO"
},
{
"input": "1\n50",
"output": "NO"
}
] | 1,629,965,530
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 248
| 12,390,400
|
n = int(input())
arr = list(map(int, input().split()))
n25 = 0
n50 = 0
poss = True
for i in range(n):
if (arr[i] == 25):
n25 += 25
elif (arr[i] == 50):
n50 += 50
n25 -= 25
else:
if (n50 > 0):
n25 -= 25
n50 -= 50
else:
n25 -= 75
if (n25 < 0 or n50 < 0):
poss = False
print("NO")
exit()
print("YES")
|
Title: Cinema Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
Output Specification:
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
Demo Input:
['4\n25 25 50 50\n', '2\n25 100\n', '4\n50 50 25 25\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
n = int(input())
arr = list(map(int, input().split()))
n25 = 0
n50 = 0
poss = True
for i in range(n):
if (arr[i] == 25):
n25 += 25
elif (arr[i] == 50):
n50 += 50
n25 -= 25
else:
if (n50 > 0):
n25 -= 25
n50 -= 50
else:
n25 -= 75
if (n25 < 0 or n50 < 0):
poss = False
print("NO")
exit()
print("YES")
```
| 3
|
|
13
|
A
|
Numbers
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Numbers
|
1
|
64
|
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
|
Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000).
|
Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.
|
[
"5\n",
"3\n"
] |
[
"7/3\n",
"2/1\n"
] |
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
| 0
|
[
{
"input": "5",
"output": "7/3"
},
{
"input": "3",
"output": "2/1"
},
{
"input": "1000",
"output": "90132/499"
},
{
"input": "927",
"output": "155449/925"
},
{
"input": "260",
"output": "6265/129"
},
{
"input": "131",
"output": "3370/129"
},
{
"input": "386",
"output": "857/12"
},
{
"input": "277",
"output": "2864/55"
},
{
"input": "766",
"output": "53217/382"
},
{
"input": "28",
"output": "85/13"
},
{
"input": "406",
"output": "7560/101"
},
{
"input": "757",
"output": "103847/755"
},
{
"input": "6",
"output": "9/4"
},
{
"input": "239",
"output": "10885/237"
},
{
"input": "322",
"output": "2399/40"
},
{
"input": "98",
"output": "317/16"
},
{
"input": "208",
"output": "4063/103"
},
{
"input": "786",
"output": "55777/392"
},
{
"input": "879",
"output": "140290/877"
},
{
"input": "702",
"output": "89217/700"
},
{
"input": "948",
"output": "7369/43"
},
{
"input": "537",
"output": "52753/535"
},
{
"input": "984",
"output": "174589/982"
},
{
"input": "934",
"output": "157951/932"
},
{
"input": "726",
"output": "95491/724"
},
{
"input": "127",
"output": "3154/125"
},
{
"input": "504",
"output": "23086/251"
},
{
"input": "125",
"output": "3080/123"
},
{
"input": "604",
"output": "33178/301"
},
{
"input": "115",
"output": "2600/113"
},
{
"input": "27",
"output": "167/25"
},
{
"input": "687",
"output": "85854/685"
},
{
"input": "880",
"output": "69915/439"
},
{
"input": "173",
"output": "640/19"
},
{
"input": "264",
"output": "6438/131"
},
{
"input": "785",
"output": "111560/783"
},
{
"input": "399",
"output": "29399/397"
},
{
"input": "514",
"output": "6031/64"
},
{
"input": "381",
"output": "26717/379"
},
{
"input": "592",
"output": "63769/590"
},
{
"input": "417",
"output": "32002/415"
},
{
"input": "588",
"output": "62723/586"
},
{
"input": "852",
"output": "131069/850"
},
{
"input": "959",
"output": "5059/29"
},
{
"input": "841",
"output": "127737/839"
},
{
"input": "733",
"output": "97598/731"
},
{
"input": "692",
"output": "87017/690"
},
{
"input": "69",
"output": "983/67"
},
{
"input": "223",
"output": "556/13"
},
{
"input": "93",
"output": "246/13"
},
{
"input": "643",
"output": "75503/641"
},
{
"input": "119",
"output": "2833/117"
},
{
"input": "498",
"output": "1459/16"
},
{
"input": "155",
"output": "4637/153"
},
{
"input": "305",
"output": "17350/303"
},
{
"input": "454",
"output": "37893/452"
},
{
"input": "88",
"output": "1529/86"
},
{
"input": "850",
"output": "32645/212"
},
{
"input": "474",
"output": "20581/236"
},
{
"input": "309",
"output": "17731/307"
},
{
"input": "762",
"output": "105083/760"
},
{
"input": "591",
"output": "63761/589"
},
{
"input": "457",
"output": "38317/455"
},
{
"input": "141",
"output": "3832/139"
},
{
"input": "385",
"output": "27232/383"
},
{
"input": "387",
"output": "27628/385"
},
{
"input": "469",
"output": "40306/467"
},
{
"input": "624",
"output": "35285/311"
},
{
"input": "330",
"output": "487/8"
},
{
"input": "31",
"output": "222/29"
},
{
"input": "975",
"output": "171679/973"
},
{
"input": "584",
"output": "62183/582"
},
{
"input": "668",
"output": "81127/666"
},
{
"input": "331",
"output": "20297/329"
},
{
"input": "189",
"output": "6789/187"
},
{
"input": "251",
"output": "11939/249"
},
{
"input": "876",
"output": "69196/437"
},
{
"input": "615",
"output": "68987/613"
},
{
"input": "451",
"output": "37258/449"
},
{
"input": "499",
"output": "45727/497"
},
{
"input": "699",
"output": "89117/697"
},
{
"input": "619",
"output": "70019/617"
},
{
"input": "413",
"output": "10515/137"
},
{
"input": "197",
"output": "7399/195"
},
{
"input": "794",
"output": "14281/99"
},
{
"input": "659",
"output": "79403/657"
},
{
"input": "653",
"output": "77695/651"
},
{
"input": "23",
"output": "45/7"
},
{
"input": "430",
"output": "16985/214"
},
{
"input": "249",
"output": "11659/247"
},
{
"input": "837",
"output": "126869/835"
},
{
"input": "258",
"output": "12373/256"
},
{
"input": "995",
"output": "59665/331"
},
{
"input": "102",
"output": "504/25"
},
{
"input": "989",
"output": "177124/987"
},
{
"input": "376",
"output": "13008/187"
},
{
"input": "657",
"output": "15715/131"
},
{
"input": "746",
"output": "50509/372"
},
{
"input": "602",
"output": "13177/120"
}
] | 1,519,606,184
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 99
| 124
| 5,632,000
|
#k=int(input())
#n,m=map(int,input().split())
#a=list(map(int,input().split()))
#b=list(map(int,input().split()))
def gcd(a,b):
while(b):
r=a%b
a=b
b=r
return a
def sumdig(a,base):
cnt=0
while(a):
cnt+=a%base
a//=base
return cnt
n=int(input())
s=0
for i in range(2,n):
s+=sumdig(n,i)
print(str(s//gcd(s,n-1-2+1))+'/'+str((n-1-2+1)//gcd(s,n-1-2+1)))
|
Title: Numbers
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input Specification:
Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000).
Output Specification:
Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.
Demo Input:
['5\n', '3\n']
Demo Output:
['7/3\n', '2/1\n']
Note:
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
|
```python
#k=int(input())
#n,m=map(int,input().split())
#a=list(map(int,input().split()))
#b=list(map(int,input().split()))
def gcd(a,b):
while(b):
r=a%b
a=b
b=r
return a
def sumdig(a,base):
cnt=0
while(a):
cnt+=a%base
a//=base
return cnt
n=int(input())
s=0
for i in range(2,n):
s+=sumdig(n,i)
print(str(s//gcd(s,n-1-2+1))+'/'+str((n-1-2+1)//gcd(s,n-1-2+1)))
```
| 3.896038
|
946
|
A
|
Partition
|
PROGRAMMING
| 800
|
[
"greedy"
] | null | null |
You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences.
Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*?
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*.
|
Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*.
|
[
"3\n1 -2 0\n",
"6\n16 23 16 15 42 8\n"
] |
[
"3\n",
"120\n"
] |
In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3.
In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120.
| 0
|
[
{
"input": "3\n1 -2 0",
"output": "3"
},
{
"input": "6\n16 23 16 15 42 8",
"output": "120"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "100\n-100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100",
"output": "10000"
},
{
"input": "2\n-1 5",
"output": "6"
},
{
"input": "3\n-2 0 1",
"output": "3"
},
{
"input": "12\n-1 -2 -3 4 4 -6 -6 56 3 3 -3 3",
"output": "94"
},
{
"input": "4\n1 -1 1 -1",
"output": "4"
},
{
"input": "4\n100 -100 100 -100",
"output": "400"
},
{
"input": "3\n-2 -5 10",
"output": "17"
},
{
"input": "5\n1 -2 3 -4 5",
"output": "15"
},
{
"input": "3\n-100 100 -100",
"output": "300"
},
{
"input": "6\n1 -1 1 -1 1 -1",
"output": "6"
},
{
"input": "6\n2 -2 2 -2 2 -2",
"output": "12"
},
{
"input": "9\n12 93 -2 0 0 0 3 -3 -9",
"output": "122"
},
{
"input": "6\n-1 2 4 -5 -3 55",
"output": "70"
},
{
"input": "6\n-12 8 68 -53 1 -15",
"output": "157"
},
{
"input": "2\n-2 1",
"output": "3"
},
{
"input": "3\n100 -100 100",
"output": "300"
},
{
"input": "5\n100 100 -1 -100 2",
"output": "303"
},
{
"input": "6\n-5 -4 -3 -2 -1 0",
"output": "15"
},
{
"input": "6\n4 4 4 -3 -3 2",
"output": "20"
},
{
"input": "2\n-1 2",
"output": "3"
},
{
"input": "1\n100",
"output": "100"
},
{
"input": "5\n-1 -2 3 1 2",
"output": "9"
},
{
"input": "5\n100 -100 100 -100 100",
"output": "500"
},
{
"input": "5\n1 -1 1 -1 1",
"output": "5"
},
{
"input": "4\n0 0 0 -1",
"output": "1"
},
{
"input": "5\n100 -100 -1 2 100",
"output": "303"
},
{
"input": "2\n75 0",
"output": "75"
},
{
"input": "4\n55 56 -59 -58",
"output": "228"
},
{
"input": "2\n9 71",
"output": "80"
},
{
"input": "2\n9 70",
"output": "79"
},
{
"input": "2\n9 69",
"output": "78"
},
{
"input": "2\n100 -100",
"output": "200"
},
{
"input": "4\n-9 4 -9 5",
"output": "27"
},
{
"input": "42\n91 -27 -79 -56 80 -93 -23 10 80 94 61 -89 -64 81 34 99 31 -32 -69 92 79 -9 73 66 -8 64 99 99 58 -19 -40 21 1 -33 93 -23 -62 27 55 41 57 36",
"output": "2348"
},
{
"input": "7\n-1 2 2 2 -1 2 -1",
"output": "11"
},
{
"input": "6\n-12 8 17 -69 7 -88",
"output": "201"
},
{
"input": "3\n1 -2 5",
"output": "8"
},
{
"input": "6\n-2 3 -4 5 6 -1",
"output": "21"
},
{
"input": "2\n-5 1",
"output": "6"
},
{
"input": "4\n2 2 -2 4",
"output": "10"
},
{
"input": "68\n21 47 -75 -25 64 83 83 -21 89 24 43 44 -35 34 -42 92 -96 -52 -66 64 14 -87 25 -61 -78 83 -96 -18 95 83 -93 -28 75 49 87 65 -93 -69 -2 95 -24 -36 -61 -71 88 -53 -93 -51 -81 -65 -53 -46 -56 6 65 58 19 100 57 61 -53 44 -58 48 -8 80 -88 72",
"output": "3991"
},
{
"input": "5\n5 5 -10 -1 1",
"output": "22"
},
{
"input": "3\n-1 2 3",
"output": "6"
},
{
"input": "76\n57 -38 -48 -81 93 -32 96 55 -44 2 38 -46 42 64 71 -73 95 31 -39 -62 -1 75 -17 57 28 52 12 -11 82 -84 59 -86 73 -97 34 97 -57 -85 -6 39 -5 -54 95 24 -44 35 -18 9 91 7 -22 -61 -80 54 -40 74 -90 15 -97 66 -52 -49 -24 65 21 -93 -29 -24 -4 -1 76 -93 7 -55 -53 1",
"output": "3787"
},
{
"input": "5\n-1 -2 1 2 3",
"output": "9"
},
{
"input": "4\n2 2 -2 -2",
"output": "8"
},
{
"input": "6\n100 -100 100 -100 100 -100",
"output": "600"
},
{
"input": "100\n-59 -33 34 0 69 24 -22 58 62 -36 5 45 -19 -73 61 -9 95 42 -73 -64 91 -96 2 53 -8 82 -79 16 18 -5 -53 26 71 38 -31 12 -33 -1 -65 -6 3 -89 22 33 -27 -36 41 11 -47 -32 47 -56 -38 57 -63 -41 23 41 29 78 16 -65 90 -58 -12 6 -60 42 -36 -52 -54 -95 -10 29 70 50 -94 1 93 48 -71 -77 -16 54 56 -60 66 76 31 8 44 -61 -74 23 37 38 18 -18 29 41",
"output": "4362"
},
{
"input": "2\n-1 1",
"output": "2"
},
{
"input": "3\n1 -2 100",
"output": "103"
},
{
"input": "5\n1 -2 3 1 2",
"output": "9"
},
{
"input": "10\n100 -10 -100 10 10 10 10 10 10 10",
"output": "280"
},
{
"input": "4\n2 0 -2 4",
"output": "8"
},
{
"input": "4\n3 -3 1 -1",
"output": "8"
},
{
"input": "3\n1 -1 1",
"output": "3"
},
{
"input": "4\n2 5 -2 4",
"output": "13"
},
{
"input": "2\n-2 2",
"output": "4"
},
{
"input": "3\n1 -2 1",
"output": "4"
},
{
"input": "5\n-1 -2 1 1 -1",
"output": "6"
},
{
"input": "4\n-2 0 2 4",
"output": "8"
},
{
"input": "8\n-42 7 87 -16 -5 65 -88 1",
"output": "311"
},
{
"input": "3\n1 -3 4",
"output": "8"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "3\n-1 2 -1",
"output": "4"
},
{
"input": "18\n-21 12 65 66 -24 62 82 35 -45 -47 28 37 5 -32 22 -14 -69 -95",
"output": "761"
},
{
"input": "4\n-1 1 -1 1",
"output": "4"
},
{
"input": "5\n-1 2 1 1 1",
"output": "6"
},
{
"input": "3\n1 1 1",
"output": "3"
}
] | 1,645,065,313
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 72
| 46
| 0
|
def particion(lista):
lista.sort()
suma_1,suma_2=0,0
for n in lista:
if n>=0:
suma_1+=n
else:
suma_2+=n
return suma_1-suma_2
def main():
number=int(input())
lista=[int(x) for x in input().split()]
print(particion(lista))
main()
|
Title: Partition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences.
Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*.
Output Specification:
Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*.
Demo Input:
['3\n1 -2 0\n', '6\n16 23 16 15 42 8\n']
Demo Output:
['3\n', '120\n']
Note:
In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3.
In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120.
|
```python
def particion(lista):
lista.sort()
suma_1,suma_2=0,0
for n in lista:
if n>=0:
suma_1+=n
else:
suma_2+=n
return suma_1-suma_2
def main():
number=int(input())
lista=[int(x) for x in input().split()]
print(particion(lista))
main()
```
| 3
|
|
886
|
A
|
ACM ICPC
|
PROGRAMMING
| 1,000
|
[
"brute force"
] | null | null |
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
|
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
|
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
|
[
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
| 500
|
[
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
"input": "353 313 327 470 597 31",
"output": "NO"
},
{
"input": "835 638 673 624 232 266",
"output": "NO"
},
{
"input": "936 342 19 398 247 874",
"output": "NO"
},
{
"input": "417 666 978 553 271 488",
"output": "NO"
},
{
"input": "71 66 124 199 67 147",
"output": "YES"
},
{
"input": "54 26 0 171 239 12",
"output": "YES"
},
{
"input": "72 8 186 92 267 69",
"output": "YES"
},
{
"input": "180 179 188 50 75 214",
"output": "YES"
},
{
"input": "16 169 110 136 404 277",
"output": "YES"
},
{
"input": "101 400 9 200 300 10",
"output": "YES"
},
{
"input": "101 400 200 9 300 10",
"output": "YES"
},
{
"input": "101 200 400 9 300 10",
"output": "YES"
},
{
"input": "101 400 200 300 9 10",
"output": "YES"
},
{
"input": "101 200 400 300 9 10",
"output": "YES"
},
{
"input": "4 4 4 4 5 4",
"output": "NO"
},
{
"input": "2 2 2 2 2 1",
"output": "NO"
},
{
"input": "1000 1000 999 1000 1000 1000",
"output": "NO"
},
{
"input": "129 1 10 29 8 111",
"output": "NO"
},
{
"input": "1000 1000 1000 999 999 1000",
"output": "YES"
},
{
"input": "101 200 300 400 9 10",
"output": "YES"
},
{
"input": "101 400 200 300 10 9",
"output": "YES"
},
{
"input": "101 200 400 300 10 9",
"output": "YES"
},
{
"input": "101 200 300 400 10 9",
"output": "YES"
},
{
"input": "101 200 300 10 400 9",
"output": "YES"
},
{
"input": "1 1 1 1 1 5",
"output": "NO"
},
{
"input": "8 1 1 3 3 0",
"output": "NO"
},
{
"input": "1 1 2 2 3 3",
"output": "YES"
},
{
"input": "1 2 2 5 2 5",
"output": "NO"
},
{
"input": "1 2 3 6 6 6",
"output": "NO"
},
{
"input": "36 91 7 86 51 89",
"output": "NO"
},
{
"input": "10 1 1 1 23 24",
"output": "NO"
},
{
"input": "1 1 1 10 23 24",
"output": "NO"
},
{
"input": "20 10 1 2 3 44",
"output": "NO"
},
{
"input": "7 0 14 11 8 6",
"output": "NO"
},
{
"input": "100 496 1 1 1 1",
"output": "NO"
},
{
"input": "5 4 2 5 11 3",
"output": "NO"
},
{
"input": "1 3 7 8 8 9",
"output": "YES"
},
{
"input": "1 3 4 5 18 19",
"output": "YES"
},
{
"input": "5 5 1 2 2 15",
"output": "NO"
},
{
"input": "2 1 0 0 0 5",
"output": "NO"
},
{
"input": "1 6 6 1 20 2",
"output": "NO"
},
{
"input": "2 10 0 0 0 0",
"output": "NO"
},
{
"input": "1 1 3 1 1 11",
"output": "NO"
},
{
"input": "10 10 1 1 1 37",
"output": "NO"
},
{
"input": "1 1 0 0 0 4",
"output": "NO"
},
{
"input": "1 1 10 1 1 28",
"output": "NO"
},
{
"input": "1 5 5 5 6 8",
"output": "YES"
},
{
"input": "0 2 3 4 4 5",
"output": "YES"
}
] | 1,530,391,369
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 108
| 0
|
a = [i for i in map(int, input().split())]
total = sum(a)
ans = False
for i in range(6):
for j in range(i + 1, 6):
for k in range(j + 1, 6):
if j + k + i == total / 2:
ans = True
print("YES" if ans else "NO")
|
Title: ACM ICPC
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
|
```python
a = [i for i in map(int, input().split())]
total = sum(a)
ans = False
for i in range(6):
for j in range(i + 1, 6):
for k in range(j + 1, 6):
if j + k + i == total / 2:
ans = True
print("YES" if ans else "NO")
```
| 0
|
|
216
|
B
|
Forming Teams
|
PROGRAMMING
| 1,700
|
[
"dfs and similar",
"implementation"
] | null | null |
One day *n* students come to the stadium. They want to play football, and for that they need to split into teams, the teams must have an equal number of people.
We know that this group of people has archenemies. Each student has at most two archenemies. Besides, if student *A* is an archenemy to student *B*, then student *B* is an archenemy to student *A*.
The students want to split so as no two archenemies were in one team. If splitting in the required manner is impossible, some students will have to sit on the bench.
Determine the minimum number of students you will have to send to the bench in order to form the two teams in the described manner and begin the game at last.
|
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=100) — the number of students and the number of pairs of archenemies correspondingly.
Next *m* lines describe enmity between students. Each enmity is described as two numbers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) — the indexes of the students who are enemies to each other. Each enmity occurs in the list exactly once. It is guaranteed that each student has no more than two archenemies.
You can consider the students indexed in some manner with distinct integers from 1 to *n*.
|
Print a single integer — the minimum number of students you will have to send to the bench in order to start the game.
|
[
"5 4\n1 2\n2 4\n5 3\n1 4\n",
"6 2\n1 4\n3 4\n",
"6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4\n"
] |
[
"1",
"0",
"2"
] |
none
| 1,500
|
[
{
"input": "5 4\n1 2\n2 4\n5 3\n1 4",
"output": "1"
},
{
"input": "6 2\n1 4\n3 4",
"output": "0"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4",
"output": "2"
},
{
"input": "5 1\n1 2",
"output": "1"
},
{
"input": "8 8\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 1",
"output": "0"
},
{
"input": "28 3\n15 3\n10 19\n17 25",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "3 1\n2 3",
"output": "1"
},
{
"input": "3 2\n1 2\n3 2",
"output": "1"
},
{
"input": "3 3\n1 2\n1 3\n2 3",
"output": "1"
},
{
"input": "4 1\n1 4",
"output": "0"
},
{
"input": "4 2\n4 1\n2 1",
"output": "0"
},
{
"input": "4 3\n1 3\n3 2\n2 4",
"output": "0"
},
{
"input": "4 3\n3 2\n4 2\n4 3",
"output": "2"
},
{
"input": "5 3\n4 2\n3 4\n5 1",
"output": "1"
},
{
"input": "10 7\n8 9\n3 6\n2 4\n4 1\n1 3\n2 7\n7 10",
"output": "0"
},
{
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"output": "1"
},
{
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"output": "0"
},
{
"input": "89 30\n86 72\n43 16\n32 80\n17 79\n29 8\n89 37\n84 65\n3 41\n55 79\n33 56\n60 40\n43 45\n59 38\n26 23\n66 61\n81 30\n65 25\n13 71\n25 8\n56 59\n46 13\n22 30\n87 3\n26 32\n75 44\n48 87\n47 4\n63 21\n36 6\n42 86",
"output": "1"
},
{
"input": "100 1\n3 87",
"output": "0"
},
{
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"output": "0"
},
{
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"output": "0"
},
{
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"output": "4"
},
{
"input": "50 48\n33 21\n1 46\n43 37\n1 48\n42 32\n31 45\n14 29\n34 28\n38 19\n46 48\n49 31\n8 3\n27 23\n26 37\n15 9\n27 17\n9 35\n18 7\n35 15\n32 4\n23 17\n36 22\n16 33\n39 6\n40 13\n11 6\n21 16\n10 40\n30 36\n20 5\n24 3\n43 26\n22 30\n41 20\n50 38\n25 29\n5 41\n34 44\n12 7\n8 24\n44 28\n25 14\n12 18\n39 11\n42 4\n45 49\n50 19\n13 10",
"output": "16"
},
{
"input": "19 16\n2 16\n7 10\n17 16\n17 14\n1 5\n19 6\n11 13\n15 19\n7 9\n13 5\n4 6\n1 11\n12 9\n10 12\n2 14\n4 15",
"output": "1"
},
{
"input": "70 70\n27 54\n45 23\n67 34\n66 25\n64 38\n30 68\n51 65\n19 4\n15 33\n47 14\n3 9\n42 29\n69 56\n10 50\n34 58\n51 23\n55 14\n18 53\n27 68\n17 6\n48 6\n8 5\n46 37\n37 33\n21 36\n69 24\n16 13\n50 12\n59 31\n63 38\n22 11\n46 28\n67 62\n63 26\n70 31\n7 59\n55 52\n28 43\n18 35\n53 3\n16 60\n43 40\n61 9\n20 44\n47 41\n35 1\n32 4\n13 54\n30 60\n45 19\n39 42\n2 20\n2 26\n52 8\n12 25\n5 41\n21 10\n58 48\n29 11\n7 56\n49 57\n65 32\n15 40\n66 36\n64 44\n22 57\n1 61\n39 49\n24 70\n62 17",
"output": "10"
},
{
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"output": "1"
},
{
"input": "10 8\n8 3\n9 7\n6 1\n10 9\n2 6\n2 1\n3 4\n4 8",
"output": "2"
},
{
"input": "20 12\n16 20\n8 3\n20 5\n5 10\n17 7\n13 2\n18 9\n17 18\n1 6\n14 4\n11 12\n10 16",
"output": "0"
},
{
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"output": "1"
},
{
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"output": "3"
},
{
"input": "77 54\n18 56\n72 2\n6 62\n58 52\n5 70\n24 4\n67 66\n65 47\n43 77\n61 66\n24 51\n70 7\n48 39\n46 11\n77 28\n65 76\n15 6\n22 13\n34 75\n33 42\n59 37\n7 31\n50 23\n28 9\n17 29\n1 14\n11 45\n36 46\n32 39\n59 21\n22 34\n53 21\n29 47\n16 44\n69 4\n62 16\n36 3\n68 75\n51 69\n49 43\n30 55\n40 20\n57 60\n45 3\n38 33\n49 9\n71 19\n73 20\n48 32\n63 67\n8 54\n42 38\n26 12\n5 74",
"output": "5"
},
{
"input": "93 72\n3 87\n88 60\n73 64\n45 35\n61 85\n68 80\n54 29\n4 88\n19 91\n82 48\n50 2\n40 53\n56 8\n66 82\n83 81\n62 8\n79 30\n89 26\n77 10\n65 15\n27 47\n15 51\n70 6\n59 85\n63 20\n64 92\n7 1\n93 52\n74 38\n71 23\n83 12\n86 52\n46 56\n34 36\n37 84\n18 16\n11 42\n69 72\n53 20\n78 84\n54 91\n14 5\n65 49\n90 19\n42 39\n68 57\n75 27\n57 32\n44 9\n79 74\n48 66\n43 93\n31 30\n58 24\n80 67\n6 60\n39 5\n23 17\n25 1\n18 36\n32 67\n10 9\n14 11\n63 21\n92 73\n13 43\n28 78\n33 51\n4 70\n75 45\n37 28\n62 46",
"output": "5"
},
{
"input": "100 72\n2 88\n55 80\n22 20\n78 52\n66 74\n91 82\n59 77\n97 93\n46 44\n99 35\n73 62\n58 24\n6 16\n47 41\n98 86\n23 19\n39 68\n32 28\n85 29\n37 40\n16 62\n19 61\n84 72\n17 15\n76 96\n37 31\n67 35\n48 15\n80 85\n90 47\n79 36\n39 54\n57 87\n42 60\n34 56\n23 61\n92 2\n88 63\n20 42\n27 81\n65 84\n6 73\n64 100\n76 95\n43 4\n65 86\n21 46\n11 64\n72 98\n63 92\n7 50\n14 22\n89 30\n31 40\n8 57\n90 70\n53 59\n69 24\n96 49\n67 99\n51 70\n18 66\n91 3\n26 38\n13 58\n51 41\n9 11\n5 74\n3 25\n4 32\n28 43\n71 56",
"output": "6"
},
{
"input": "6 5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "2"
},
{
"input": "6 4\n1 2\n1 3\n4 5\n4 6",
"output": "0"
},
{
"input": "16 16\n1 2\n2 3\n1 3\n4 5\n5 6\n4 6\n7 8\n8 9\n9 10\n10 11\n11 7\n12 13\n13 14\n14 15\n15 16\n16 12",
"output": "4"
},
{
"input": "4 4\n1 2\n4 3\n1 4\n2 3",
"output": "0"
},
{
"input": "9 9\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4\n7 8\n8 9\n9 7",
"output": "3"
},
{
"input": "20 11\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 1",
"output": "2"
},
{
"input": "4 3\n1 2\n3 4\n1 3",
"output": "0"
},
{
"input": "4 2\n2 4\n3 4",
"output": "0"
},
{
"input": "10 10\n1 2\n2 3\n3 4\n4 5\n5 1\n6 7\n7 8\n8 9\n9 10\n10 6",
"output": "2"
},
{
"input": "6 5\n2 1\n3 4\n2 3\n4 5\n5 6",
"output": "0"
},
{
"input": "8 5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "2"
},
{
"input": "6 5\n1 2\n2 3\n3 4\n4 5\n1 5",
"output": "2"
},
{
"input": "8 8\n1 2\n2 3\n3 4\n1 4\n5 6\n6 7\n7 8\n5 8",
"output": "0"
},
{
"input": "6 5\n1 3\n1 2\n2 4\n5 3\n5 4",
"output": "2"
}
] | 1,692,092,920
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
print("_RANDOM_GUESS_1692092920.1371987")# 1692092920.137215
|
Title: Forming Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* students come to the stadium. They want to play football, and for that they need to split into teams, the teams must have an equal number of people.
We know that this group of people has archenemies. Each student has at most two archenemies. Besides, if student *A* is an archenemy to student *B*, then student *B* is an archenemy to student *A*.
The students want to split so as no two archenemies were in one team. If splitting in the required manner is impossible, some students will have to sit on the bench.
Determine the minimum number of students you will have to send to the bench in order to form the two teams in the described manner and begin the game at last.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=100) — the number of students and the number of pairs of archenemies correspondingly.
Next *m* lines describe enmity between students. Each enmity is described as two numbers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) — the indexes of the students who are enemies to each other. Each enmity occurs in the list exactly once. It is guaranteed that each student has no more than two archenemies.
You can consider the students indexed in some manner with distinct integers from 1 to *n*.
Output Specification:
Print a single integer — the minimum number of students you will have to send to the bench in order to start the game.
Demo Input:
['5 4\n1 2\n2 4\n5 3\n1 4\n', '6 2\n1 4\n3 4\n', '6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4\n']
Demo Output:
['1', '0', '2']
Note:
none
|
```python
print("_RANDOM_GUESS_1692092920.1371987")# 1692092920.137215
```
| 0
|
|
689
|
A
|
Mike and Cellphone
|
PROGRAMMING
| 1,400
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bought a cheap replacement phone with an old-fashioned keyboard. The keyboard has only ten digital equal-sized keys, located in the following way:
Together with his old phone, he lost all his contacts and now he can only remember the way his fingers moved when he put some number in. One can formally consider finger movements as a sequence of vectors connecting centers of keys pressed consecutively to put in a number. For example, the finger movements for number "586" are the same as finger movements for number "253":
Mike has already put in a number by his "finger memory" and started calling it, so he is now worrying, can he be sure that he is calling the correct number? In other words, is there any other number, that has the same finger movements?
|
The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=9) — the number of digits in the phone number that Mike put in.
The second line contains the string consisting of *n* digits (characters from '0' to '9') representing the number that Mike put in.
|
If there is no other phone number with the same finger movements and Mike can be sure he is calling the correct number, print "YES" (without quotes) in the only line.
Otherwise print "NO" (without quotes) in the first line.
|
[
"3\n586\n",
"2\n09\n",
"9\n123456789\n",
"3\n911\n"
] |
[
"NO\n",
"NO\n",
"YES\n",
"YES\n"
] |
You can find the picture clarifying the first sample case in the statement above.
| 500
|
[
{
"input": "3\n586",
"output": "NO"
},
{
"input": "2\n09",
"output": "NO"
},
{
"input": "9\n123456789",
"output": "YES"
},
{
"input": "3\n911",
"output": "YES"
},
{
"input": "3\n089",
"output": "NO"
},
{
"input": "3\n159",
"output": "YES"
},
{
"input": "9\n000000000",
"output": "NO"
},
{
"input": "4\n0874",
"output": "NO"
},
{
"input": "6\n235689",
"output": "NO"
},
{
"input": "2\n10",
"output": "YES"
},
{
"input": "3\n358",
"output": "NO"
},
{
"input": "6\n123456",
"output": "NO"
},
{
"input": "1\n0",
"output": "NO"
},
{
"input": "4\n0068",
"output": "NO"
},
{
"input": "6\n021149",
"output": "YES"
},
{
"input": "5\n04918",
"output": "YES"
},
{
"input": "2\n05",
"output": "NO"
},
{
"input": "4\n0585",
"output": "NO"
},
{
"input": "4\n0755",
"output": "NO"
},
{
"input": "2\n08",
"output": "NO"
},
{
"input": "4\n0840",
"output": "NO"
},
{
"input": "9\n103481226",
"output": "YES"
},
{
"input": "4\n1468",
"output": "NO"
},
{
"input": "7\n1588216",
"output": "NO"
},
{
"input": "9\n188758557",
"output": "NO"
},
{
"input": "1\n2",
"output": "NO"
},
{
"input": "2\n22",
"output": "NO"
},
{
"input": "8\n23482375",
"output": "YES"
},
{
"input": "9\n246112056",
"output": "YES"
},
{
"input": "9\n256859223",
"output": "NO"
},
{
"input": "6\n287245",
"output": "NO"
},
{
"input": "8\n28959869",
"output": "NO"
},
{
"input": "9\n289887167",
"output": "YES"
},
{
"input": "4\n3418",
"output": "NO"
},
{
"input": "4\n3553",
"output": "NO"
},
{
"input": "2\n38",
"output": "NO"
},
{
"input": "6\n386126",
"output": "NO"
},
{
"input": "6\n392965",
"output": "NO"
},
{
"input": "1\n4",
"output": "NO"
},
{
"input": "6\n423463",
"output": "NO"
},
{
"input": "4\n4256",
"output": "NO"
},
{
"input": "8\n42937903",
"output": "YES"
},
{
"input": "1\n5",
"output": "NO"
},
{
"input": "8\n50725390",
"output": "YES"
},
{
"input": "9\n515821866",
"output": "NO"
},
{
"input": "2\n56",
"output": "NO"
},
{
"input": "2\n57",
"output": "NO"
},
{
"input": "7\n5740799",
"output": "NO"
},
{
"input": "9\n582526521",
"output": "NO"
},
{
"input": "9\n585284126",
"output": "NO"
},
{
"input": "1\n6",
"output": "NO"
},
{
"input": "3\n609",
"output": "NO"
},
{
"input": "2\n63",
"output": "NO"
},
{
"input": "3\n633",
"output": "NO"
},
{
"input": "7\n6668940",
"output": "NO"
},
{
"input": "5\n66883",
"output": "NO"
},
{
"input": "2\n68",
"output": "NO"
},
{
"input": "5\n69873",
"output": "YES"
},
{
"input": "1\n7",
"output": "NO"
},
{
"input": "4\n7191",
"output": "YES"
},
{
"input": "9\n722403540",
"output": "YES"
},
{
"input": "9\n769554547",
"output": "NO"
},
{
"input": "3\n780",
"output": "NO"
},
{
"input": "5\n78248",
"output": "NO"
},
{
"input": "4\n7844",
"output": "NO"
},
{
"input": "4\n7868",
"output": "NO"
},
{
"input": "1\n8",
"output": "NO"
},
{
"input": "6\n817332",
"output": "YES"
},
{
"input": "7\n8465393",
"output": "YES"
},
{
"input": "7\n8526828",
"output": "NO"
},
{
"input": "8\n85812664",
"output": "NO"
},
{
"input": "8\n93008225",
"output": "YES"
},
{
"input": "7\n9454566",
"output": "NO"
},
{
"input": "4\n9625",
"output": "NO"
},
{
"input": "8\n97862407",
"output": "YES"
},
{
"input": "3\n993",
"output": "NO"
},
{
"input": "3\n267",
"output": "YES"
},
{
"input": "3\n249",
"output": "YES"
},
{
"input": "3\n672",
"output": "YES"
},
{
"input": "3\n176",
"output": "YES"
},
{
"input": "3\n123",
"output": "NO"
},
{
"input": "3\n367",
"output": "YES"
},
{
"input": "2\n12",
"output": "NO"
},
{
"input": "4\n2580",
"output": "YES"
},
{
"input": "2\n20",
"output": "YES"
},
{
"input": "3\n492",
"output": "YES"
},
{
"input": "3\n167",
"output": "YES"
},
{
"input": "3\n970",
"output": "NO"
},
{
"input": "3\n460",
"output": "NO"
},
{
"input": "4\n4268",
"output": "NO"
},
{
"input": "4\n9394",
"output": "YES"
},
{
"input": "2\n13",
"output": "NO"
},
{
"input": "3\n729",
"output": "YES"
},
{
"input": "6\n456893",
"output": "YES"
}
] | 1,467,823,588
| 688
|
Python 3
|
OK
|
TESTS
| 93
| 62
| 0
|
n = int(input())
s = input()
haszero = 0
for x in s:
if(x=='0'):
haszero = 1
hasup = 0
hasleft = 0
hasright = 0
hasdown = 0
for x in s:
if(x=='1')or(x=='2')or(x=='3'):
hasup = 1
if(x=='1')or(x=='4')or(x=='7'):
hasleft = 1
if(x=='3')or(x=='6')or(x=='9'):
hasright = 1
if(x=='7')or(x=='0')or(x=='9'):
hasdown = 1
if hasup and hasleft and hasright and hasdown:
print('YES')
else:
if hasup and haszero:
print('YES')
else:
print('NO')
|
Title: Mike and Cellphone
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bought a cheap replacement phone with an old-fashioned keyboard. The keyboard has only ten digital equal-sized keys, located in the following way:
Together with his old phone, he lost all his contacts and now he can only remember the way his fingers moved when he put some number in. One can formally consider finger movements as a sequence of vectors connecting centers of keys pressed consecutively to put in a number. For example, the finger movements for number "586" are the same as finger movements for number "253":
Mike has already put in a number by his "finger memory" and started calling it, so he is now worrying, can he be sure that he is calling the correct number? In other words, is there any other number, that has the same finger movements?
Input Specification:
The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=9) — the number of digits in the phone number that Mike put in.
The second line contains the string consisting of *n* digits (characters from '0' to '9') representing the number that Mike put in.
Output Specification:
If there is no other phone number with the same finger movements and Mike can be sure he is calling the correct number, print "YES" (without quotes) in the only line.
Otherwise print "NO" (without quotes) in the first line.
Demo Input:
['3\n586\n', '2\n09\n', '9\n123456789\n', '3\n911\n']
Demo Output:
['NO\n', 'NO\n', 'YES\n', 'YES\n']
Note:
You can find the picture clarifying the first sample case in the statement above.
|
```python
n = int(input())
s = input()
haszero = 0
for x in s:
if(x=='0'):
haszero = 1
hasup = 0
hasleft = 0
hasright = 0
hasdown = 0
for x in s:
if(x=='1')or(x=='2')or(x=='3'):
hasup = 1
if(x=='1')or(x=='4')or(x=='7'):
hasleft = 1
if(x=='3')or(x=='6')or(x=='9'):
hasright = 1
if(x=='7')or(x=='0')or(x=='9'):
hasdown = 1
if hasup and hasleft and hasright and hasdown:
print('YES')
else:
if hasup and haszero:
print('YES')
else:
print('NO')
```
| 3
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,659,419,971
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 92
| 0
|
n=int(input())
l=list(map(int,input().split()))
s=l[:]
def myFunc(e):
return e%2
l.sort(key=myFunc)
if (l[1]%2!=0):
print(s.index(l[0])+1)
else:
print(s.index(l[-1])+1)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n=int(input())
l=list(map(int,input().split()))
s=l[:]
def myFunc(e):
return e%2
l.sort(key=myFunc)
if (l[1]%2!=0):
print(s.index(l[0])+1)
else:
print(s.index(l[-1])+1)
```
| 3.977
|
867
|
A
|
Between the Offices
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
|
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
|
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
|
[
"4\nFSSF\n",
"2\nSF\n",
"10\nFFFFFFFFFF\n",
"10\nSSFFSFFSFF\n"
] |
[
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] |
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
| 500
|
[
{
"input": "4\nFSSF",
"output": "NO"
},
{
"input": "2\nSF",
"output": "YES"
},
{
"input": "10\nFFFFFFFFFF",
"output": "NO"
},
{
"input": "10\nSSFFSFFSFF",
"output": "YES"
},
{
"input": "20\nSFSFFFFSSFFFFSSSSFSS",
"output": "NO"
},
{
"input": "20\nSSFFFFFSFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "20\nSSFSFSFSFSFSFSFSSFSF",
"output": "YES"
},
{
"input": "20\nSSSSFSFSSFSFSSSSSSFS",
"output": "NO"
},
{
"input": "100\nFFFSFSFSFSSFSFFSSFFFFFSSSSFSSFFFFSFFFFFSFFFSSFSSSFFFFSSFFSSFSFFSSFSSSFSFFSFSFFSFSFFSSFFSFSSSSFSFSFSS",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFSS",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFSFFFFFFFFFSFSSFFFFFFFFFFFFFFFFFFFFFFSFFSFFFFFSFFFFFFFFSFFFFFFFFFFFFFSFFFFFFFFSFFFFFFFSF",
"output": "NO"
},
{
"input": "100\nSFFSSFFFFFFSSFFFSSFSFFFFFSSFFFSFFFFFFSFSSSFSFSFFFFSFSSFFFFFFFFSFFFFFSFFFFFSSFFFSFFSFSFFFFSFFSFFFFFFF",
"output": "YES"
},
{
"input": "100\nFFFFSSSSSFFSSSFFFSFFFFFSFSSFSFFSFFSSFFSSFSFFFFFSFSFSFSFFFFFFFFFSFSFFSFFFFSFSFFFFFFFFFFFFSFSSFFSSSSFF",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFSSFFFFSFSFFFSFSSSFSSSSSFSSSSFFSSFFFSFSFSSFFFSSSFFSFSFSSFSFSSFSFFFSFFFFFSSFSFFFSSSFSSSFFS",
"output": "NO"
},
{
"input": "100\nFFFSSSFSFSSSSFSSFSFFSSSFFSSFSSFFSSFFSFSSSSFFFSFFFSFSFSSSFSSFSFSFSFFSSSSSFSSSFSFSFFSSFSFSSFFSSFSFFSFS",
"output": "NO"
},
{
"input": "100\nFFSSSSFSSSFSSSSFSSSFFSFSSFFSSFSSSFSSSFFSFFSSSSSSSSSSSSFSSFSSSSFSFFFSSFFFFFFSFSFSSSSSSFSSSFSFSSFSSFSS",
"output": "NO"
},
{
"input": "100\nSSSFFFSSSSFFSSSSSFSSSSFSSSFSSSSSFSSSSSSSSFSFFSSSFFSSFSSSSFFSSSSSSFFSSSSFSSSSSSFSSSFSSSSSSSFSSSSFSSSS",
"output": "NO"
},
{
"input": "100\nFSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSSSSSSFSSSSSSSSSSSSSFSSFSSSSSFSSFSSSSSSSSSFFSSSSSFSFSSSFFSSSSSSSSSSSSS",
"output": "NO"
},
{
"input": "100\nSSSSSSSSSSSSSFSSSSSSSSSSSSFSSSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSFS",
"output": "NO"
},
{
"input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS",
"output": "NO"
},
{
"input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFSFFFFFFFFFFFSFSFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSFFFFFFFFFFFFSSFFFFSFFFFFFFFFFFFFFFFFFFSFFFSSFFFFSFSFFFSFFFFFFFFFFFFFFFSSFFFFFFFFSSFFFFFFFFFFFFFFSFF",
"output": "YES"
},
{
"input": "100\nSFFSSSFFSFSFSFFFFSSFFFFSFFFFFFFFSFSFFFSFFFSFFFSFFFFSFSFFFFFFFSFFFFFFFFFFSFFSSSFFSSFFFFSFFFFSFFFFSFFF",
"output": "YES"
},
{
"input": "100\nSFFFSFFFFSFFFSSFFFSFSFFFSFFFSSFSFFFFFSFFFFFFFFSFSFSFFSFFFSFSSFSFFFSFSFFSSFSFSSSFFFFFFSSFSFFSFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSSSSFFFFSFFFFFFFSFFFFSFSFFFFSSFFFFFFFFFSFFSSFFFFFFSFSFSSFSSSFFFFFFFSFSFFFSSSFFFFFFFSFFFSSFFFFSSFFFSF",
"output": "YES"
},
{
"input": "100\nSSSFSSFFFSFSSSSFSSFSSSSFSSFFFFFSFFSSSSFFSSSFSSSFSSSSFSSSSFSSSSSSSFSFSSFFFSSFFSFFSSSSFSSFFSFSSFSFFFSF",
"output": "YES"
},
{
"input": "100\nSFFSFSSSSSSSFFSSSFSSSSFSFSSFFFSSSSSSFSSSSFSSFSSSFSSSSSSSFSSFSFFFSSFSSFSFSFSSSSSSSSSSSSFFFFSSSSSFSFFF",
"output": "YES"
},
{
"input": "100\nSSSFSFFSFSFFSSSSSFSSSFSSSFFFSSSSSSSSSFSFSFSSSSFSFSSFFFFFSSSSSSSSSSSSSSSSSSSFFSSSSSFSFSSSSFFSSSSFSSSF",
"output": "YES"
},
{
"input": "100\nSSSFSSSSSSSSSSFSSSSFSSSSSSFSSSSSSFSSSSSSSSSSSSSSFSSSFSSSFSSSSSSSSSSSFSSSSSSFSFSSSSFSSSSSSFSSSSSSSSFF",
"output": "YES"
},
{
"input": "100\nSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFFSSSSSSSSSFSSSSSSSSSSSSSSSSSF",
"output": "YES"
},
{
"input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSF",
"output": "YES"
},
{
"input": "2\nSS",
"output": "NO"
}
] | 1,698,064,869
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 34
| 62
| 0
|
n = int(input())
str = input()
cnt1 = 0
cnt2 = 0
for i in range(1, n):
if str[i - 1] == 'F' and str[i] == 'S':
cnt1 += 1
elif str[i - 1] == 'S' and str[i] == 'F':
cnt2 += 1
if cnt2 > cnt1:
print("YES")
else:
print("NO")# 1698064869.6158943
|
Title: Between the Offices
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
Input Specification:
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output Specification:
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['4\nFSSF\n', '2\nSF\n', '10\nFFFFFFFFFF\n', '10\nSSFFSFFSFF\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
|
```python
n = int(input())
str = input()
cnt1 = 0
cnt2 = 0
for i in range(1, n):
if str[i - 1] == 'F' and str[i] == 'S':
cnt1 += 1
elif str[i - 1] == 'S' and str[i] == 'F':
cnt2 += 1
if cnt2 > cnt1:
print("YES")
else:
print("NO")# 1698064869.6158943
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.
|
The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order.
|
If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
|
[
"4 4\n1 3 5 7\n",
"10 8\n10 13 15 16 17 19 20 22 24 25\n",
"3 1\n2 5 10\n"
] |
[
"0.5\n",
"0.875\n",
"-1\n"
] |
In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 0
|
[
{
"input": "4 4\n1 3 5 7",
"output": "0.5"
},
{
"input": "10 8\n10 13 15 16 17 19 20 22 24 25",
"output": "0.875"
},
{
"input": "3 1\n2 5 10",
"output": "-1"
},
{
"input": "5 3\n4 6 8 9 10",
"output": "0.5"
},
{
"input": "10 128\n110 121 140 158 174 188 251 271 272 277",
"output": "0.86554621848739499157"
},
{
"input": "20 17\n104 107 121 131 138 140 143 144 178 192 193 198 201 206 238 242 245 248 255 265",
"output": "0.92857142857142860315"
},
{
"input": "30 23\n102 104 105 107 108 109 110 111 116 118 119 122 127 139 140 142 145 157 166 171 173 174 175 181 187 190 191 193 195 196",
"output": "0.95652173913043481157"
},
{
"input": "50 64\n257 258 350 375 1014 1017 1051 1097 1169 1177 1223 1836 1942 1983 2111 2131 2341 2418 2593 2902 2948 3157 3243 3523 3566 4079 4499 4754 5060 5624 6279 6976 7011 7071 7278 7366 7408 7466 7526 7837 7934 8532 8577 8680 9221 9271 9327 9411 9590 9794",
"output": "0.91891891891891896993"
},
{
"input": "5 2\n4 6 8 9 10",
"output": "0.5"
},
{
"input": "10 2\n110 121 140 158 174 188 251 271 272 277",
"output": "-1"
},
{
"input": "30 5\n102 104 105 107 108 109 110 111 116 118 119 122 127 139 140 142 145 157 166 171 173 174 175 181 187 190 191 193 195 196",
"output": "0.80000000000000004441"
},
{
"input": "10 6\n110 121 140 158 174 188 251 271 272 277",
"output": "0.83333333333333337034"
},
{
"input": "20 4\n104 107 121 131 138 140 143 144 178 192 193 198 201 206 238 242 245 248 255 265",
"output": "0.25"
},
{
"input": "3 1000000000\n1 2 1000000000",
"output": "0.99999999900000002828"
},
{
"input": "3 1\n1 2 3",
"output": "-1"
},
{
"input": "5 1000000000\n1 2 3 999999999 1000000000",
"output": "0.99999999900000002828"
},
{
"input": "10 199\n1 3 190 191 193 195 196 197 199 200",
"output": "0.98994974874371854945"
},
{
"input": "10 300\n80 100 103 140 146 159 392 393 396 398",
"output": "0.98993288590604022747"
},
{
"input": "10 92\n44 119 252 281 303 323 351 363 377 392",
"output": "0.77528089887640450062"
},
{
"input": "4 2\n1 3 5 7",
"output": "-1"
},
{
"input": "8 2\n1 3 7 9 15 17 23 25",
"output": "-1"
},
{
"input": "3 5\n1 2 10",
"output": "-1"
},
{
"input": "4 7\n1 5 8 9",
"output": "0.42857142857142854764"
}
] | 1,521,922,859
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <iostream>
#include <vector>
using namespace std;
int bin_l(vector <double> num, double x) {
int r = num.size();
int l = -1;
while (r - l > 1) {
int m = (l + r) / 2;
if (num[m] <= x) {
l = m;
}
else {
r = m;
}
}
return l;
}
int main() {
int n, u;
cin >> n >> u;
vector <double> num(n);
for (int i = 0; i < n; i++) {
cin >> num[i];
}
double ans = -1;
int fl = 0;
for (int i = 0; i < n - 2; i++) {
if (num[i + 2] - num[i] <= u) {
fl = 1;
break;
}
}
if (fl) {
for (int j = 1; j < n - 1; j++) {
double x = num[j - 1] + u;
int k = bin_l(num, x);
int i = j - 1;
if (i != k) {
if ((double) (num[k] - num[j]) / (num[k] - num[i]) > ans) {
ans = (double) (num[k] - num[j]) / (num[k] - num[i]);
}
}
}
cout << ans;
}
else {
cout << -1;
}
return 0;
}
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.
Input Specification:
The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order.
Output Specification:
If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
Demo Input:
['4 4\n1 3 5 7\n', '10 8\n10 13 15 16 17 19 20 22 24 25\n', '3 1\n2 5 10\n']
Demo Output:
['0.5\n', '0.875\n', '-1\n']
Note:
In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
#include <iostream>
#include <vector>
using namespace std;
int bin_l(vector <double> num, double x) {
int r = num.size();
int l = -1;
while (r - l > 1) {
int m = (l + r) / 2;
if (num[m] <= x) {
l = m;
}
else {
r = m;
}
}
return l;
}
int main() {
int n, u;
cin >> n >> u;
vector <double> num(n);
for (int i = 0; i < n; i++) {
cin >> num[i];
}
double ans = -1;
int fl = 0;
for (int i = 0; i < n - 2; i++) {
if (num[i + 2] - num[i] <= u) {
fl = 1;
break;
}
}
if (fl) {
for (int j = 1; j < n - 1; j++) {
double x = num[j - 1] + u;
int k = bin_l(num, x);
int i = j - 1;
if (i != k) {
if ((double) (num[k] - num[j]) / (num[k] - num[i]) > ans) {
ans = (double) (num[k] - num[j]) / (num[k] - num[i]);
}
}
}
cout << ans;
}
else {
cout << -1;
}
return 0;
}
```
| -1
|
|
75
|
A
|
Life Without Zeros
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Life Without Zeros
|
2
|
256
|
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
|
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
|
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
|
[
"101\n102\n",
"105\n106\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "101\n102",
"output": "YES"
},
{
"input": "105\n106",
"output": "NO"
},
{
"input": "544\n397",
"output": "YES"
},
{
"input": "822\n280",
"output": "NO"
},
{
"input": "101\n413",
"output": "NO"
},
{
"input": "309\n139",
"output": "NO"
},
{
"input": "693\n970",
"output": "NO"
},
{
"input": "981\n1",
"output": "YES"
},
{
"input": "352\n276",
"output": "YES"
},
{
"input": "164\n691",
"output": "YES"
},
{
"input": "110036\n43",
"output": "YES"
},
{
"input": "100\n1000",
"output": "NO"
},
{
"input": "1000000000\n1000000000",
"output": "YES"
},
{
"input": "999999999\n999999999",
"output": "YES"
},
{
"input": "6\n4",
"output": "NO"
},
{
"input": "123456\n876543",
"output": "YES"
},
{
"input": "1234567\n9876543",
"output": "NO"
},
{
"input": "1111111\n1119111",
"output": "NO"
},
{
"input": "10502405\n40403403",
"output": "YES"
},
{
"input": "1\n1",
"output": "YES"
},
{
"input": "5\n4",
"output": "YES"
},
{
"input": "1\n6",
"output": "YES"
},
{
"input": "40\n17",
"output": "NO"
},
{
"input": "35\n81",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "2081\n7450",
"output": "NO"
},
{
"input": "84355694\n373629503",
"output": "NO"
},
{
"input": "885992048\n510468672",
"output": "NO"
},
{
"input": "540144752\n647307841",
"output": "NO"
},
{
"input": "341781108\n784147010",
"output": "NO"
},
{
"input": "960440938\n572334653",
"output": "NO"
},
{
"input": "657414510\n768771903",
"output": "NO"
},
{
"input": "501871728\n725074574",
"output": "NO"
},
{
"input": "968\n851",
"output": "YES"
},
{
"input": "943\n939",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "611\n41",
"output": "YES"
},
{
"input": "737\n289",
"output": "NO"
},
{
"input": "168\n571",
"output": "YES"
},
{
"input": "868\n463",
"output": "YES"
},
{
"input": "815732007\n913882008",
"output": "NO"
},
{
"input": "459039901\n453033850",
"output": "NO"
},
{
"input": "975305404\n484601256",
"output": "NO"
}
] | 1,530,553,962
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 218
| 0
|
a = input()
b = input()
ans1 = int(a)+int(b)
ans1 = str(ans1)
c = ''
for i in range(len(ans1)):
if(ans1[i]!='0'):
c+=ans1[i]
c = int(c)
s1 = ''
s2 = ''
for i in range(len(a)):
if(a[i]!='0'):
s1+=a[i]
for i in range(len(b)):
if(b[i]!='0'):
s2+=b[i]
ans2 = int(s1)+int(s2)
if(ans2==c):
print('YES')
else:
print('NO')
|
Title: Life Without Zeros
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
Input Specification:
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
Output Specification:
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Demo Input:
['101\n102\n', '105\n106\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
a = input()
b = input()
ans1 = int(a)+int(b)
ans1 = str(ans1)
c = ''
for i in range(len(ans1)):
if(ans1[i]!='0'):
c+=ans1[i]
c = int(c)
s1 = ''
s2 = ''
for i in range(len(a)):
if(a[i]!='0'):
s1+=a[i]
for i in range(len(b)):
if(b[i]!='0'):
s2+=b[i]
ans2 = int(s1)+int(s2)
if(ans2==c):
print('YES')
else:
print('NO')
```
| 3.9455
|
786
|
C
|
Till I Collapse
|
PROGRAMMING
| 2,400
|
[
"data structures",
"divide and conquer"
] | null | null |
Rick and Morty want to find MR. PBH and they can't do it alone. So they need of Mr. Meeseeks. They Have generated *n* Mr. Meeseeks, standing in a line numbered from 1 to *n*. Each of them has his own color. *i*-th Mr. Meeseeks' color is *a**i*.
Rick and Morty are gathering their army and they want to divide Mr. Meeseeks into some squads. They don't want their squads to be too colorful, so each squad should have Mr. Meeseeks of at most *k* different colors. Also each squad should be a continuous subarray of Mr. Meeseeks in the line. Meaning that for each 1<=≤<=*i*<=≤<=*e*<=≤<=*j*<=≤<=*n*, if Mr. Meeseeks number *i* and Mr. Meeseeks number *j* are in the same squad then Mr. Meeseeks number *e* should be in that same squad.
Also, each squad needs its own presidio, and building a presidio needs money, so they want the total number of squads to be minimized.
Rick and Morty haven't finalized the exact value of *k*, so in order to choose it, for each *k* between 1 and *n* (inclusive) need to know the minimum number of presidios needed.
|
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105) — number of Mr. Meeseeks.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* separated by spaces (1<=≤<=*a**i*<=≤<=*n*) — colors of Mr. Meeseeks in order they standing in a line.
|
In the first and only line of input print *n* integers separated by spaces. *i*-th integer should be the minimum number of presidios needed if the value of *k* is *i*.
|
[
"5\n1 3 4 3 3\n",
"8\n1 5 7 8 1 7 6 1\n"
] |
[
"4 2 1 1 1 \n",
"8 4 3 2 1 1 1 1 \n"
] |
For the first sample testcase, some optimal ways of dividing army into squads for each *k* are:
1. [1], [3], [4], [3, 3] 1. [1], [3, 4, 3, 3] 1. [1, 3, 4, 3, 3] 1. [1, 3, 4, 3, 3] 1. [1, 3, 4, 3, 3]
For the second testcase, some optimal ways of dividing army into squads for each *k* are:
1. [1], [5], [7], [8], [1], [7], [6], [1] 1. [1, 5], [7, 8], [1, 7], [6, 1] 1. [1, 5, 7], [8], [1, 7, 6, 1] 1. [1, 5, 7, 8], [1, 7, 6, 1] 1. [1, 5, 7, 8, 1, 7, 6, 1] 1. [1, 5, 7, 8, 1, 7, 6, 1] 1. [1, 5, 7, 8, 1, 7, 6, 1] 1. [1, 5, 7, 8, 1, 7, 6, 1]
| 1,500
|
[
{
"input": "5\n1 3 4 3 3",
"output": "4 2 1 1 1 "
},
{
"input": "8\n1 5 7 8 1 7 6 1",
"output": "8 4 3 2 1 1 1 1 "
},
{
"input": "10\n4 1 2 6 8 5 3 9 3 9",
"output": "10 4 3 2 2 2 2 1 1 1 "
},
{
"input": "85\n23 11 69 1 49 10 7 13 66 35 81 4 51 2 62 55 31 18 85 34 59 44 20 28 27 5 6 79 43 78 45 64 61 56 12 40 54 52 24 14 26 65 75 72 30 46 67 80 38 70 25 60 50 8 17 84 41 71 58 76 19 47 73 29 3 48 82 33 39 63 15 37 83 36 9 32 16 57 68 53 21 77 22 42 74",
"output": "85 43 29 22 17 15 13 11 10 9 8 8 7 7 6 6 5 5 5 5 5 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 "
},
{
"input": "100\n39 78 71 61 54 13 17 81 30 33 83 98 44 10 45 87 75 47 70 84 41 86 49 94 85 91 37 64 5 56 67 79 28 89 50 53 77 93 81 14 97 67 58 6 48 60 89 62 29 3 38 8 88 19 66 63 100 17 43 97 21 12 58 76 2 78 25 73 99 11 27 18 57 46 4 72 68 45 74 18 2 80 51 52 42 59 55 35 1 95 42 92 36 40 59 15 7 21 97 53",
"output": "100 50 34 25 20 17 15 13 11 10 10 9 8 7 7 6 6 6 6 5 5 5 5 4 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "1\n1",
"output": "1 "
}
] | 1,684,477,699
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 7
| 2,000
| 13,619,200
|
if __name__ == '__main__':
n = int(input())
a = list(map(int, input().split()))
check = [0] * (n + 1)
vis = [0] * n
Clocks = 0
def get(k: int)->int:
if check[k] != 0: return check[k]
ans = 0
i = 0
while i < n:
global Clocks
Clocks += 1
j, len = i, 0
while j < n:
if vis[a[j] - 1] != Clocks:
vis[a[j] - 1] = Clocks
len += 1
if len > k: break
j += 1
ans += 1
i = j - 1
i += 1
check[k] = ans
return ans
ans = [0] * n
for i in range(1, n + 1):
low, high = i, n + 1
while low < high:
mid = low + high >> 1
if get(mid) == get(i): low = mid + 1
else: high = mid
for j in range(i, low): ans[j - 1] = get(i)
i = low - 1
for x in ans: print(x, end=' ')
|
Title: Till I Collapse
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Rick and Morty want to find MR. PBH and they can't do it alone. So they need of Mr. Meeseeks. They Have generated *n* Mr. Meeseeks, standing in a line numbered from 1 to *n*. Each of them has his own color. *i*-th Mr. Meeseeks' color is *a**i*.
Rick and Morty are gathering their army and they want to divide Mr. Meeseeks into some squads. They don't want their squads to be too colorful, so each squad should have Mr. Meeseeks of at most *k* different colors. Also each squad should be a continuous subarray of Mr. Meeseeks in the line. Meaning that for each 1<=≤<=*i*<=≤<=*e*<=≤<=*j*<=≤<=*n*, if Mr. Meeseeks number *i* and Mr. Meeseeks number *j* are in the same squad then Mr. Meeseeks number *e* should be in that same squad.
Also, each squad needs its own presidio, and building a presidio needs money, so they want the total number of squads to be minimized.
Rick and Morty haven't finalized the exact value of *k*, so in order to choose it, for each *k* between 1 and *n* (inclusive) need to know the minimum number of presidios needed.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105) — number of Mr. Meeseeks.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* separated by spaces (1<=≤<=*a**i*<=≤<=*n*) — colors of Mr. Meeseeks in order they standing in a line.
Output Specification:
In the first and only line of input print *n* integers separated by spaces. *i*-th integer should be the minimum number of presidios needed if the value of *k* is *i*.
Demo Input:
['5\n1 3 4 3 3\n', '8\n1 5 7 8 1 7 6 1\n']
Demo Output:
['4 2 1 1 1 \n', '8 4 3 2 1 1 1 1 \n']
Note:
For the first sample testcase, some optimal ways of dividing army into squads for each *k* are:
1. [1], [3], [4], [3, 3] 1. [1], [3, 4, 3, 3] 1. [1, 3, 4, 3, 3] 1. [1, 3, 4, 3, 3] 1. [1, 3, 4, 3, 3]
For the second testcase, some optimal ways of dividing army into squads for each *k* are:
1. [1], [5], [7], [8], [1], [7], [6], [1] 1. [1, 5], [7, 8], [1, 7], [6, 1] 1. [1, 5, 7], [8], [1, 7, 6, 1] 1. [1, 5, 7, 8], [1, 7, 6, 1] 1. [1, 5, 7, 8, 1, 7, 6, 1] 1. [1, 5, 7, 8, 1, 7, 6, 1] 1. [1, 5, 7, 8, 1, 7, 6, 1] 1. [1, 5, 7, 8, 1, 7, 6, 1]
|
```python
if __name__ == '__main__':
n = int(input())
a = list(map(int, input().split()))
check = [0] * (n + 1)
vis = [0] * n
Clocks = 0
def get(k: int)->int:
if check[k] != 0: return check[k]
ans = 0
i = 0
while i < n:
global Clocks
Clocks += 1
j, len = i, 0
while j < n:
if vis[a[j] - 1] != Clocks:
vis[a[j] - 1] = Clocks
len += 1
if len > k: break
j += 1
ans += 1
i = j - 1
i += 1
check[k] = ans
return ans
ans = [0] * n
for i in range(1, n + 1):
low, high = i, n + 1
while low < high:
mid = low + high >> 1
if get(mid) == get(i): low = mid + 1
else: high = mid
for j in range(i, low): ans[j - 1] = get(i)
i = low - 1
for x in ans: print(x, end=' ')
```
| 0
|
|
729
|
B
|
Spotlights
|
PROGRAMMING
| 1,200
|
[
"dp",
"implementation"
] | null | null |
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
|
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
|
Print one integer — the number of good positions for placing the spotlight.
|
[
"2 4\n0 1 0 0\n1 0 1 0\n",
"4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n"
] |
[
"9\n",
"20\n"
] |
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example.
| 1,000
|
[
{
"input": "2 4\n0 1 0 0\n1 0 1 0",
"output": "9"
},
{
"input": "4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0",
"output": "20"
},
{
"input": "1 5\n1 1 0 0 0",
"output": "3"
},
{
"input": "2 10\n0 0 0 0 0 0 0 1 0 0\n1 0 0 0 0 0 0 0 0 0",
"output": "20"
},
{
"input": "3 1\n1\n0\n0",
"output": "2"
},
{
"input": "5 7\n0 0 0 0 0 0 1\n0 0 0 0 0 0 1\n0 0 0 1 0 0 0\n0 0 0 0 0 0 0\n0 0 0 0 0 0 0",
"output": "25"
},
{
"input": "10 20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "5 7\n0 0 0 0 0 0 0\n0 1 1 1 1 0 1\n0 1 1 1 1 1 1\n0 0 0 0 0 0 0\n0 1 1 1 1 0 1",
"output": "26"
},
{
"input": "10 20\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "121"
},
{
"input": "1 2\n0 1",
"output": "1"
},
{
"input": "1 2\n1 0",
"output": "1"
},
{
"input": "1 2\n1 1",
"output": "0"
},
{
"input": "2 1\n1\n0",
"output": "1"
},
{
"input": "2 1\n0\n1",
"output": "1"
},
{
"input": "2 1\n1\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "4 4\n1 1 1 1\n1 0 0 1\n1 0 0 1\n1 1 1 1",
"output": "16"
}
] | 1,565,347,284
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 72
| 483
| 14,540,800
|
from sys import stdin,stdout
input = stdin.readline
n,m = map(int, input().split())
arr = []
for i in range(n):
arr.append(list(map(int, input().split())))
dp = [[0 for i in range(m)] for j in range(n)]
for i in range(n):
t = 0
for j in range(m):
if arr[i][j] == 0: dp[i][j] += t
else: t = 1
t = 0
for j in range(m-1,-1,-1):
if arr[i][j] == 0: dp[i][j] += t
else: t = 1
for i in range(m):
t = 0
for j in range(n):
if arr[j][i] == 0: dp[j][i] += t
else: t = 1
t = 0
for j in range(n-1,-1,-1):
if arr[j][i] == 0: dp[j][i] += t
else: t = 1
ans = 0
for i in dp:
ans += sum(i)
print(ans)
|
Title: Spotlights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
Output Specification:
Print one integer — the number of good positions for placing the spotlight.
Demo Input:
['2 4\n0 1 0 0\n1 0 1 0\n', '4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n']
Demo Output:
['9\n', '20\n']
Note:
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example.
|
```python
from sys import stdin,stdout
input = stdin.readline
n,m = map(int, input().split())
arr = []
for i in range(n):
arr.append(list(map(int, input().split())))
dp = [[0 for i in range(m)] for j in range(n)]
for i in range(n):
t = 0
for j in range(m):
if arr[i][j] == 0: dp[i][j] += t
else: t = 1
t = 0
for j in range(m-1,-1,-1):
if arr[i][j] == 0: dp[i][j] += t
else: t = 1
for i in range(m):
t = 0
for j in range(n):
if arr[j][i] == 0: dp[j][i] += t
else: t = 1
t = 0
for j in range(n-1,-1,-1):
if arr[j][i] == 0: dp[j][i] += t
else: t = 1
ans = 0
for i in dp:
ans += sum(i)
print(ans)
```
| 3
|
|
18
|
C
|
Stripe
|
PROGRAMMING
| 1,200
|
[
"data structures",
"implementation"
] |
C. Stripe
|
2
|
64
|
Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?
|
The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.
|
Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.
|
[
"9\n1 5 -6 7 9 -16 0 -2 2\n",
"3\n1 1 1\n",
"2\n0 0\n"
] |
[
"3\n",
"0\n",
"1\n"
] |
none
| 0
|
[
{
"input": "9\n1 5 -6 7 9 -16 0 -2 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "0"
},
{
"input": "2\n0 0",
"output": "1"
},
{
"input": "4\n100 1 10 111",
"output": "1"
},
{
"input": "10\n0 4 -3 0 -2 2 -3 -3 2 5",
"output": "3"
},
{
"input": "10\n0 -1 2 2 -1 1 0 0 0 2",
"output": "0"
},
{
"input": "10\n-1 -1 1 -1 0 1 0 1 1 1",
"output": "1"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "9"
},
{
"input": "50\n-4 -3 3 4 -1 0 2 -4 -3 -4 1 4 3 0 4 1 0 -3 4 -3 -2 2 2 1 0 -4 -4 -5 3 2 -1 4 5 -3 -3 4 4 -5 2 -3 4 -5 2 5 -4 4 1 -2 -4 3",
"output": "3"
},
{
"input": "15\n0 4 0 3 -1 4 -2 -2 -4 -4 3 2 4 -1 -3",
"output": "0"
},
{
"input": "10\n3 -1 -3 -1 3 -2 0 3 1 -2",
"output": "0"
},
{
"input": "100\n-4 2 4 4 1 3 -3 -3 2 1 -4 0 0 2 3 -1 -4 -3 4 -2 -3 -3 -3 -1 -2 -3 -1 -4 0 4 0 -1 4 0 -4 -4 4 -4 -2 1 -4 1 -3 -2 3 -4 4 0 -1 3 -1 4 -1 4 -1 3 -3 -3 -2 -2 4 -3 -3 4 -3 -2 -1 0 -2 4 0 -3 -1 -2 -3 1 -4 1 -3 -3 -3 -2 -3 0 1 -2 -2 -4 -3 -1 2 3 -1 1 1 0 3 -3 -1 -2",
"output": "1"
},
{
"input": "100\n-2 -1 1 0 -2 -1 2 2 0 0 2 1 0 2 0 2 1 0 -1 -1 -1 0 -2 -1 2 -1 -2 2 -2 2 -2 -2 2 1 1 1 -2 2 0 0 2 -1 2 2 2 0 -1 -1 -1 1 -2 2 2 2 -2 0 0 -2 0 -2 -2 0 -1 -1 -2 -1 1 2 -2 -1 1 -2 -1 0 -2 2 2 -1 -2 2 0 0 0 1 0 1 2 1 -2 -2 -1 -1 -2 -2 -2 -1 2 2 2 -2",
"output": "6"
},
{
"input": "100\n2 7 6 0 8 -2 0 1 8 -1 7 -4 -1 1 0 3 4 -4 3 7 8 -4 -6 6 4 -2 -5 7 4 6 1 6 3 8 -2 6 -6 -3 0 -1 -7 -8 -2 5 8 6 -2 -2 -5 -4 -7 -3 -5 -3 -3 -1 5 8 4 0 -7 -6 8 -4 -1 7 2 6 6 4 -5 -4 -5 -2 2 -2 -7 -1 5 -8 -6 -2 -5 4 2 8 -6 7 -8 -1 -5 8 6 -3 -1 7 -1 2 -8 -8",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n10000",
"output": "0"
},
{
"input": "1\n-10000",
"output": "0"
}
] | 1,695,094,228
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 186
| 3,072,000
|
import sys; R = sys.stdin.readline
S = lambda: map(int,R().split())
n = int(R())
s = [0]+[*S()]
m = sum(s)
if m&1: print(0); exit()
r = 0
for j in range(1,n):
s[j] += s[j-1]
r += s[j]==k//2
print(r)
|
Title: Stripe
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.
Output Specification:
Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.
Demo Input:
['9\n1 5 -6 7 9 -16 0 -2 2\n', '3\n1 1 1\n', '2\n0 0\n']
Demo Output:
['3\n', '0\n', '1\n']
Note:
none
|
```python
import sys; R = sys.stdin.readline
S = lambda: map(int,R().split())
n = int(R())
s = [0]+[*S()]
m = sum(s)
if m&1: print(0); exit()
r = 0
for j in range(1,n):
s[j] += s[j-1]
r += s[j]==k//2
print(r)
```
| -1
|
831
|
A
|
Unimodal Array
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Array of integers is unimodal, if:
- it is strictly increasing in the beginning; - after that it is constant; - after that it is strictly decreasing.
The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.
For example, the following three arrays are unimodal: [5,<=7,<=11,<=11,<=2,<=1], [4,<=4,<=2], [7], but the following three are not unimodal: [5,<=5,<=6,<=6,<=1], [1,<=2,<=1,<=2], [4,<=5,<=5,<=6].
Write a program that checks if an array is unimodal.
|
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*<=≤<=1<=000) — the elements of the array.
|
Print "YES" if the given array is unimodal. Otherwise, print "NO".
You can output each letter in any case (upper or lower).
|
[
"6\n1 5 5 5 4 2\n",
"5\n10 20 30 20 10\n",
"4\n1 2 1 2\n",
"7\n3 3 3 3 3 3 3\n"
] |
[
"YES\n",
"YES\n",
"NO\n",
"YES\n"
] |
In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively).
| 500
|
[
{
"input": "6\n1 5 5 5 4 2",
"output": "YES"
},
{
"input": "5\n10 20 30 20 10",
"output": "YES"
},
{
"input": "4\n1 2 1 2",
"output": "NO"
},
{
"input": "7\n3 3 3 3 3 3 3",
"output": "YES"
},
{
"input": "6\n5 7 11 11 2 1",
"output": "YES"
},
{
"input": "1\n7",
"output": "YES"
},
{
"input": "100\n527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527",
"output": "YES"
},
{
"input": "5\n5 5 6 6 1",
"output": "NO"
},
{
"input": "3\n4 4 2",
"output": "YES"
},
{
"input": "4\n4 5 5 6",
"output": "NO"
},
{
"input": "3\n516 516 515",
"output": "YES"
},
{
"input": "5\n502 503 508 508 507",
"output": "YES"
},
{
"input": "10\n538 538 538 538 538 538 538 538 538 538",
"output": "YES"
},
{
"input": "15\n452 454 455 455 450 448 443 442 439 436 433 432 431 428 426",
"output": "YES"
},
{
"input": "20\n497 501 504 505 509 513 513 513 513 513 513 513 513 513 513 513 513 513 513 513",
"output": "YES"
},
{
"input": "50\n462 465 465 465 463 459 454 449 444 441 436 435 430 429 426 422 421 418 417 412 408 407 406 403 402 399 395 392 387 386 382 380 379 376 374 371 370 365 363 359 358 354 350 349 348 345 342 341 338 337",
"output": "YES"
},
{
"input": "70\n290 292 294 297 299 300 303 305 310 312 313 315 319 320 325 327 328 333 337 339 340 341 345 350 351 354 359 364 367 372 374 379 381 382 383 384 389 393 395 397 398 400 402 405 409 411 416 417 422 424 429 430 434 435 440 442 445 449 451 453 458 460 465 470 474 477 482 482 482 479",
"output": "YES"
},
{
"input": "99\n433 435 439 444 448 452 457 459 460 464 469 470 471 476 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 479 478 477 476 474 469 468 465 460 457 453 452 450 445 443 440 438 433 432 431 430 428 425 421 418 414 411 406 402 397 396 393",
"output": "YES"
},
{
"input": "100\n537 538 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543",
"output": "YES"
},
{
"input": "100\n524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 521",
"output": "YES"
},
{
"input": "100\n235 239 243 245 246 251 254 259 260 261 264 269 272 275 277 281 282 285 289 291 292 293 298 301 302 303 305 307 308 310 315 317 320 324 327 330 334 337 342 346 347 348 353 357 361 366 370 373 376 378 379 384 386 388 390 395 398 400 405 408 413 417 420 422 424 429 434 435 438 441 443 444 445 450 455 457 459 463 465 468 471 473 475 477 481 486 491 494 499 504 504 504 504 504 504 504 504 504 504 504",
"output": "YES"
},
{
"input": "100\n191 196 201 202 207 212 216 219 220 222 224 227 230 231 234 235 238 242 246 250 253 254 259 260 263 267 269 272 277 280 284 287 288 290 295 297 300 305 307 312 316 320 324 326 327 332 333 334 338 343 347 351 356 358 363 368 370 374 375 380 381 386 390 391 394 396 397 399 402 403 405 410 414 419 422 427 429 433 437 442 443 447 448 451 455 459 461 462 464 468 473 478 481 484 485 488 492 494 496 496",
"output": "YES"
},
{
"input": "100\n466 466 466 466 466 464 459 455 452 449 446 443 439 436 435 433 430 428 425 424 420 419 414 412 407 404 401 396 394 391 386 382 379 375 374 369 364 362 360 359 356 351 350 347 342 340 338 337 333 330 329 326 321 320 319 316 311 306 301 297 292 287 286 281 278 273 269 266 261 257 256 255 253 252 250 245 244 242 240 238 235 230 225 220 216 214 211 209 208 206 203 198 196 194 192 190 185 182 177 173",
"output": "YES"
},
{
"input": "100\n360 362 367 369 374 377 382 386 389 391 396 398 399 400 405 410 413 416 419 420 423 428 431 436 441 444 445 447 451 453 457 459 463 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 465 460 455 453 448 446 443 440 436 435 430 425 420 415 410 405 404 403 402 399 394 390 387 384 382 379 378 373 372 370 369 366 361 360 355 353 349 345 344 342 339 338 335 333",
"output": "YES"
},
{
"input": "1\n1000",
"output": "YES"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1",
"output": "YES"
},
{
"input": "100\n1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "100\n998 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 999",
"output": "NO"
},
{
"input": "100\n537 538 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 691 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543",
"output": "NO"
},
{
"input": "100\n527 527 527 527 527 527 527 527 872 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527",
"output": "NO"
},
{
"input": "100\n524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 208 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 521",
"output": "NO"
},
{
"input": "100\n235 239 243 245 246 251 254 259 260 261 264 269 272 275 277 281 282 285 289 291 292 293 298 301 302 303 305 307 308 310 315 317 320 324 327 330 334 337 342 921 347 348 353 357 361 366 370 373 376 378 379 384 386 388 390 395 398 400 405 408 413 417 420 422 424 429 434 435 438 441 443 444 445 450 455 457 459 463 465 468 471 473 475 477 481 486 491 494 499 504 504 504 504 504 504 504 504 504 504 504",
"output": "NO"
},
{
"input": "100\n191 196 201 202 207 212 216 219 220 222 224 227 230 231 234 235 238 242 246 250 253 254 259 260 263 267 269 272 277 280 284 287 288 290 295 297 300 305 307 312 316 320 324 326 327 332 333 334 338 343 347 351 356 358 119 368 370 374 375 380 381 386 390 391 394 396 397 399 402 403 405 410 414 419 422 427 429 433 437 442 443 447 448 451 455 459 461 462 464 468 473 478 481 484 485 488 492 494 496 496",
"output": "NO"
},
{
"input": "100\n466 466 466 466 466 464 459 455 452 449 446 443 439 436 435 433 430 428 425 424 420 419 414 412 407 404 401 396 394 391 386 382 379 375 374 369 364 362 360 359 356 335 350 347 342 340 338 337 333 330 329 326 321 320 319 316 311 306 301 297 292 287 286 281 278 273 269 266 261 257 256 255 253 252 250 245 244 242 240 238 235 230 225 220 216 214 211 209 208 206 203 198 196 194 192 190 185 182 177 173",
"output": "NO"
},
{
"input": "100\n360 362 367 369 374 377 382 386 389 391 396 398 399 400 405 410 413 416 419 420 423 428 525 436 441 444 445 447 451 453 457 459 463 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 465 460 455 453 448 446 443 440 436 435 430 425 420 415 410 405 404 403 402 399 394 390 387 384 382 379 378 373 372 370 369 366 361 360 355 353 349 345 344 342 339 338 335 333",
"output": "NO"
},
{
"input": "3\n1 2 3",
"output": "YES"
},
{
"input": "3\n3 2 1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "NO"
},
{
"input": "3\n2 1 1",
"output": "NO"
},
{
"input": "3\n2 1 2",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "NO"
},
{
"input": "3\n1 3 2",
"output": "YES"
},
{
"input": "100\n395 399 402 403 405 408 413 415 419 424 426 431 434 436 439 444 447 448 449 454 457 459 461 462 463 464 465 469 470 473 477 480 482 484 485 487 492 494 496 497 501 504 505 508 511 506 505 503 500 499 494 490 488 486 484 481 479 474 472 471 470 465 462 458 453 452 448 445 440 436 433 430 428 426 424 421 419 414 413 408 404 403 399 395 393 388 384 379 377 375 374 372 367 363 360 356 353 351 350 346",
"output": "YES"
},
{
"input": "100\n263 268 273 274 276 281 282 287 288 292 294 295 296 300 304 306 308 310 311 315 319 322 326 330 333 336 339 341 342 347 351 353 356 358 363 365 369 372 374 379 383 387 389 391 392 395 396 398 403 404 407 411 412 416 419 421 424 428 429 430 434 436 440 443 444 448 453 455 458 462 463 464 469 473 477 481 486 489 492 494 499 503 506 509 510 512 514 515 511 510 507 502 499 498 494 491 486 482 477 475",
"output": "YES"
},
{
"input": "100\n482 484 485 489 492 496 499 501 505 509 512 517 520 517 515 513 509 508 504 503 498 496 493 488 486 481 478 476 474 470 468 466 463 459 456 453 452 449 445 444 439 438 435 432 428 427 424 423 421 419 417 413 408 405 402 399 397 393 388 385 380 375 370 366 363 361 360 355 354 352 349 345 340 336 335 331 329 327 324 319 318 317 315 314 310 309 307 304 303 300 299 295 291 287 285 282 280 278 273 271",
"output": "YES"
},
{
"input": "100\n395 399 402 403 405 408 413 415 419 424 426 431 434 436 439 444 447 448 449 454 457 459 461 462 463 464 465 469 470 473 477 480 482 484 485 487 492 494 496 32 501 504 505 508 511 506 505 503 500 499 494 490 488 486 484 481 479 474 472 471 470 465 462 458 453 452 448 445 440 436 433 430 428 426 424 421 419 414 413 408 404 403 399 395 393 388 384 379 377 375 374 372 367 363 360 356 353 351 350 346",
"output": "NO"
},
{
"input": "100\n263 268 273 274 276 281 282 287 288 292 294 295 296 300 304 306 308 310 311 315 319 322 326 330 247 336 339 341 342 347 351 353 356 358 363 365 369 372 374 379 383 387 389 391 392 395 396 398 403 404 407 411 412 416 419 421 424 428 429 430 434 436 440 443 444 448 453 455 458 462 463 464 469 473 477 481 486 489 492 494 499 503 506 509 510 512 514 515 511 510 507 502 499 498 494 491 486 482 477 475",
"output": "NO"
},
{
"input": "100\n482 484 485 489 492 496 499 501 505 509 512 517 520 517 515 513 509 508 504 503 497 496 493 488 486 481 478 476 474 470 468 466 463 459 456 453 452 449 445 444 439 438 435 432 428 427 424 423 421 419 417 413 408 405 402 399 397 393 388 385 380 375 370 366 363 361 360 355 354 352 349 345 340 336 335 331 329 327 324 319 318 317 315 314 310 309 307 304 303 300 299 295 291 287 285 282 280 278 273 271",
"output": "YES"
},
{
"input": "2\n1 3",
"output": "YES"
},
{
"input": "2\n1 2",
"output": "YES"
},
{
"input": "5\n2 2 1 1 1",
"output": "NO"
},
{
"input": "4\n1 3 2 2",
"output": "NO"
},
{
"input": "6\n1 2 1 2 2 1",
"output": "NO"
},
{
"input": "2\n4 2",
"output": "YES"
},
{
"input": "3\n3 2 2",
"output": "NO"
},
{
"input": "9\n1 2 2 3 3 4 3 2 1",
"output": "NO"
},
{
"input": "4\n5 5 4 4",
"output": "NO"
},
{
"input": "2\n2 1",
"output": "YES"
},
{
"input": "5\n5 4 3 2 1",
"output": "YES"
},
{
"input": "7\n4 3 3 3 3 3 3",
"output": "NO"
},
{
"input": "5\n1 2 3 4 5",
"output": "YES"
},
{
"input": "3\n2 2 1",
"output": "YES"
},
{
"input": "3\n4 3 3",
"output": "NO"
},
{
"input": "7\n1 5 5 4 3 3 1",
"output": "NO"
},
{
"input": "6\n3 3 1 2 2 1",
"output": "NO"
},
{
"input": "5\n1 2 1 2 1",
"output": "NO"
},
{
"input": "2\n5 1",
"output": "YES"
},
{
"input": "9\n1 2 3 4 4 3 2 2 1",
"output": "NO"
},
{
"input": "3\n2 2 3",
"output": "NO"
},
{
"input": "2\n5 4",
"output": "YES"
},
{
"input": "5\n1 3 3 2 2",
"output": "NO"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 99",
"output": "YES"
},
{
"input": "4\n1 2 3 4",
"output": "YES"
},
{
"input": "3\n5 5 2",
"output": "YES"
},
{
"input": "4\n1 4 2 3",
"output": "NO"
},
{
"input": "2\n3 2",
"output": "YES"
},
{
"input": "5\n1 2 2 1 1",
"output": "NO"
},
{
"input": "4\n3 3 2 2",
"output": "NO"
},
{
"input": "5\n1 2 3 2 2",
"output": "NO"
},
{
"input": "5\n5 6 6 5 5",
"output": "NO"
},
{
"input": "4\n2 2 1 1",
"output": "NO"
},
{
"input": "5\n5 4 3 3 2",
"output": "NO"
},
{
"input": "7\n1 3 3 3 2 1 1",
"output": "NO"
},
{
"input": "9\n5 6 6 5 5 4 4 3 3",
"output": "NO"
},
{
"input": "6\n1 5 5 3 2 2",
"output": "NO"
},
{
"input": "5\n2 1 3 3 1",
"output": "NO"
},
{
"input": "2\n4 3",
"output": "YES"
},
{
"input": "5\n3 2 2 1 1",
"output": "NO"
},
{
"input": "4\n5 4 3 2",
"output": "YES"
},
{
"input": "4\n4 4 1 1",
"output": "NO"
},
{
"input": "4\n3 3 1 1",
"output": "NO"
},
{
"input": "4\n4 4 2 2",
"output": "NO"
},
{
"input": "5\n4 4 3 2 2",
"output": "NO"
},
{
"input": "8\n4 4 4 4 5 6 7 8",
"output": "NO"
},
{
"input": "5\n3 5 4 4 3",
"output": "NO"
},
{
"input": "6\n2 5 3 3 2 2",
"output": "NO"
},
{
"input": "4\n5 5 2 2",
"output": "NO"
},
{
"input": "5\n1 2 2 3 5",
"output": "NO"
}
] | 1,655,803,714
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
n = int(input())
array = list(map(int,input().split()))
uniModal = True
dropped = False
for i in range(1,len(array)):
if(array[i]<array[i-1] and not dropped) :
dropped = True
print(array[i])
if(dropped):
if(array[i]<array[i-1]):
pass
else:
uniModal = False
if uniModal:print("YES")
else:print("NO")
|
Title: Unimodal Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Array of integers is unimodal, if:
- it is strictly increasing in the beginning; - after that it is constant; - after that it is strictly decreasing.
The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.
For example, the following three arrays are unimodal: [5,<=7,<=11,<=11,<=2,<=1], [4,<=4,<=2], [7], but the following three are not unimodal: [5,<=5,<=6,<=6,<=1], [1,<=2,<=1,<=2], [4,<=5,<=5,<=6].
Write a program that checks if an array is unimodal.
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*<=≤<=1<=000) — the elements of the array.
Output Specification:
Print "YES" if the given array is unimodal. Otherwise, print "NO".
You can output each letter in any case (upper or lower).
Demo Input:
['6\n1 5 5 5 4 2\n', '5\n10 20 30 20 10\n', '4\n1 2 1 2\n', '7\n3 3 3 3 3 3 3\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively).
|
```python
n = int(input())
array = list(map(int,input().split()))
uniModal = True
dropped = False
for i in range(1,len(array)):
if(array[i]<array[i-1] and not dropped) :
dropped = True
print(array[i])
if(dropped):
if(array[i]<array[i-1]):
pass
else:
uniModal = False
if uniModal:print("YES")
else:print("NO")
```
| 0
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integer — the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,694,092,339
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
x1, x2, x3=map(int, input().split())
max_=max(x1, x2, x3)
min_=min(x1, x2, x3)
mid=(x1+x2+x3)-(max_+min_)
y1=x1-mid
y2=x2-mid
y3=x3-mid
res=y1+y2+y3
print(res)
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
x1, x2, x3=map(int, input().split())
max_=max(x1, x2, x3)
min_=min(x1, x2, x3)
mid=(x1+x2+x3)-(max_+min_)
y1=x1-mid
y2=x2-mid
y3=x3-mid
res=y1+y2+y3
print(res)
```
| 0
|
|
344
|
A
|
Magnets
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
|
On the single line of the output print the number of groups of magnets.
|
[
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] |
[
"3\n",
"2\n"
] |
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
| 500
|
[
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",
"output": "3"
},
{
"input": "1\n01",
"output": "1"
},
{
"input": "2\n01\n01",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "2"
},
{
"input": "3\n01\n01\n01",
"output": "1"
},
{
"input": "3\n10\n10\n01",
"output": "2"
},
{
"input": "3\n01\n10\n10",
"output": "2"
},
{
"input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10",
"output": "55"
}
] | 1,692,683,275
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
n = int(input())
li=[]
i=0
while i<n:
number = input()
li.append(number)
i+=1
print(len(li)//2)
|
Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
|
```python
n = int(input())
li=[]
i=0
while i<n:
number = input()
li.append(number)
i+=1
print(len(li)//2)
```
| 0
|
|
420
|
A
|
Start Up
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
|
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
|
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
|
[
"AHA\n",
"Z\n",
"XO\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "AHA",
"output": "YES"
},
{
"input": "Z",
"output": "NO"
},
{
"input": "XO",
"output": "NO"
},
{
"input": "AAA",
"output": "YES"
},
{
"input": "AHHA",
"output": "YES"
},
{
"input": "BAB",
"output": "NO"
},
{
"input": "OMMMAAMMMO",
"output": "YES"
},
{
"input": "YYHUIUGYI",
"output": "NO"
},
{
"input": "TT",
"output": "YES"
},
{
"input": "UUU",
"output": "YES"
},
{
"input": "WYYW",
"output": "YES"
},
{
"input": "MITIM",
"output": "YES"
},
{
"input": "VO",
"output": "NO"
},
{
"input": "WWS",
"output": "NO"
},
{
"input": "VIYMAXXAVM",
"output": "NO"
},
{
"input": "OVWIHIWVYXMVAAAATOXWOIUUHYXHIHHVUIOOXWHOXTUUMUUVHVWWYUTIAUAITAOMHXWMTTOIVMIVOTHOVOIOHYHAOXWAUVWAVIVM",
"output": "NO"
},
{
"input": "CC",
"output": "NO"
},
{
"input": "QOQ",
"output": "NO"
},
{
"input": "AEEA",
"output": "NO"
},
{
"input": "OQQQO",
"output": "NO"
},
{
"input": "HNCMEEMCNH",
"output": "NO"
},
{
"input": "QDPINBMCRFWXPDBFGOZVVOCEMJRUCTOADEWEGTVBVBFWWRPGYEEYGPRWWFBVBVTGEWEDAOTCURJMECOVVZOGFBDPXWFRCMBNIPDQ",
"output": "NO"
},
{
"input": "A",
"output": "YES"
},
{
"input": "B",
"output": "NO"
},
{
"input": "C",
"output": "NO"
},
{
"input": "D",
"output": "NO"
},
{
"input": "E",
"output": "NO"
},
{
"input": "F",
"output": "NO"
},
{
"input": "G",
"output": "NO"
},
{
"input": "H",
"output": "YES"
},
{
"input": "I",
"output": "YES"
},
{
"input": "J",
"output": "NO"
},
{
"input": "K",
"output": "NO"
},
{
"input": "L",
"output": "NO"
},
{
"input": "M",
"output": "YES"
},
{
"input": "N",
"output": "NO"
},
{
"input": "O",
"output": "YES"
},
{
"input": "P",
"output": "NO"
},
{
"input": "Q",
"output": "NO"
},
{
"input": "R",
"output": "NO"
},
{
"input": "S",
"output": "NO"
},
{
"input": "T",
"output": "YES"
},
{
"input": "U",
"output": "YES"
},
{
"input": "V",
"output": "YES"
},
{
"input": "W",
"output": "YES"
},
{
"input": "X",
"output": "YES"
},
{
"input": "Y",
"output": "YES"
},
{
"input": "JL",
"output": "NO"
},
{
"input": "AAAKTAAA",
"output": "NO"
},
{
"input": "AKA",
"output": "NO"
},
{
"input": "AAJAA",
"output": "NO"
},
{
"input": "ABA",
"output": "NO"
},
{
"input": "AAAAAABAAAAAA",
"output": "NO"
},
{
"input": "ZZ",
"output": "NO"
},
{
"input": "ADA",
"output": "NO"
},
{
"input": "N",
"output": "NO"
},
{
"input": "P",
"output": "NO"
},
{
"input": "LAL",
"output": "NO"
},
{
"input": "AABAA",
"output": "NO"
},
{
"input": "AZA",
"output": "NO"
},
{
"input": "V",
"output": "YES"
},
{
"input": "SSS",
"output": "NO"
},
{
"input": "NNN",
"output": "NO"
},
{
"input": "S",
"output": "NO"
},
{
"input": "I",
"output": "YES"
},
{
"input": "SS",
"output": "NO"
},
{
"input": "E",
"output": "NO"
}
] | 1,563,915,076
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 80
| 171
| 4,710,400
|
mirror_letter = 'AHIMOTUVWXY'
s = input()
left = s[:len(s) // 2]
right = s[:-len(s) // 2 - (len(s) % 2 == 0):-1]
if all([c in mirror_letter for c in s]) and left == right:
print('YES')
else:
print('NO')
|
Title: Start Up
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input Specification:
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
Output Specification:
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Demo Input:
['AHA\n', 'Z\n', 'XO\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
mirror_letter = 'AHIMOTUVWXY'
s = input()
left = s[:len(s) // 2]
right = s[:-len(s) // 2 - (len(s) % 2 == 0):-1]
if all([c in mirror_letter for c in s]) and left == right:
print('YES')
else:
print('NO')
```
| 3
|
|
259
|
A
|
Little Elephant and Chess
|
PROGRAMMING
| 1,000
|
[
"brute force",
"strings"
] | null | null |
The Little Elephant loves chess very much.
One day the Little Elephant and his friend decided to play chess. They've got the chess pieces but the board is a problem. They've got an 8<=×<=8 checkered board, each square is painted either black or white. The Little Elephant and his friend know that a proper chessboard doesn't have any side-adjacent cells with the same color and the upper left cell is white. To play chess, they want to make the board they have a proper chessboard. For that the friends can choose any row of the board and cyclically shift the cells of the chosen row, that is, put the last (rightmost) square on the first place in the row and shift the others one position to the right. You can run the described operation multiple times (or not run it at all).
For example, if the first line of the board looks like that "BBBBBBWW" (the white cells of the line are marked with character "W", the black cells are marked with character "B"), then after one cyclic shift it will look like that "WBBBBBBW".
Help the Little Elephant and his friend to find out whether they can use any number of the described operations to turn the board they have into a proper chessboard.
|
The input consists of exactly eight lines. Each line contains exactly eight characters "W" or "B" without any spaces: the *j*-th character in the *i*-th line stands for the color of the *j*-th cell of the *i*-th row of the elephants' board. Character "W" stands for the white color, character "B" stands for the black color.
Consider the rows of the board numbered from 1 to 8 from top to bottom, and the columns — from 1 to 8 from left to right. The given board can initially be a proper chessboard.
|
In a single line print "YES" (without the quotes), if we can make the board a proper chessboard and "NO" (without the quotes) otherwise.
|
[
"WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\n",
"WBWBWBWB\nWBWBWBWB\nBBWBWWWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWWW\nBWBWBWBW\nBWBWBWBW\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you should shift the following lines one position to the right: the 3-rd, the 6-th, the 7-th and the 8-th.
In the second sample there is no way you can achieve the goal.
| 500
|
[
{
"input": "WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB",
"output": "YES"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nBBWBWWWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWWW\nBWBWBWBW\nBWBWBWBW",
"output": "NO"
},
{
"input": "BWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB",
"output": "YES"
},
{
"input": "BWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB",
"output": "YES"
},
{
"input": "WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW",
"output": "YES"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB\nBWWWBWBW",
"output": "NO"
},
{
"input": "BBBBBWWW\nWBBWBWWB\nWWWWWBWW\nBWBWWBWW\nBBBWWBWW\nBBBBBWBW\nWBBBWBWB\nWBWBWWWB",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nBWWWWWBB\nBBWBWBWB\nWBWBWBWB\nWWBWWBWW\nBWBWBWBW\nWBWWBBBB",
"output": "NO"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWBWWBWBB",
"output": "NO"
},
{
"input": "WBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW",
"output": "YES"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW",
"output": "YES"
},
{
"input": "WWWWBWWB\nBWBWBWBW\nBWBWBWBW\nWWBWBBBB\nBBWWBBBB\nBBBWWBBW\nBWWWWWWB\nBWWBBBWW",
"output": "NO"
},
{
"input": "WBBWWBWB\nBBWBWBWB\nBWBWBWBW\nBWBWBWBW\nWBWBWBBW\nWBWBBBBW\nBWWWWBWB\nBBBBBBBW",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nBBWWWBBB\nWBBBBBWW\nWBBBBWBB\nWBWBWBWB\nWBWWBWWB\nWBBWBBWW",
"output": "NO"
},
{
"input": "WBBBBBWB\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBBBBBWBB\nWBBWWBWB\nBWBWBWBW",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nWBBWWBWB",
"output": "NO"
},
{
"input": "BWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWWWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBBW",
"output": "NO"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "YES"
},
{
"input": "BWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "YES"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "YES"
},
{
"input": "WWBBWWBB\nBWWBBWWB\nBWBWBWBW\nWWBBWWWB\nWBWWWWBB\nWBWWBBWB\nBWBBWBWW\nBWBWWWWW",
"output": "NO"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nWWBBWBBB\nWBWBWBWB\nWWWWBWWB\nWBBBBWWW\nBWBWWWBW\nWWWBWBBB",
"output": "NO"
},
{
"input": "WBWBWBWB\nBWWBWWWW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWWBBBBBW\nWWWBWWBW\nWWBBBBWW",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBBBWWB\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "YES"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW",
"output": "NO"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB",
"output": "NO"
},
{
"input": "BWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB",
"output": "NO"
},
{
"input": "WWBWWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "NO"
},
{
"input": "WWWWWWWW\nBBBBBBBB\nWWWWWWWW\nBBBBBBBB\nWWWWWWWW\nBBBBBBBB\nWWWWWWWW\nBBBBBBBB",
"output": "NO"
},
{
"input": "BBBBBBBB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB",
"output": "NO"
},
{
"input": "BBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW",
"output": "NO"
},
{
"input": "BBBWWWWW\nWWWBBBBB\nBBBWWWWW\nWWWBBBBB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB",
"output": "NO"
}
] | 1,652,553,769
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 15
| 124
| 409,600
|
import re
flag = True
for i in range(8):
l = input()
if re.match(r"^(WB|BW){4}$", l) == None:
flag = False
if flag:
print("YES")
else:
print("NO")
|
Title: Little Elephant and Chess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves chess very much.
One day the Little Elephant and his friend decided to play chess. They've got the chess pieces but the board is a problem. They've got an 8<=×<=8 checkered board, each square is painted either black or white. The Little Elephant and his friend know that a proper chessboard doesn't have any side-adjacent cells with the same color and the upper left cell is white. To play chess, they want to make the board they have a proper chessboard. For that the friends can choose any row of the board and cyclically shift the cells of the chosen row, that is, put the last (rightmost) square on the first place in the row and shift the others one position to the right. You can run the described operation multiple times (or not run it at all).
For example, if the first line of the board looks like that "BBBBBBWW" (the white cells of the line are marked with character "W", the black cells are marked with character "B"), then after one cyclic shift it will look like that "WBBBBBBW".
Help the Little Elephant and his friend to find out whether they can use any number of the described operations to turn the board they have into a proper chessboard.
Input Specification:
The input consists of exactly eight lines. Each line contains exactly eight characters "W" or "B" without any spaces: the *j*-th character in the *i*-th line stands for the color of the *j*-th cell of the *i*-th row of the elephants' board. Character "W" stands for the white color, character "B" stands for the black color.
Consider the rows of the board numbered from 1 to 8 from top to bottom, and the columns — from 1 to 8 from left to right. The given board can initially be a proper chessboard.
Output Specification:
In a single line print "YES" (without the quotes), if we can make the board a proper chessboard and "NO" (without the quotes) otherwise.
Demo Input:
['WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\n', 'WBWBWBWB\nWBWBWBWB\nBBWBWWWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWWW\nBWBWBWBW\nBWBWBWBW\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should shift the following lines one position to the right: the 3-rd, the 6-th, the 7-th and the 8-th.
In the second sample there is no way you can achieve the goal.
|
```python
import re
flag = True
for i in range(8):
l = input()
if re.match(r"^(WB|BW){4}$", l) == None:
flag = False
if flag:
print("YES")
else:
print("NO")
```
| 0
|
|
295
|
B
|
Greg and Graph
|
PROGRAMMING
| 1,700
|
[
"dp",
"graphs",
"shortest paths"
] | null | null |
Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes.
|
Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
|
[
"1\n0\n1\n",
"2\n0 5\n4 0\n1 2\n",
"4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n"
] |
[
"0 ",
"9 0 ",
"17 23 404 0 "
] |
none
| 1,000
|
[
{
"input": "1\n0\n1",
"output": "0 "
},
{
"input": "2\n0 5\n4 0\n1 2",
"output": "9 0 "
},
{
"input": "4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3",
"output": "17 23 404 0 "
},
{
"input": "4\n0 57148 51001 13357\n71125 0 98369 67226\n49388 90852 0 66291\n39573 38165 97007 0\n2 3 1 4",
"output": "723897 306638 52930 0 "
},
{
"input": "5\n0 27799 15529 16434 44291\n47134 0 90227 26873 52252\n41605 21269 0 9135 55784\n70744 17563 79061 0 73981\n70529 35681 91073 52031 0\n5 2 3 1 4",
"output": "896203 429762 232508 87178 0 "
},
{
"input": "6\n0 72137 71041 29217 96749 46417\n40199 0 55907 57677 68590 78796\n83463 50721 0 30963 31779 28646\n94529 47831 98222 0 61665 73941\n24397 66286 2971 81613 0 52501\n26285 3381 51438 45360 20160 0\n6 3 2 4 5 1",
"output": "1321441 1030477 698557 345837 121146 0 "
},
{
"input": "7\n0 34385 31901 51111 10191 14089 95685\n11396 0 8701 33277 1481 517 46253\n51313 2255 0 5948 66085 37201 65310\n21105 60985 10748 0 89271 42883 77345\n34686 29401 73565 47795 0 13793 66997\n70279 49576 62900 40002 70943 0 89601\n65045 1681 28239 12023 40414 89585 0\n3 5 7 6 1 2 4",
"output": "1108867 1016339 729930 407114 206764 94262 0 "
},
{
"input": "8\n0 74961 47889 4733 72876 21399 63105 48239\n15623 0 9680 89133 57989 63401 26001 29608\n42369 82390 0 32866 46171 11871 67489 54070\n23425 80027 18270 0 28105 42657 40876 29267\n78793 18701 7655 94798 0 88885 71424 86914\n44835 76636 11553 46031 13617 0 16971 51915\n33037 53719 43116 52806 56897 71241 0 11629\n2119 62373 93265 69513 5770 90751 36619 0\n3 7 6 5 8 1 2 4",
"output": "1450303 1188349 900316 531281 383344 219125 169160 0 "
},
{
"input": "9\n0 85236 27579 82251 69479 24737 87917 15149 52311\n59640 0 74687 34711 3685 30121 4961 7552 83399\n33376 68733 0 81357 18042 74297 15466 29476 5865\n7493 5601 3321 0 20263 55901 45756 55361 87633\n26751 17161 76681 40376 0 39745 50717 56887 90055\n18885 76353 47089 43601 21561 0 60571 33551 53753\n74595 877 71853 93156 97499 70876 0 22713 63961\n67725 25309 56358 92376 40641 35433 39781 0 97482\n81818 12561 85961 81445 3941 76799 31701 43725 0\n6 2 9 3 5 7 1 4 8",
"output": "2106523 1533575 1645151 1255230 946667 618567 287636 147737 0 "
}
] | 1,601,623,401
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 248
| 0
|
#!/usr/bin/env python
# coding: utf-8
# In[2]:
### 4 0 3 1 1 6 0 400 1 2 4 0 1 1 1 1 0 4 1 2 3
# In[3]:
n = int(input())
a = []
for i in range(n):
sor = list(map(int, input().split()))
a+=[sor]
x = []
x = list(map(int, input().split()))
for i in range(n):
x[i]=x[i]-1
# In[4]:
def copy(matrix):
n=len(matrix)
copymatrix = []
for i in range(n):
sor=[]
for j in range(n):
sor+=[matrix[i][j]]
copymatrix+=[sor]
return copymatrix
# In[9]:
m=copy(a)
summs=[]
for i in range(n):
newnode = int(x[i])
d=copy(m)
for j in range(n):
for k in range(n):
d[j][k] = min(d[j][k],d[j][newnode]+d[newnode][k])
summ=0
for j in range(n):
for k in range(n):
summ += d[j][k]
for j in range(n):
for k in range(n):
m[j][newnode]=0
m[newnode][k]=0
summs+=[summ]
for i in range(len(summs)):
print(summs[i],end=" ")
# In[ ]:
|
Title: Greg and Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes.
Output Specification:
Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Demo Input:
['1\n0\n1\n', '2\n0 5\n4 0\n1 2\n', '4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n']
Demo Output:
['0 ', '9 0 ', '17 23 404 0 ']
Note:
none
|
```python
#!/usr/bin/env python
# coding: utf-8
# In[2]:
### 4 0 3 1 1 6 0 400 1 2 4 0 1 1 1 1 0 4 1 2 3
# In[3]:
n = int(input())
a = []
for i in range(n):
sor = list(map(int, input().split()))
a+=[sor]
x = []
x = list(map(int, input().split()))
for i in range(n):
x[i]=x[i]-1
# In[4]:
def copy(matrix):
n=len(matrix)
copymatrix = []
for i in range(n):
sor=[]
for j in range(n):
sor+=[matrix[i][j]]
copymatrix+=[sor]
return copymatrix
# In[9]:
m=copy(a)
summs=[]
for i in range(n):
newnode = int(x[i])
d=copy(m)
for j in range(n):
for k in range(n):
d[j][k] = min(d[j][k],d[j][newnode]+d[newnode][k])
summ=0
for j in range(n):
for k in range(n):
summ += d[j][k]
for j in range(n):
for k in range(n):
m[j][newnode]=0
m[newnode][k]=0
summs+=[summ]
for i in range(len(summs)):
print(summs[i],end=" ")
# In[ ]:
```
| 0
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,603,249,878
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 218
| 0
|
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 21 11:04:13 2020
@author: Cui Shiao
"""
n=int(input())
s=list(map(int,input().split()))
a=b=c=d=0
for x in range(n):
if s[x]%2:
a+=1;c=x
else:
b+=1;d=x
if a>b:
print(d+1)
else:
print(c+1)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 21 11:04:13 2020
@author: Cui Shiao
"""
n=int(input())
s=list(map(int,input().split()))
a=b=c=d=0
for x in range(n):
if s[x]%2:
a+=1;c=x
else:
b+=1;d=x
if a>b:
print(d+1)
else:
print(c+1)
```
| 3.9455
|
285
|
C
|
Building Permutation
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
|
Print a single number — the minimum number of moves.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"2\n3 0\n",
"3\n-1 -1 2\n"
] |
[
"2\n",
"6\n"
] |
In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).
In the second sample you need 6 moves to build permutation (1, 3, 2).
| 1,500
|
[
{
"input": "2\n3 0",
"output": "2"
},
{
"input": "3\n-1 -1 2",
"output": "6"
},
{
"input": "5\n-3 5 -3 3 3",
"output": "10"
},
{
"input": "10\n9 6 -2 4 1 1 1 9 6 2",
"output": "18"
},
{
"input": "9\n2 0 0 6 5 4 1 9 3",
"output": "15"
},
{
"input": "100\n-77 57 -95 -23 53 -28 82 -83 38 -73 85 28 25 6 -43 4 -10 -30 -9 -92 14 34 -93 61 36 -100 90 -68 28 16 100 -3 97 30 36 -55 62 -62 53 74 -50 -23 67 11 22 -30 -19 83 7 84 43 90 -65 -75 -15 97 90 15 66 2 13 -91 91 -44 46 51 51 -58 95 77 20 30 76 79 91 60 76 2 82 42 -93 94 -57 88 65 -95 -66 100 -9 33 -67 54 -99 97 53 13 54 66 60 -48",
"output": "3459"
}
] | 1,553,426,881
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 451
| 25,497,600
|
n=int(input())
moves=0
l1=list(map(int,input().split()))
l1.sort()
for i in range(0,n):
moves+=abs(l1[i]-(i+1))
print(moves)
|
Title: Building Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single number — the minimum number of moves.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['2\n3 0\n', '3\n-1 -1 2\n']
Demo Output:
['2\n', '6\n']
Note:
In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).
In the second sample you need 6 moves to build permutation (1, 3, 2).
|
```python
n=int(input())
moves=0
l1=list(map(int,input().split()))
l1.sort()
for i in range(0,n):
moves+=abs(l1[i]-(i+1))
print(moves)
```
| 3
|
|
400
|
B
|
Inna and New Matrix of Candies
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation",
"schedules"
] | null | null |
Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload".
The field for the new game is a rectangle table of size *n*<=×<=*m*. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs:
- some dwarf in one of the chosen lines is located in the rightmost cell of his row; - some dwarf in the chosen lines is located in the cell with the candy.
The point of the game is to transport all the dwarves to the candy cells.
Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game.
|
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 2<=≤<=*m*<=≤<=1000).
Next *n* lines each contain *m* characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S".
|
In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field.
|
[
"3 4\n*G*S\nG**S\n*G*S\n",
"1 3\nS*G\n"
] |
[
"2\n",
"-1\n"
] |
none
| 1,000
|
[
{
"input": "3 4\n*G*S\nG**S\n*G*S",
"output": "2"
},
{
"input": "1 3\nS*G",
"output": "-1"
},
{
"input": "10 10\nG********S\n*G*******S\n**G******S\n***G*****S\n****G****S\n*****G***S\n******G**S\n*******G*S\n********GS\nG********S",
"output": "9"
},
{
"input": "5 10\nG***S*****\nG****S****\n***GS*****\nG*S*******\nG***S*****",
"output": "4"
},
{
"input": "4 8\nG*S*****\n****G*S*\nG*****S*\n**G***S*",
"output": "3"
},
{
"input": "4 10\n***G****S*\n*****GS***\nG****S****\nG*******S*",
"output": "3"
},
{
"input": "1 2\nSG",
"output": "-1"
},
{
"input": "1 2\nGS",
"output": "1"
},
{
"input": "1 4\nSG**",
"output": "-1"
}
] | 1,595,990,619
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 34
| 202
| 23,859,200
|
from sys import stdin, stdout
def main():
n, m = map(int, stdin.readline().split())
s = set()
for _ in range(n):
l = stdin.readline().strip()
dwarf = -1
candy = -1
for i in range(m):
if l[i] == "G":
dwarf = i
if l[i] == "S":
candy = i
if candy < dwarf:
print(-1)
return
s.add(candy-dwarf)
print(len(s))
return
main()
|
Title: Inna and New Matrix of Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload".
The field for the new game is a rectangle table of size *n*<=×<=*m*. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs:
- some dwarf in one of the chosen lines is located in the rightmost cell of his row; - some dwarf in the chosen lines is located in the cell with the candy.
The point of the game is to transport all the dwarves to the candy cells.
Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 2<=≤<=*m*<=≤<=1000).
Next *n* lines each contain *m* characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S".
Output Specification:
In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field.
Demo Input:
['3 4\n*G*S\nG**S\n*G*S\n', '1 3\nS*G\n']
Demo Output:
['2\n', '-1\n']
Note:
none
|
```python
from sys import stdin, stdout
def main():
n, m = map(int, stdin.readline().split())
s = set()
for _ in range(n):
l = stdin.readline().strip()
dwarf = -1
candy = -1
for i in range(m):
if l[i] == "G":
dwarf = i
if l[i] == "S":
candy = i
if candy < dwarf:
print(-1)
return
s.add(candy-dwarf)
print(len(s))
return
main()
```
| 3
|
|
579
|
A
|
Raising Bacteria
|
PROGRAMMING
| 1,000
|
[
"bitmasks"
] | null | null |
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
|
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
|
The only line containing one integer: the answer.
|
[
"5\n",
"8\n"
] |
[
"2\n",
"1\n"
] |
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
| 250
|
[
{
"input": "5",
"output": "2"
},
{
"input": "8",
"output": "1"
},
{
"input": "536870911",
"output": "29"
},
{
"input": "1",
"output": "1"
},
{
"input": "343000816",
"output": "14"
},
{
"input": "559980448",
"output": "12"
},
{
"input": "697681824",
"output": "14"
},
{
"input": "41313494",
"output": "14"
},
{
"input": "673935585",
"output": "13"
},
{
"input": "117422204",
"output": "19"
},
{
"input": "954746654",
"output": "15"
},
{
"input": "536838144",
"output": "14"
},
{
"input": "1000000000",
"output": "13"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "6",
"output": "2"
},
{
"input": "7",
"output": "3"
},
{
"input": "9",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "999999999",
"output": "21"
},
{
"input": "999999998",
"output": "20"
},
{
"input": "999999997",
"output": "20"
},
{
"input": "999999996",
"output": "19"
},
{
"input": "999999995",
"output": "20"
},
{
"input": "999999994",
"output": "19"
},
{
"input": "999999993",
"output": "19"
},
{
"input": "999999992",
"output": "18"
},
{
"input": "999999991",
"output": "20"
},
{
"input": "999999990",
"output": "19"
},
{
"input": "536870910",
"output": "28"
},
{
"input": "536870912",
"output": "1"
},
{
"input": "536870913",
"output": "2"
},
{
"input": "536870914",
"output": "2"
},
{
"input": "22",
"output": "3"
},
{
"input": "14",
"output": "3"
}
] | 1,679,097,909
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
print(bin(int(input().count('0'))))
|
Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
|
```python
print(bin(int(input().count('0'))))
```
| 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": "0\n0\n3\n3\n3\n0\n3"
},
{
"input": "1\n653\n9\n903\n980\n80\n770\n965\n874\n381\n657\n969",
"output": "1\n1\n0\n1\n1\n1\n0\n1\n1"
},
{
"input": "12\n35345 58181 32223 84621 35905 73863 99537 30666 67771 39229 36847 29038\n23\n55052824\n82504840\n35160556\n78141700\n73401989\n86305919\n39430705\n31939373\n23501765\n4406029\n61436920\n14295390\n34275309\n28028753\n85724689\n70158847\n2396455\n66994588\n84024224\n30175981\n65622319\n76517111\n76605341",
"output": "12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12"
},
{
"input": "4\n698 193 1584 76192\n4\n553640\n310518\n274280\n429192",
"output": "4\n4\n4\n4"
},
{
"input": "1\n1\n1\n1000000000",
"output": "1"
},
{
"input": "1\n1\n1\n100005",
"output": "1"
},
{
"input": "3\n1 2 3\n1\n100005",
"output": "3"
},
{
"input": "2\n1 2\n3\n100004\n100005\n100006",
"output": "2\n2\n2"
},
{
"input": "3\n1 1 1\n1\n1",
"output": "3"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "3\n1 2 3\n1\n100000",
"output": "3"
},
{
"input": "5\n1 1 1 1 10\n2\n1\n10",
"output": "4\n5"
},
{
"input": "5\n3 10 8 6 11\n1\n10000000",
"output": "5"
}
] | 1,699,519,683
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 9
| 2,000
| 1,433,600
|
#买饮料，用count函数超时。思考怎样用一种单次遍历而不查找的方法。算法复杂度2n？
n=int(input())
l=list(map(int,input().split()))
dp=[0]*max(l)
for i in l:
dp[i-1]+=1#先标记出可能有的价格。
for i in range(1,max(l)+1):
if i in l and i!=1:
dp[i-1]=dp[i-1]+dp[i-2]
elif i!=1:
dp[i-1]=dp[i-2]
s=int(input())
for i in range(s):
t=int(input())
if t>=max(l):
print(dp[max(l)-1])
else:
print(dp[t-1])
|
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
#买饮料，用count函数超时。思考怎样用一种单次遍历而不查找的方法。算法复杂度2n？
n=int(input())
l=list(map(int,input().split()))
dp=[0]*max(l)
for i in l:
dp[i-1]+=1#先标记出可能有的价格。
for i in range(1,max(l)+1):
if i in l and i!=1:
dp[i-1]=dp[i-1]+dp[i-2]
elif i!=1:
dp[i-1]=dp[i-2]
s=int(input())
for i in range(s):
t=int(input())
if t>=max(l):
print(dp[max(l)-1])
else:
print(dp[t-1])
```
| 0
|
|
234
|
A
|
Lefthanders and Righthanders
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there *n* pupils who sat at the desks, two people at each desk. Vasily quickly realized that number *n* is even. Like all true mathematicians, Vasily has all students numbered from 1 to *n*.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
|
The first input line contains a single even integer *n* (4<=≤<=*n*<=≤<=100) — the number of students in the class. The second line contains exactly *n* capital English letters "L" and "R". If the *i*-th letter at the second line equals "L", then the student number *i* is a lefthander, otherwise he is a righthander.
|
Print integer pairs, one pair per line. In the *i*-th line print the numbers of students that will sit at the *i*-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
|
[
"6\nLLRLLL\n",
"4\nRRLL\n"
] |
[
"1 4\n2 5\n6 3\n",
"3 1\n4 2\n"
] |
none
| 0
|
[
{
"input": "6\nLLRLLL",
"output": "1 4\n2 5\n6 3"
},
{
"input": "4\nRRLL",
"output": "3 1\n4 2"
},
{
"input": "4\nLLRR",
"output": "1 3\n2 4"
},
{
"input": "6\nRLLRRL",
"output": "1 4\n2 5\n3 6"
},
{
"input": "8\nLRLRLLLR",
"output": "1 5\n6 2\n3 7\n4 8"
},
{
"input": "10\nRLLRLRRRLL",
"output": "1 6\n2 7\n3 8\n9 4\n5 10"
},
{
"input": "12\nLRRRRRLRRRRL",
"output": "1 7\n2 8\n3 9\n4 10\n5 11\n12 6"
},
{
"input": "14\nRLLRLLLLRLLLRL",
"output": "8 1\n2 9\n3 10\n11 4\n5 12\n6 13\n7 14"
},
{
"input": "16\nLLLRRRLRRLLRRLLL",
"output": "1 9\n2 10\n3 11\n4 12\n5 13\n14 6\n7 15\n16 8"
},
{
"input": "18\nRRRLLLLRRRLRLRLLRL",
"output": "1 10\n11 2\n3 12\n4 13\n5 14\n6 15\n7 16\n8 17\n18 9"
},
{
"input": "20\nRLRLLRLRRLLRRRRRRLRL",
"output": "11 1\n2 12\n3 13\n4 14\n5 15\n6 16\n7 17\n18 8\n9 19\n10 20"
},
{
"input": "22\nRLLLRLLLRRLRRRLRLLLLLL",
"output": "1 12\n2 13\n3 14\n4 15\n5 16\n6 17\n7 18\n8 19\n20 9\n21 10\n11 22"
},
{
"input": "24\nLRRRLRLLRLRRRRLLLLRRLRLR",
"output": "1 13\n2 14\n15 3\n16 4\n5 17\n18 6\n7 19\n8 20\n21 9\n10 22\n23 11\n12 24"
},
{
"input": "26\nRLRRLLRRLLRLRRLLRLLRRLRLRR",
"output": "1 14\n2 15\n16 3\n4 17\n5 18\n6 19\n7 20\n8 21\n9 22\n10 23\n24 11\n12 25\n13 26"
},
{
"input": "28\nLLLRRRRRLRRLRRRLRLRLRRLRLRRL",
"output": "1 15\n2 16\n3 17\n18 4\n5 19\n20 6\n7 21\n8 22\n9 23\n10 24\n25 11\n12 26\n13 27\n28 14"
},
{
"input": "30\nLRLLRLRRLLRLRLLRRRRRLRLRLRLLLL",
"output": "1 16\n2 17\n3 18\n4 19\n5 20\n6 21\n7 22\n23 8\n9 24\n10 25\n11 26\n12 27\n28 13\n14 29\n15 30"
},
{
"input": "32\nRLRLLRRLLRRLRLLRLRLRLLRLRRRLLRRR",
"output": "17 1\n2 18\n19 3\n4 20\n5 21\n22 6\n7 23\n8 24\n9 25\n10 26\n11 27\n12 28\n29 13\n14 30\n15 31\n16 32"
},
{
"input": "34\nLRRLRLRLLRRRRLLRLRRLRRLRLRRLRRRLLR",
"output": "1 18\n2 19\n20 3\n4 21\n5 22\n6 23\n7 24\n8 25\n9 26\n10 27\n28 11\n12 29\n13 30\n14 31\n15 32\n33 16\n17 34"
},
{
"input": "36\nRRLLLRRRLLLRRLLLRRLLRLLRLRLLRLRLRLLL",
"output": "19 1\n20 2\n3 21\n4 22\n5 23\n6 24\n25 7\n8 26\n9 27\n10 28\n11 29\n30 12\n13 31\n14 32\n15 33\n16 34\n35 17\n36 18"
},
{
"input": "38\nLLRRRLLRRRLRRLRLRRLRRLRLRLLRRRRLLLLRLL",
"output": "1 20\n2 21\n22 3\n4 23\n24 5\n6 25\n7 26\n27 8\n9 28\n10 29\n11 30\n12 31\n32 13\n14 33\n34 15\n16 35\n17 36\n37 18\n19 38"
},
{
"input": "40\nLRRRRRLRLLRRRLLRRLRLLRLRRLRRLLLRRLRRRLLL",
"output": "1 21\n2 22\n23 3\n4 24\n5 25\n26 6\n7 27\n8 28\n9 29\n10 30\n31 11\n12 32\n13 33\n14 34\n15 35\n16 36\n17 37\n18 38\n39 19\n20 40"
},
{
"input": "42\nRLRRLLLLLLLRRRLRLLLRRRLRLLLRLRLRLLLRLRLRRR",
"output": "1 22\n2 23\n3 24\n25 4\n5 26\n6 27\n7 28\n8 29\n9 30\n10 31\n11 32\n33 12\n34 13\n35 14\n15 36\n37 16\n17 38\n18 39\n19 40\n20 41\n21 42"
},
{
"input": "44\nLLLLRRLLRRLLRRLRLLRRRLRLRLLRLRLRRLLRLRRLLLRR",
"output": "1 23\n2 24\n3 25\n4 26\n27 5\n6 28\n7 29\n8 30\n31 9\n10 32\n11 33\n12 34\n35 13\n14 36\n15 37\n16 38\n17 39\n18 40\n41 19\n42 20\n21 43\n22 44"
},
{
"input": "46\nRRRLLLLRRLRLRRRRRLRLLRLRRLRLLLLLLLLRRLRLRLRLLL",
"output": "1 24\n2 25\n26 3\n4 27\n5 28\n6 29\n7 30\n31 8\n32 9\n10 33\n34 11\n12 35\n13 36\n14 37\n38 15\n16 39\n40 17\n18 41\n42 19\n20 43\n21 44\n45 22\n23 46"
},
{
"input": "48\nLLLLRRLRRRRLRRRLRLLLLLRRLLRLLRLLRRLRRLLRLRLRRRRL",
"output": "1 25\n2 26\n3 27\n4 28\n29 5\n6 30\n7 31\n32 8\n9 33\n10 34\n35 11\n12 36\n13 37\n38 14\n39 15\n16 40\n41 17\n18 42\n19 43\n20 44\n21 45\n22 46\n23 47\n48 24"
},
{
"input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 26\n2 27\n3 28\n4 29\n5 30\n6 31\n7 32\n8 33\n9 34\n10 35\n11 36\n12 37\n13 38\n14 39\n15 40\n16 41\n17 42\n18 43\n19 44\n20 45\n21 46\n22 47\n23 48\n24 49\n25 50"
},
{
"input": "52\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 27\n2 28\n3 29\n4 30\n5 31\n6 32\n7 33\n8 34\n9 35\n10 36\n11 37\n12 38\n13 39\n14 40\n15 41\n16 42\n17 43\n18 44\n19 45\n20 46\n21 47\n22 48\n23 49\n24 50\n25 51\n26 52"
},
{
"input": "54\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 28\n2 29\n3 30\n4 31\n5 32\n6 33\n7 34\n8 35\n9 36\n10 37\n11 38\n12 39\n13 40\n14 41\n15 42\n16 43\n17 44\n18 45\n19 46\n20 47\n21 48\n22 49\n23 50\n24 51\n25 52\n26 53\n27 54"
},
{
"input": "56\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 29\n2 30\n3 31\n4 32\n5 33\n6 34\n7 35\n8 36\n9 37\n10 38\n11 39\n12 40\n13 41\n14 42\n15 43\n16 44\n17 45\n18 46\n19 47\n20 48\n21 49\n22 50\n23 51\n24 52\n25 53\n26 54\n27 55\n28 56"
},
{
"input": "58\nRRRLLLRLLLLRRLRRRLLRLLRLRLLRLRRRRLLLLLLRLRRLRLRRRLRLRRLRRL",
"output": "1 30\n2 31\n3 32\n4 33\n5 34\n6 35\n36 7\n8 37\n9 38\n10 39\n11 40\n41 12\n13 42\n14 43\n44 15\n16 45\n46 17\n18 47\n19 48\n20 49\n21 50\n22 51\n52 23\n24 53\n25 54\n26 55\n27 56\n28 57\n29 58"
},
{
"input": "60\nRLLLLRRLLRRRLLLLRRRRRLRRRLRRRLLLRLLLRLRRRLRLLLRLLRRLLRRRRRLL",
"output": "31 1\n2 32\n3 33\n4 34\n5 35\n36 6\n7 37\n8 38\n9 39\n10 40\n11 41\n42 12\n13 43\n14 44\n15 45\n16 46\n17 47\n48 18\n49 19\n20 50\n21 51\n22 52\n53 23\n24 54\n25 55\n26 56\n27 57\n28 58\n59 29\n30 60"
},
{
"input": "62\nLRRLRLRLLLLRRLLLLRRRLRLLLLRRRLLLLLLRRRLLLLRRLRRLRLLLLLLLLRRLRR",
"output": "1 32\n33 2\n34 3\n4 35\n5 36\n6 37\n7 38\n8 39\n9 40\n10 41\n11 42\n12 43\n13 44\n14 45\n15 46\n16 47\n17 48\n18 49\n50 19\n51 20\n21 52\n53 22\n23 54\n24 55\n25 56\n26 57\n27 58\n28 59\n60 29\n30 61\n31 62"
},
{
"input": "64\nRLLLLRRRLRLLRRRRLRLLLRRRLLLRRRLLRLLRLRLRRRLLRRRRLRLRRRLLLLRRLLLL",
"output": "1 33\n2 34\n3 35\n4 36\n5 37\n6 38\n39 7\n8 40\n9 41\n10 42\n11 43\n12 44\n13 45\n14 46\n15 47\n16 48\n17 49\n18 50\n19 51\n20 52\n21 53\n22 54\n55 23\n56 24\n25 57\n26 58\n27 59\n28 60\n61 29\n62 30\n31 63\n32 64"
},
{
"input": "66\nLLRRRLLRLRLLRRRRRRRLLLLRRLLLLLLRLLLRLLLLLLRRRLRRLLRRRRRLRLLRLLLLRR",
"output": "1 34\n2 35\n3 36\n37 4\n38 5\n6 39\n7 40\n41 8\n9 42\n10 43\n11 44\n12 45\n46 13\n14 47\n15 48\n49 16\n50 17\n18 51\n19 52\n20 53\n21 54\n22 55\n23 56\n24 57\n58 25\n26 59\n27 60\n28 61\n29 62\n30 63\n31 64\n32 65\n33 66"
},
{
"input": "68\nRRLRLRLLRLRLRRRRRRLRRRLLLLRLLRLRLRLRRRRLRLRLLRRRRLRRLLRLRRLLRLRRLRRL",
"output": "35 1\n2 36\n3 37\n4 38\n5 39\n40 6\n7 41\n8 42\n9 43\n10 44\n45 11\n12 46\n13 47\n14 48\n15 49\n50 16\n17 51\n18 52\n19 53\n54 20\n21 55\n56 22\n23 57\n24 58\n25 59\n26 60\n27 61\n28 62\n29 63\n30 64\n31 65\n32 66\n33 67\n68 34"
},
{
"input": "70\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 36\n2 37\n3 38\n4 39\n5 40\n6 41\n7 42\n8 43\n9 44\n10 45\n11 46\n12 47\n13 48\n14 49\n15 50\n16 51\n17 52\n18 53\n19 54\n20 55\n21 56\n22 57\n23 58\n24 59\n25 60\n26 61\n27 62\n28 63\n29 64\n30 65\n31 66\n32 67\n33 68\n34 69\n35 70"
},
{
"input": "72\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 37\n2 38\n3 39\n4 40\n5 41\n6 42\n7 43\n8 44\n9 45\n10 46\n11 47\n12 48\n13 49\n14 50\n15 51\n16 52\n17 53\n18 54\n19 55\n20 56\n21 57\n22 58\n23 59\n24 60\n25 61\n26 62\n27 63\n28 64\n29 65\n30 66\n31 67\n32 68\n33 69\n34 70\n35 71\n36 72"
},
{
"input": "74\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 38\n2 39\n3 40\n4 41\n5 42\n6 43\n7 44\n8 45\n9 46\n10 47\n11 48\n12 49\n13 50\n14 51\n15 52\n16 53\n17 54\n18 55\n19 56\n20 57\n21 58\n22 59\n23 60\n24 61\n25 62\n26 63\n27 64\n28 65\n29 66\n30 67\n31 68\n32 69\n33 70\n34 71\n35 72\n36 73\n37 74"
},
{
"input": "76\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 39\n2 40\n3 41\n4 42\n5 43\n6 44\n7 45\n8 46\n9 47\n10 48\n11 49\n12 50\n13 51\n14 52\n15 53\n16 54\n17 55\n18 56\n19 57\n20 58\n21 59\n22 60\n23 61\n24 62\n25 63\n26 64\n27 65\n28 66\n29 67\n30 68\n31 69\n32 70\n33 71\n34 72\n35 73\n36 74\n37 75\n38 76"
},
{
"input": "78\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 40\n2 41\n3 42\n4 43\n5 44\n6 45\n7 46\n8 47\n9 48\n10 49\n11 50\n12 51\n13 52\n14 53\n15 54\n16 55\n17 56\n18 57\n19 58\n20 59\n21 60\n22 61\n23 62\n24 63\n25 64\n26 65\n27 66\n28 67\n29 68\n30 69\n31 70\n32 71\n33 72\n34 73\n35 74\n36 75\n37 76\n38 77\n39 78"
},
{
"input": "80\nLRLRRRRLRRRRLLLLRLLRLRLLRRLRLLLRRLLLLRLLLRLRLLRRRLRRRLRLRRRRRLRLLRLLRRLLLRLRRRLL",
"output": "1 41\n2 42\n3 43\n4 44\n45 5\n46 6\n7 47\n8 48\n9 49\n50 10\n11 51\n12 52\n13 53\n14 54\n15 55\n16 56\n17 57\n18 58\n19 59\n20 60\n21 61\n62 22\n23 63\n24 64\n65 25\n26 66\n27 67\n68 28\n29 69\n30 70\n31 71\n72 32\n73 33\n34 74\n35 75\n36 76\n37 77\n38 78\n39 79\n40 80"
},
{
"input": "82\nRLRRLLRLRLRLLLRLLLRRLLRRLRRRRLLRLLLLRRRRRLLLRRRLLLLRLRRLRRRLRLLLLRRRLRLRLLLRLLLLLR",
"output": "42 1\n2 43\n44 3\n4 45\n5 46\n6 47\n48 7\n8 49\n50 9\n10 51\n11 52\n12 53\n13 54\n14 55\n56 15\n16 57\n17 58\n18 59\n60 19\n20 61\n21 62\n22 63\n64 23\n65 24\n25 66\n26 67\n27 68\n69 28\n29 70\n30 71\n31 72\n73 32\n33 74\n34 75\n35 76\n36 77\n78 37\n79 38\n80 39\n81 40\n41 82"
},
{
"input": "84\nLRLRRRRRRLLLRLRLLLLLRRLRLRLRRRLLRLLLRLRLLLRRRLRLRRLRLRLLLLLLLLRRRRRRLLLRRLRLRLLLRLRR",
"output": "1 43\n2 44\n3 45\n46 4\n5 47\n48 6\n7 49\n8 50\n51 9\n10 52\n11 53\n12 54\n55 13\n14 56\n57 15\n16 58\n17 59\n18 60\n19 61\n20 62\n21 63\n22 64\n23 65\n24 66\n25 67\n26 68\n27 69\n70 28\n71 29\n30 72\n31 73\n32 74\n33 75\n34 76\n35 77\n36 78\n79 37\n38 80\n39 81\n40 82\n41 83\n42 84"
},
{
"input": "86\nRRRLLLRLLRLLRLRLRLLLRLRLRRLLRLLLRLLLLLLRRRLRLLRLLLRRRLRLLLLRLLRLRRLLRLLLRRRLLRLRLLRLLR",
"output": "1 44\n45 2\n46 3\n4 47\n5 48\n6 49\n50 7\n8 51\n9 52\n10 53\n11 54\n12 55\n56 13\n14 57\n58 15\n16 59\n17 60\n18 61\n19 62\n20 63\n64 21\n22 65\n23 66\n24 67\n68 25\n26 69\n27 70\n28 71\n72 29\n30 73\n31 74\n32 75\n76 33\n34 77\n35 78\n36 79\n37 80\n38 81\n39 82\n40 83\n84 41\n85 42\n43 86"
},
{
"input": "88\nLLRLRLRLLLLRRRRRRLRRLLLLLRRLRRLLLLLRLRLRLLLLLRLRLRRLRLRRLRLLRRLRLLLRLLLLRRLLRRLRLRLRRLRR",
"output": "1 45\n2 46\n47 3\n4 48\n49 5\n6 50\n7 51\n8 52\n9 53\n10 54\n11 55\n12 56\n57 13\n14 58\n59 15\n60 16\n17 61\n18 62\n63 19\n20 64\n21 65\n22 66\n23 67\n24 68\n25 69\n70 26\n71 27\n28 72\n29 73\n30 74\n31 75\n32 76\n33 77\n34 78\n35 79\n36 80\n37 81\n38 82\n39 83\n40 84\n41 85\n42 86\n43 87\n44 88"
},
{
"input": "90\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 46\n2 47\n3 48\n4 49\n5 50\n6 51\n7 52\n8 53\n9 54\n10 55\n11 56\n12 57\n13 58\n14 59\n15 60\n16 61\n17 62\n18 63\n19 64\n20 65\n21 66\n22 67\n23 68\n24 69\n25 70\n26 71\n27 72\n28 73\n29 74\n30 75\n31 76\n32 77\n33 78\n34 79\n35 80\n36 81\n37 82\n38 83\n39 84\n40 85\n41 86\n42 87\n43 88\n44 89\n45 90"
},
{
"input": "92\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 47\n2 48\n3 49\n4 50\n5 51\n6 52\n7 53\n8 54\n9 55\n10 56\n11 57\n12 58\n13 59\n14 60\n15 61\n16 62\n17 63\n18 64\n19 65\n20 66\n21 67\n22 68\n23 69\n24 70\n25 71\n26 72\n27 73\n28 74\n29 75\n30 76\n31 77\n32 78\n33 79\n34 80\n35 81\n36 82\n37 83\n38 84\n39 85\n40 86\n41 87\n42 88\n43 89\n44 90\n45 91\n46 92"
},
{
"input": "94\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 48\n2 49\n3 50\n4 51\n5 52\n6 53\n7 54\n8 55\n9 56\n10 57\n11 58\n12 59\n13 60\n14 61\n15 62\n16 63\n17 64\n18 65\n19 66\n20 67\n21 68\n22 69\n23 70\n24 71\n25 72\n26 73\n27 74\n28 75\n29 76\n30 77\n31 78\n32 79\n33 80\n34 81\n35 82\n36 83\n37 84\n38 85\n39 86\n40 87\n41 88\n42 89\n43 90\n44 91\n45 92\n46 93\n47 94"
},
{
"input": "96\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 49\n2 50\n3 51\n4 52\n5 53\n6 54\n7 55\n8 56\n9 57\n10 58\n11 59\n12 60\n13 61\n14 62\n15 63\n16 64\n17 65\n18 66\n19 67\n20 68\n21 69\n22 70\n23 71\n24 72\n25 73\n26 74\n27 75\n28 76\n29 77\n30 78\n31 79\n32 80\n33 81\n34 82\n35 83\n36 84\n37 85\n38 86\n39 87\n40 88\n41 89\n42 90\n43 91\n44 92\n45 93\n46 94\n47 95\n48 96"
},
{
"input": "98\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 50\n2 51\n3 52\n4 53\n5 54\n6 55\n7 56\n8 57\n9 58\n10 59\n11 60\n12 61\n13 62\n14 63\n15 64\n16 65\n17 66\n18 67\n19 68\n20 69\n21 70\n22 71\n23 72\n24 73\n25 74\n26 75\n27 76\n28 77\n29 78\n30 79\n31 80\n32 81\n33 82\n34 83\n35 84\n36 85\n37 86\n38 87\n39 88\n40 89\n41 90\n42 91\n43 92\n44 93\n45 94\n46 95\n47 96\n48 97\n49 98"
},
{
"input": "100\nRLRRRRLLLLRRRRLRRRRRRRRLRLRRLLRRRRRRRRLRRRRLLLLRRRRLRRLRLRRRLLRRLRRLLLRLRRLLLLLLRLRLRLRRLRLRLRRRLLLR",
"output": "1 51\n2 52\n3 53\n4 54\n55 5\n6 56\n7 57\n8 58\n9 59\n10 60\n61 11\n62 12\n13 63\n14 64\n15 65\n16 66\n17 67\n68 18\n69 19\n70 20\n21 71\n72 22\n23 73\n24 74\n75 25\n26 76\n77 27\n78 28\n29 79\n30 80\n31 81\n82 32\n33 83\n84 34\n35 85\n86 36\n37 87\n38 88\n39 89\n40 90\n91 41\n42 92\n93 43\n44 94\n45 95\n46 96\n47 97\n98 48\n99 49\n50 100"
},
{
"input": "100\nLRLLLLRLLLLRRRRRLRRRRLRRLRRLRLLRRLRRRRLLRRRLLLRLLLRRRRLLRLRLRRLRLLRRLLRRLRRLRRRRRLRRLRLRLRLLLLLLLLRL",
"output": "1 51\n2 52\n3 53\n4 54\n5 55\n6 56\n7 57\n8 58\n9 59\n10 60\n11 61\n12 62\n63 13\n14 64\n65 15\n66 16\n17 67\n18 68\n69 19\n70 20\n21 71\n22 72\n73 23\n24 74\n25 75\n76 26\n27 77\n28 78\n29 79\n30 80\n31 81\n82 32\n33 83\n34 84\n85 35\n36 86\n87 37\n38 88\n39 89\n40 90\n91 41\n92 42\n93 43\n44 94\n45 95\n46 96\n97 47\n48 98\n49 99\n50 100"
},
{
"input": "100\nLLLRRLLRLRLLLRLLLRLRLLRRRLRRLLLRLRLRRLLRLRRRLLLRRLLRLLRRLLRRRRRLRLRRLRLRRLRLRRLLRLRLLRLLLRLLRLLLLRLL",
"output": "1 51\n2 52\n3 53\n54 4\n5 55\n6 56\n7 57\n58 8\n9 59\n10 60\n11 61\n12 62\n13 63\n64 14\n15 65\n16 66\n17 67\n18 68\n19 69\n20 70\n21 71\n22 72\n23 73\n74 24\n25 75\n26 76\n27 77\n28 78\n29 79\n30 80\n31 81\n82 32\n33 83\n84 34\n35 85\n36 86\n87 37\n38 88\n39 89\n40 90\n41 91\n92 42\n43 93\n94 44\n45 95\n46 96\n47 97\n48 98\n99 49\n50 100"
},
{
"input": "100\nRLLLLRRLLLLRRRRLLRLRRRLLLRLLRLLLLLRRLLLLLLRRLRRRRRLRLLRLRRRLLLRLRLRLLLRRRLLLLLRRRRRLRRLLLLRLLLRRLLLL",
"output": "51 1\n2 52\n3 53\n4 54\n5 55\n56 6\n7 57\n8 58\n9 59\n10 60\n11 61\n62 12\n13 63\n64 14\n15 65\n16 66\n17 67\n68 18\n19 69\n70 20\n21 71\n22 72\n23 73\n24 74\n25 75\n76 26\n27 77\n28 78\n29 79\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n40 90\n41 91\n42 92\n93 43\n94 44\n45 95\n46 96\n97 47\n98 48\n99 49\n100 50"
},
{
"input": "100\nRLRRLRLRRLRLLRLLRRRLRRLLLLLRLRLRRRRRRRLLRRRLLRLRLLLRRRLLRRRLLRLRLLLLRRLRLLRLLRLLLLRRLRLRRLRLLLLRLRRR",
"output": "51 1\n2 52\n3 53\n4 54\n5 55\n56 6\n7 57\n8 58\n9 59\n10 60\n61 11\n12 62\n13 63\n14 64\n15 65\n16 66\n67 17\n68 18\n19 69\n20 70\n71 21\n22 72\n23 73\n24 74\n25 75\n26 76\n27 77\n28 78\n29 79\n80 30\n31 81\n82 32\n33 83\n34 84\n85 35\n36 86\n87 37\n38 88\n39 89\n40 90\n41 91\n92 42\n93 43\n44 94\n45 95\n46 96\n47 97\n48 98\n49 99\n50 100"
},
{
"input": "100\nLRRLRLRRRRRRLRRLRRLLLLLLRRLLRRLLRLLLLLLRRRLLRLRRRLLRLLRRLRRRLLRLRLLRRLRRRLLLRRRRLLRRRLLLRRRRRLLLLLLR",
"output": "1 51\n2 52\n53 3\n4 54\n5 55\n6 56\n57 7\n8 58\n9 59\n10 60\n61 11\n62 12\n13 63\n64 14\n15 65\n16 66\n67 17\n18 68\n19 69\n20 70\n21 71\n22 72\n23 73\n24 74\n75 25\n76 26\n27 77\n28 78\n29 79\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n40 90\n41 91\n42 92\n43 93\n44 94\n95 45\n46 96\n97 47\n98 48\n99 49\n50 100"
},
{
"input": "100\nRRLRRLRLRLRRRRLLRRLLRLRRLLRRRLLRLRRLRLRRLLLRRLLRRRRRRLLLRRRLLRRLLLLLLRLLLLLLRLLLRRRLRLLRRRRRLLRLLRRR",
"output": "1 51\n2 52\n3 53\n54 4\n55 5\n6 56\n7 57\n8 58\n9 59\n10 60\n61 11\n12 62\n13 63\n64 14\n15 65\n16 66\n67 17\n68 18\n19 69\n20 70\n71 21\n22 72\n73 23\n74 24\n25 75\n26 76\n27 77\n78 28\n79 29\n30 80\n31 81\n32 82\n33 83\n84 34\n35 85\n36 86\n87 37\n38 88\n39 89\n40 90\n41 91\n42 92\n43 93\n94 44\n45 95\n46 96\n47 97\n48 98\n49 99\n50 100"
},
{
"input": "100\nRRLLLRLRRLRLLRRLRRRLLRRRLRRLLLLLLLLLRRRLLRLRRLRRLRRLRRLRLLLLRLLRRRLLLLRLRRRLLRRRRLRRLLRRRRLRRRLRLLLR",
"output": "1 51\n52 2\n3 53\n4 54\n5 55\n6 56\n7 57\n58 8\n59 9\n10 60\n11 61\n12 62\n13 63\n14 64\n15 65\n16 66\n67 17\n68 18\n69 19\n20 70\n21 71\n72 22\n23 73\n24 74\n25 75\n76 26\n77 27\n28 78\n29 79\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n40 90\n41 91\n42 92\n43 93\n44 94\n95 45\n46 96\n97 47\n98 48\n49 99\n50 100"
},
{
"input": "100\nLLLLLRRLRRRRRRRLLRRRRRLRRLRLRLLRLRRLLLRRRRLLRRLRLLRLLLRLRLLRRRRRRRRRLRLLLRLRLLLLLRLRRRRLRLLRLRLRLRRL",
"output": "1 51\n2 52\n3 53\n4 54\n5 55\n56 6\n7 57\n8 58\n59 9\n10 60\n11 61\n12 62\n13 63\n14 64\n15 65\n16 66\n17 67\n18 68\n69 19\n20 70\n71 21\n72 22\n23 73\n24 74\n75 25\n26 76\n77 27\n28 78\n79 29\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n90 40\n91 41\n42 92\n43 93\n44 94\n95 45\n46 96\n47 97\n48 98\n49 99\n50 100"
},
{
"input": "100\nLLRRRLLLRLLLLRLRLRLRRRLLLRRRLRLLRLLLRRRRRLRRLRRLRRRLRRLRRLLLRLRLLRRRRLRLRRRRRLRRLRLLRRRRLLLRRRRRLLLL",
"output": "1 51\n2 52\n3 53\n4 54\n55 5\n6 56\n7 57\n8 58\n59 9\n10 60\n11 61\n12 62\n13 63\n64 14\n15 65\n16 66\n17 67\n18 68\n19 69\n70 20\n21 71\n72 22\n23 73\n24 74\n25 75\n26 76\n27 77\n78 28\n29 79\n30 80\n31 81\n32 82\n83 33\n34 84\n35 85\n36 86\n37 87\n38 88\n89 39\n90 40\n91 41\n42 92\n43 93\n44 94\n45 95\n46 96\n97 47\n48 98\n99 49\n100 50"
}
] | 1,666,049,433
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 60
| 46
| 0
|
def solve():
with open("input.txt", 'r') as f:
n = int(f.readline())
s = f.readline()
outputfile = open("output.txt", 'w')
for i in range(int(n/2)):
if s[i]=='L':
outputfile.write(f"{i+1} {int(n/2+i+1)}\n")
else:
outputfile.write(f"{int(n/2+i+1)} {i+1}\n")
def init():
T=1
# T=int(input())
while T:
solve()
# print(solve())
T -=1
init()
|
Title: Lefthanders and Righthanders
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there *n* pupils who sat at the desks, two people at each desk. Vasily quickly realized that number *n* is even. Like all true mathematicians, Vasily has all students numbered from 1 to *n*.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
Input Specification:
The first input line contains a single even integer *n* (4<=≤<=*n*<=≤<=100) — the number of students in the class. The second line contains exactly *n* capital English letters "L" and "R". If the *i*-th letter at the second line equals "L", then the student number *i* is a lefthander, otherwise he is a righthander.
Output Specification:
Print integer pairs, one pair per line. In the *i*-th line print the numbers of students that will sit at the *i*-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
Demo Input:
['6\nLLRLLL\n', '4\nRRLL\n']
Demo Output:
['1 4\n2 5\n6 3\n', '3 1\n4 2\n']
Note:
none
|
```python
def solve():
with open("input.txt", 'r') as f:
n = int(f.readline())
s = f.readline()
outputfile = open("output.txt", 'w')
for i in range(int(n/2)):
if s[i]=='L':
outputfile.write(f"{i+1} {int(n/2+i+1)}\n")
else:
outputfile.write(f"{int(n/2+i+1)} {i+1}\n")
def init():
T=1
# T=int(input())
while T:
solve()
# print(solve())
T -=1
init()
```
| 3
|
|
774
|
E
|
Big Number and Remainder
|
PROGRAMMING
| 2,300
|
[
"*special",
"math",
"number theory"
] | null | null |
Stepan has a very big positive integer.
Let's consider all cyclic shifts of Stepan's integer (if we look at his integer like at a string) which are also integers (i.e. they do not have leading zeros). Let's call such shifts as good shifts. For example, for the integer 10203 the good shifts are the integer itself 10203 and integers 20310 and 31020.
Stepan wants to know the minimum remainder of the division by the given number *m* among all good shifts. Your task is to determine the minimum remainder of the division by *m*.
|
The first line contains the integer which Stepan has. The length of Stepan's integer is between 2 and 200<=000 digits, inclusive. It is guaranteed that Stepan's integer does not contain leading zeros.
The second line contains the integer *m* (2<=≤<=*m*<=≤<=108) — the number by which Stepan divides good shifts of his integer.
|
Print the minimum remainder which Stepan can get if he divides all good shifts of his integer by the given number *m*.
|
[
"521\n3\n",
"1001\n5\n",
"5678901234567890123456789\n10000\n"
] |
[
"2\n",
"0\n",
"123\n"
] |
In the first example all good shifts of the integer 521 (good shifts are equal to 521, 215 and 152) has same remainder 2 when dividing by 3.
In the second example there are only two good shifts: the Stepan's integer itself and the shift by one position to the right. The integer itself is 1001 and the remainder after dividing it by 5 equals 1. The shift by one position to the right equals to 1100 and the remainder after dividing it by 5 equals 0, which is the minimum possible remainder.
| 0
|
[
{
"input": "521\n3",
"output": "2"
},
{
"input": "1001\n5",
"output": "0"
},
{
"input": "5678901234567890123456789\n10000",
"output": "123"
},
{
"input": "552352155\n13",
"output": "2"
},
{
"input": "11533077525260\n193983",
"output": "22331"
},
{
"input": "15\n19",
"output": "13"
},
{
"input": "2342341\n2342340",
"output": "1"
},
{
"input": "12345\n12344",
"output": "1"
},
{
"input": "23457\n23456",
"output": "1"
},
{
"input": "79\n60847671",
"output": "79"
},
{
"input": "456\n79575973",
"output": "456"
},
{
"input": "1908\n86567928",
"output": "1908"
},
{
"input": "30866\n55357692",
"output": "30866"
},
{
"input": "980552\n32506042",
"output": "298055"
},
{
"input": "4865180\n11799246",
"output": "1804865"
},
{
"input": "63014535\n5261418",
"output": "948834"
},
{
"input": "260597722\n10577910",
"output": "38629"
},
{
"input": "8460711003592660299377170851036141857177105821455679225940496548902563144980707248030407967017317182\n13315371",
"output": "16290"
}
] | 1,491,407,667
| 1,167
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 9
| 3,000
| 6,348,800
|
s = input()
m = int(input())
mn = m
sz = len(s)
step = 10 ** (sz - 1)
n = int(s)
for i in range(0, sz):
a = n % 10
n //= 10
n += step * a
if (a != 0):
mn = min(n % m, mn)
print(mn)
|
Title: Big Number and Remainder
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Stepan has a very big positive integer.
Let's consider all cyclic shifts of Stepan's integer (if we look at his integer like at a string) which are also integers (i.e. they do not have leading zeros). Let's call such shifts as good shifts. For example, for the integer 10203 the good shifts are the integer itself 10203 and integers 20310 and 31020.
Stepan wants to know the minimum remainder of the division by the given number *m* among all good shifts. Your task is to determine the minimum remainder of the division by *m*.
Input Specification:
The first line contains the integer which Stepan has. The length of Stepan's integer is between 2 and 200<=000 digits, inclusive. It is guaranteed that Stepan's integer does not contain leading zeros.
The second line contains the integer *m* (2<=≤<=*m*<=≤<=108) — the number by which Stepan divides good shifts of his integer.
Output Specification:
Print the minimum remainder which Stepan can get if he divides all good shifts of his integer by the given number *m*.
Demo Input:
['521\n3\n', '1001\n5\n', '5678901234567890123456789\n10000\n']
Demo Output:
['2\n', '0\n', '123\n']
Note:
In the first example all good shifts of the integer 521 (good shifts are equal to 521, 215 and 152) has same remainder 2 when dividing by 3.
In the second example there are only two good shifts: the Stepan's integer itself and the shift by one position to the right. The integer itself is 1001 and the remainder after dividing it by 5 equals 1. The shift by one position to the right equals to 1100 and the remainder after dividing it by 5 equals 0, which is the minimum possible remainder.
|
```python
s = input()
m = int(input())
mn = m
sz = len(s)
step = 10 ** (sz - 1)
n = int(s)
for i in range(0, sz):
a = n % 10
n //= 10
n += step * a
if (a != 0):
mn = min(n % m, mn)
print(mn)
```
| 0
|
|
791
|
A
|
Bear and Big Brother
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
|
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
|
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
|
[
"4 7\n",
"4 9\n",
"1 1\n"
] |
[
"2\n",
"3\n",
"1\n"
] |
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then.
| 500
|
[
{
"input": "4 7",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "4"
},
{
"input": "1 6",
"output": "5"
},
{
"input": "1 7",
"output": "5"
},
{
"input": "1 8",
"output": "6"
},
{
"input": "1 9",
"output": "6"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "2 4",
"output": "2"
},
{
"input": "2 5",
"output": "3"
},
{
"input": "2 6",
"output": "3"
},
{
"input": "2 7",
"output": "4"
},
{
"input": "2 8",
"output": "4"
},
{
"input": "2 9",
"output": "4"
},
{
"input": "2 10",
"output": "4"
},
{
"input": "3 3",
"output": "1"
},
{
"input": "3 4",
"output": "1"
},
{
"input": "3 5",
"output": "2"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "3 7",
"output": "3"
},
{
"input": "3 8",
"output": "3"
},
{
"input": "3 9",
"output": "3"
},
{
"input": "3 10",
"output": "3"
},
{
"input": "4 4",
"output": "1"
},
{
"input": "4 5",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "4 7",
"output": "2"
},
{
"input": "4 8",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "4 10",
"output": "3"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "5 6",
"output": "1"
},
{
"input": "5 7",
"output": "1"
},
{
"input": "5 8",
"output": "2"
},
{
"input": "5 9",
"output": "2"
},
{
"input": "5 10",
"output": "2"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "6 7",
"output": "1"
},
{
"input": "6 8",
"output": "1"
},
{
"input": "6 9",
"output": "2"
},
{
"input": "6 10",
"output": "2"
},
{
"input": "7 7",
"output": "1"
},
{
"input": "7 8",
"output": "1"
},
{
"input": "7 9",
"output": "1"
},
{
"input": "7 10",
"output": "1"
},
{
"input": "8 8",
"output": "1"
},
{
"input": "8 9",
"output": "1"
},
{
"input": "8 10",
"output": "1"
},
{
"input": "9 9",
"output": "1"
},
{
"input": "9 10",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "1 2",
"output": "2"
}
] | 1,697,891,948
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 62
| 46
| 0
|
a,b=input().split()
a=int(a)
b=int(b)
for i in (range(1,max(a,b)+1)):
if a*3>b*2:
print(i)
break
else:
a=a*3
b=b*2
|
Title: Bear and Big Brother
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
Output Specification:
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
Demo Input:
['4 7\n', '4 9\n', '1 1\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then.
|
```python
a,b=input().split()
a=int(a)
b=int(b)
for i in (range(1,max(a,b)+1)):
if a*3>b*2:
print(i)
break
else:
a=a*3
b=b*2
```
| 3
|
|
965
|
A
|
Paper Airplanes
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
|
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
|
Print a single integer — the minimum number of packs they should buy.
|
[
"5 3 2 3\n",
"5 3 100 1\n"
] |
[
"4\n",
"5\n"
] |
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets.
| 500
|
[
{
"input": "5 3 2 3",
"output": "4"
},
{
"input": "5 3 100 1",
"output": "5"
},
{
"input": "10000 10000 1 1",
"output": "100000000"
},
{
"input": "1 1 10000 10000",
"output": "1"
},
{
"input": "300 300 21 23",
"output": "196"
},
{
"input": "300 2 37 51",
"output": "6"
},
{
"input": "2 400 23 57",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10000 10000 3 2",
"output": "16670000"
},
{
"input": "1 2 1 2",
"output": "1"
},
{
"input": "1 1 2 2",
"output": "1"
},
{
"input": "1 1 10 10",
"output": "1"
},
{
"input": "5324 5439 32 13",
"output": "69622"
},
{
"input": "9000 1 2432 1",
"output": "9000"
},
{
"input": "230 1234 9124 23",
"output": "10"
},
{
"input": "11 1 1 1",
"output": "11"
},
{
"input": "6246 8489 1227 9",
"output": "4858"
},
{
"input": "9 20 5 7",
"output": "6"
}
] | 1,564,861,163
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 108
| 0
|
a=list(map(int,input().split()))
s=(a[1]//a[2])+1
s=s*a[0]
if s%a[3]==0:
print(s//a[3])
else:
print(1+(s//a[3]))
|
Title: Paper Airplanes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
Input Specification:
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
Output Specification:
Print a single integer — the minimum number of packs they should buy.
Demo Input:
['5 3 2 3\n', '5 3 100 1\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets.
|
```python
a=list(map(int,input().split()))
s=(a[1]//a[2])+1
s=s*a[0]
if s%a[3]==0:
print(s//a[3])
else:
print(1+(s//a[3]))
```
| 0
|
|
868
|
B
|
Race Against Time
|
PROGRAMMING
| 1,400
|
[
"implementation"
] | null | null |
Have you ever tried to explain to the coordinator, why it is eight hours to the contest and not a single problem has been prepared yet? Misha had. And this time he has a really strong excuse: he faced a space-time paradox! Space and time replaced each other.
The entire universe turned into an enormous clock face with three hands — hour, minute, and second. Time froze, and clocks now show the time *h* hours, *m* minutes, *s* seconds.
Last time Misha talked with the coordinator at *t*1 o'clock, so now he stands on the number *t*1 on the clock face. The contest should be ready by *t*2 o'clock. In the terms of paradox it means that Misha has to go to number *t*2 somehow. Note that he doesn't have to move forward only: in these circumstances time has no direction.
Clock hands are very long, and Misha cannot get round them. He also cannot step over as it leads to the collapse of space-time. That is, if hour clock points 12 and Misha stands at 11 then he cannot move to 1 along the top arc. He has to follow all the way round the clock center (of course, if there are no other hands on his way).
Given the hands' positions, *t*1, and *t*2, find if Misha can prepare the contest on time (or should we say on space?). That is, find if he can move from *t*1 to *t*2 by the clock face.
|
Five integers *h*, *m*, *s*, *t*1, *t*2 (1<=≤<=*h*<=≤<=12, 0<=≤<=*m*,<=*s*<=≤<=59, 1<=≤<=*t*1,<=*t*2<=≤<=12, *t*1<=≠<=*t*2).
Misha's position and the target time do not coincide with the position of any hand.
|
Print "YES" (quotes for clarity), if Misha can prepare the contest on time, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
|
[
"12 30 45 3 11\n",
"12 0 1 12 1\n",
"3 47 0 4 9\n"
] |
[
"NO\n",
"YES\n",
"YES\n"
] |
The three examples are shown on the pictures below from left to right. The starting position of Misha is shown with green, the ending position is shown with pink. Note that the positions of the hands on the pictures are not exact, but are close to the exact and the answer is the same.
| 500
|
[
{
"input": "12 30 45 3 11",
"output": "NO"
},
{
"input": "12 0 1 12 1",
"output": "YES"
},
{
"input": "3 47 0 4 9",
"output": "YES"
},
{
"input": "10 22 59 6 10",
"output": "YES"
},
{
"input": "3 1 13 12 3",
"output": "NO"
},
{
"input": "11 19 28 9 10",
"output": "YES"
},
{
"input": "9 38 22 6 1",
"output": "NO"
},
{
"input": "5 41 11 5 8",
"output": "NO"
},
{
"input": "11 2 53 10 4",
"output": "YES"
},
{
"input": "9 41 17 10 1",
"output": "YES"
},
{
"input": "6 54 48 12 6",
"output": "YES"
},
{
"input": "12 55 9 5 1",
"output": "NO"
},
{
"input": "8 55 35 9 3",
"output": "NO"
},
{
"input": "3 21 34 3 10",
"output": "YES"
},
{
"input": "2 52 1 12 3",
"output": "NO"
},
{
"input": "7 17 11 1 7",
"output": "NO"
},
{
"input": "11 6 37 6 4",
"output": "YES"
},
{
"input": "9 6 22 8 1",
"output": "NO"
},
{
"input": "3 10 5 5 9",
"output": "YES"
},
{
"input": "7 12 22 11 2",
"output": "YES"
},
{
"input": "7 19 4 7 3",
"output": "NO"
},
{
"input": "11 36 21 4 6",
"output": "NO"
},
{
"input": "10 32 49 1 3",
"output": "YES"
},
{
"input": "1 9 43 11 3",
"output": "NO"
},
{
"input": "1 8 33 4 8",
"output": "NO"
},
{
"input": "3 0 33 9 4",
"output": "NO"
},
{
"input": "7 15 9 10 3",
"output": "NO"
},
{
"input": "8 3 57 11 1",
"output": "NO"
},
{
"input": "1 33 49 5 9",
"output": "NO"
},
{
"input": "3 40 0 5 7",
"output": "YES"
},
{
"input": "5 50 9 2 7",
"output": "NO"
},
{
"input": "10 0 52 6 1",
"output": "YES"
},
{
"input": "3 10 4 1 11",
"output": "NO"
},
{
"input": "2 41 53 4 6",
"output": "YES"
},
{
"input": "10 29 30 4 7",
"output": "NO"
},
{
"input": "5 13 54 9 11",
"output": "NO"
},
{
"input": "1 0 23 3 9",
"output": "NO"
},
{
"input": "1 0 41 12 1",
"output": "NO"
},
{
"input": "6 30 30 3 9",
"output": "YES"
},
{
"input": "3 7 32 11 10",
"output": "YES"
},
{
"input": "1 0 25 12 4",
"output": "NO"
},
{
"input": "12 0 0 5 6",
"output": "YES"
},
{
"input": "1 5 4 3 2",
"output": "YES"
},
{
"input": "6 30 30 9 10",
"output": "YES"
},
{
"input": "6 0 0 2 8",
"output": "NO"
},
{
"input": "10 50 59 9 10",
"output": "YES"
},
{
"input": "12 59 59 12 6",
"output": "NO"
},
{
"input": "3 0 30 3 4",
"output": "NO"
},
{
"input": "2 10 10 1 11",
"output": "YES"
},
{
"input": "10 5 30 1 12",
"output": "YES"
},
{
"input": "5 29 31 5 10",
"output": "YES"
},
{
"input": "5 2 2 11 2",
"output": "NO"
},
{
"input": "5 15 46 3 10",
"output": "YES"
},
{
"input": "1 30 50 1 2",
"output": "NO"
},
{
"input": "5 26 14 1 12",
"output": "YES"
},
{
"input": "1 58 43 12 1",
"output": "YES"
},
{
"input": "12 0 12 11 1",
"output": "NO"
},
{
"input": "6 52 41 6 5",
"output": "YES"
},
{
"input": "5 8 2 1 3",
"output": "NO"
},
{
"input": "2 0 0 1 3",
"output": "NO"
},
{
"input": "1 5 6 2 1",
"output": "YES"
},
{
"input": "9 5 5 11 12",
"output": "YES"
},
{
"input": "12 5 19 3 4",
"output": "NO"
},
{
"input": "6 14 59 1 3",
"output": "NO"
},
{
"input": "10 38 34 4 12",
"output": "YES"
},
{
"input": "2 54 14 2 12",
"output": "YES"
},
{
"input": "5 31 0 6 7",
"output": "NO"
},
{
"input": "6 15 30 3 9",
"output": "YES"
},
{
"input": "3 54 41 8 10",
"output": "NO"
},
{
"input": "3 39 10 10 12",
"output": "YES"
},
{
"input": "1 11 50 1 2",
"output": "NO"
},
{
"input": "5 40 24 8 1",
"output": "NO"
},
{
"input": "9 5 59 1 3",
"output": "NO"
},
{
"input": "5 0 0 6 7",
"output": "YES"
},
{
"input": "4 40 59 6 8",
"output": "YES"
},
{
"input": "10 13 55 12 1",
"output": "YES"
},
{
"input": "6 50 0 5 6",
"output": "YES"
},
{
"input": "7 59 3 7 4",
"output": "YES"
},
{
"input": "6 0 1 6 7",
"output": "NO"
},
{
"input": "6 15 55 3 5",
"output": "NO"
},
{
"input": "12 9 55 10 2",
"output": "YES"
},
{
"input": "2 0 1 11 2",
"output": "NO"
},
{
"input": "8 45 17 12 9",
"output": "NO"
},
{
"input": "5 30 31 11 3",
"output": "YES"
},
{
"input": "6 43 0 10 6",
"output": "NO"
},
{
"input": "6 30 30 1 11",
"output": "YES"
},
{
"input": "11 59 59 11 12",
"output": "YES"
},
{
"input": "5 45 35 9 5",
"output": "NO"
},
{
"input": "2 43 4 9 7",
"output": "NO"
},
{
"input": "12 30 50 6 9",
"output": "NO"
},
{
"input": "1 10 1 2 3",
"output": "NO"
},
{
"input": "10 5 55 9 1",
"output": "NO"
},
{
"input": "1 59 59 2 3",
"output": "YES"
},
{
"input": "1 49 14 10 3",
"output": "NO"
},
{
"input": "3 15 15 2 4",
"output": "YES"
},
{
"input": "10 5 55 1 5",
"output": "NO"
},
{
"input": "6 33 45 12 6",
"output": "YES"
},
{
"input": "1 20 20 11 1",
"output": "YES"
},
{
"input": "2 30 45 1 11",
"output": "YES"
},
{
"input": "1 55 1 11 10",
"output": "YES"
},
{
"input": "3 0 1 11 1",
"output": "NO"
},
{
"input": "1 5 6 1 12",
"output": "YES"
},
{
"input": "12 10 5 11 4",
"output": "YES"
},
{
"input": "6 5 59 12 1",
"output": "YES"
},
{
"input": "12 0 20 11 12",
"output": "YES"
},
{
"input": "3 25 30 4 5",
"output": "YES"
},
{
"input": "2 15 18 11 1",
"output": "YES"
},
{
"input": "12 5 48 11 9",
"output": "NO"
},
{
"input": "6 30 30 10 2",
"output": "YES"
},
{
"input": "1 0 11 1 2",
"output": "NO"
},
{
"input": "10 0 1 10 11",
"output": "NO"
},
{
"input": "3 30 45 10 1",
"output": "YES"
},
{
"input": "11 21 56 3 12",
"output": "YES"
},
{
"input": "3 16 16 3 4",
"output": "YES"
},
{
"input": "11 55 55 1 12",
"output": "YES"
},
{
"input": "12 12 12 11 12",
"output": "YES"
},
{
"input": "3 5 1 1 2",
"output": "NO"
},
{
"input": "3 30 45 10 11",
"output": "YES"
},
{
"input": "10 0 1 9 10",
"output": "YES"
},
{
"input": "3 0 5 3 6",
"output": "NO"
},
{
"input": "1 1 49 1 10",
"output": "NO"
},
{
"input": "9 45 15 11 12",
"output": "YES"
},
{
"input": "12 15 59 3 8",
"output": "NO"
},
{
"input": "5 0 5 3 6",
"output": "NO"
},
{
"input": "3 30 45 1 2",
"output": "YES"
},
{
"input": "2 10 11 2 4",
"output": "YES"
},
{
"input": "2 10 10 4 3",
"output": "YES"
},
{
"input": "3 30 59 1 2",
"output": "YES"
},
{
"input": "11 59 59 12 11",
"output": "YES"
},
{
"input": "10 25 50 5 6",
"output": "NO"
},
{
"input": "6 30 30 9 3",
"output": "YES"
},
{
"input": "8 14 40 12 3",
"output": "NO"
},
{
"input": "2 10 10 1 6",
"output": "YES"
}
] | 1,507,914,170
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 133
| 62
| 0
|
h, m, s, a, b = list(map(int, input().split()))
mx = m + s/60
hx = 5 * h + mx/60
ax = 5 * a
bx = 5 * b
t = [s, mx, hx]
t.sort()
if t[0] <= ax <= t[1] and t[0] <= bx <= t[1]:
print('YES')
elif t[1] <= ax <= t[2] and t[1] <= bx <= t[2]:
print('YES')
elif ax <= t[0] and bx <= t[0]:
print('YES')
elif ax >= t[2] and bx >= t[2]:
print('YES')
elif ax >= t[2] and bx <= t[0]:
print('YES')
elif ax <= t[0] and bx >= t[2]:
print('YES')
else:
print('NO')
|
Title: Race Against Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Have you ever tried to explain to the coordinator, why it is eight hours to the contest and not a single problem has been prepared yet? Misha had. And this time he has a really strong excuse: he faced a space-time paradox! Space and time replaced each other.
The entire universe turned into an enormous clock face with three hands — hour, minute, and second. Time froze, and clocks now show the time *h* hours, *m* minutes, *s* seconds.
Last time Misha talked with the coordinator at *t*1 o'clock, so now he stands on the number *t*1 on the clock face. The contest should be ready by *t*2 o'clock. In the terms of paradox it means that Misha has to go to number *t*2 somehow. Note that he doesn't have to move forward only: in these circumstances time has no direction.
Clock hands are very long, and Misha cannot get round them. He also cannot step over as it leads to the collapse of space-time. That is, if hour clock points 12 and Misha stands at 11 then he cannot move to 1 along the top arc. He has to follow all the way round the clock center (of course, if there are no other hands on his way).
Given the hands' positions, *t*1, and *t*2, find if Misha can prepare the contest on time (or should we say on space?). That is, find if he can move from *t*1 to *t*2 by the clock face.
Input Specification:
Five integers *h*, *m*, *s*, *t*1, *t*2 (1<=≤<=*h*<=≤<=12, 0<=≤<=*m*,<=*s*<=≤<=59, 1<=≤<=*t*1,<=*t*2<=≤<=12, *t*1<=≠<=*t*2).
Misha's position and the target time do not coincide with the position of any hand.
Output Specification:
Print "YES" (quotes for clarity), if Misha can prepare the contest on time, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['12 30 45 3 11\n', '12 0 1 12 1\n', '3 47 0 4 9\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n']
Note:
The three examples are shown on the pictures below from left to right. The starting position of Misha is shown with green, the ending position is shown with pink. Note that the positions of the hands on the pictures are not exact, but are close to the exact and the answer is the same.
|
```python
h, m, s, a, b = list(map(int, input().split()))
mx = m + s/60
hx = 5 * h + mx/60
ax = 5 * a
bx = 5 * b
t = [s, mx, hx]
t.sort()
if t[0] <= ax <= t[1] and t[0] <= bx <= t[1]:
print('YES')
elif t[1] <= ax <= t[2] and t[1] <= bx <= t[2]:
print('YES')
elif ax <= t[0] and bx <= t[0]:
print('YES')
elif ax >= t[2] and bx >= t[2]:
print('YES')
elif ax >= t[2] and bx <= t[0]:
print('YES')
elif ax <= t[0] and bx >= t[2]:
print('YES')
else:
print('NO')
```
| 3
|
|
143
|
A
|
Help Vasilisa the Wise 2
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math"
] | null | null |
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
|
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
|
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
|
[
"3 7\n4 6\n5 5\n",
"11 10\n13 8\n5 16\n",
"1 2\n3 4\n5 6\n",
"10 10\n10 10\n10 10\n"
] |
[
"1 2\n3 4\n",
"4 7\n9 1\n",
"-1\n",
"-1\n"
] |
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
| 500
|
[
{
"input": "3 7\n4 6\n5 5",
"output": "1 2\n3 4"
},
{
"input": "11 10\n13 8\n5 16",
"output": "4 7\n9 1"
},
{
"input": "1 2\n3 4\n5 6",
"output": "-1"
},
{
"input": "10 10\n10 10\n10 10",
"output": "-1"
},
{
"input": "5 13\n8 10\n11 7",
"output": "3 2\n5 8"
},
{
"input": "12 17\n10 19\n13 16",
"output": "-1"
},
{
"input": "11 11\n17 5\n12 10",
"output": "9 2\n8 3"
},
{
"input": "12 11\n11 12\n16 7",
"output": "-1"
},
{
"input": "5 9\n7 7\n8 6",
"output": "3 2\n4 5"
},
{
"input": "10 7\n4 13\n11 6",
"output": "-1"
},
{
"input": "18 10\n16 12\n12 16",
"output": "-1"
},
{
"input": "13 6\n10 9\n6 13",
"output": "-1"
},
{
"input": "14 16\n16 14\n18 12",
"output": "-1"
},
{
"input": "16 10\n16 10\n12 14",
"output": "-1"
},
{
"input": "11 9\n12 8\n11 9",
"output": "-1"
},
{
"input": "5 14\n10 9\n10 9",
"output": "-1"
},
{
"input": "2 4\n1 5\n3 3",
"output": "-1"
},
{
"input": "17 16\n14 19\n18 15",
"output": "-1"
},
{
"input": "12 12\n14 10\n16 8",
"output": "9 3\n5 7"
},
{
"input": "15 11\n16 10\n9 17",
"output": "7 8\n9 2"
},
{
"input": "8 10\n9 9\n13 5",
"output": "6 2\n3 7"
},
{
"input": "13 7\n10 10\n5 15",
"output": "4 9\n6 1"
},
{
"input": "14 11\n9 16\n16 9",
"output": "-1"
},
{
"input": "12 8\n14 6\n8 12",
"output": "-1"
},
{
"input": "10 6\n6 10\n4 12",
"output": "-1"
},
{
"input": "10 8\n10 8\n4 14",
"output": "-1"
},
{
"input": "14 13\n9 18\n14 13",
"output": "-1"
},
{
"input": "9 14\n8 15\n8 15",
"output": "-1"
},
{
"input": "3 8\n2 9\n6 5",
"output": "-1"
},
{
"input": "14 17\n18 13\n15 16",
"output": "-1"
},
{
"input": "16 14\n15 15\n17 13",
"output": "9 7\n6 8"
},
{
"input": "14 11\n16 9\n13 12",
"output": "9 5\n7 4"
},
{
"input": "13 10\n11 12\n7 16",
"output": "4 9\n7 3"
},
{
"input": "14 8\n11 11\n13 9",
"output": "8 6\n3 5"
},
{
"input": "12 11\n13 10\n10 13",
"output": "-1"
},
{
"input": "6 5\n2 9\n5 6",
"output": "-1"
},
{
"input": "7 8\n8 7\n12 3",
"output": "-1"
},
{
"input": "7 11\n7 11\n6 12",
"output": "-1"
},
{
"input": "8 5\n11 2\n8 5",
"output": "-1"
},
{
"input": "10 16\n14 12\n14 12",
"output": "-1"
},
{
"input": "7 9\n4 12\n5 11",
"output": "-1"
},
{
"input": "11 13\n19 5\n12 12",
"output": "-1"
},
{
"input": "8 12\n5 15\n11 9",
"output": "2 6\n3 9"
},
{
"input": "16 5\n13 8\n10 11",
"output": "9 7\n4 1"
},
{
"input": "3 14\n8 9\n10 7",
"output": "2 1\n6 8"
},
{
"input": "12 14\n11 15\n9 17",
"output": "3 9\n8 6"
},
{
"input": "13 7\n9 11\n14 6",
"output": "8 5\n1 6"
},
{
"input": "13 8\n15 6\n11 10",
"output": "9 4\n6 2"
},
{
"input": "9 6\n5 10\n3 12",
"output": "1 8\n4 2"
},
{
"input": "12 7\n5 14\n8 11",
"output": "3 9\n2 5"
},
{
"input": "9 12\n3 17\n10 10",
"output": "-1"
},
{
"input": "3 8\n4 6\n5 5",
"output": "-1"
}
] | 1,623,841,806
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 124
| 0
|
r1,r2 = (map(int,input().split()))
c1,c2 = (map(int,input().split()))
d1,d2 = (map(int,input().split()))
for i in range(1,10):
j = r1 - i
k = c1 - i
l = d1 - i
if(i+l == d1 and j+k == d2):
if(i!=(j,k,l) and j!=(k,l) and k!=l):
if(i,j,k,l>=1 and i,j,k,l<=9):
a = [i,j]
b = [k,l]
print(*a)
print(*b)
break
else:
print('-1')
|
Title: Help Vasilisa the Wise 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
Input Specification:
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
Output Specification:
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
Demo Input:
['3 7\n4 6\n5 5\n', '11 10\n13 8\n5 16\n', '1 2\n3 4\n5 6\n', '10 10\n10 10\n10 10\n']
Demo Output:
['1 2\n3 4\n', '4 7\n9 1\n', '-1\n', '-1\n']
Note:
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
|
```python
r1,r2 = (map(int,input().split()))
c1,c2 = (map(int,input().split()))
d1,d2 = (map(int,input().split()))
for i in range(1,10):
j = r1 - i
k = c1 - i
l = d1 - i
if(i+l == d1 and j+k == d2):
if(i!=(j,k,l) and j!=(k,l) and k!=l):
if(i,j,k,l>=1 and i,j,k,l<=9):
a = [i,j]
b = [k,l]
print(*a)
print(*b)
break
else:
print('-1')
```
| 0
|
|
810
|
A
|
Straight <<A>>
|
PROGRAMMING
| 900
|
[
"implementation",
"math"
] | null | null |
Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year. However, she has to get an «A» graduation certificate in order to apply to a prestigious one.
In school, where Noora is studying, teachers are putting down marks to the online class register, which are integers from 1 to *k*. The worst mark is 1, the best is *k*. Mark that is going to the certificate, is calculated as an average of all the marks, rounded to the closest integer. If several answers are possible, rounding up is produced. For example, 7.3 is rounded to 7, but 7.5 and 7.8784 — to 8.
For instance, if Noora has marks [8,<=9], then the mark to the certificate is 9, because the average is equal to 8.5 and rounded to 9, but if the marks are [8,<=8,<=9], Noora will have graduation certificate with 8.
To graduate with «A» certificate, Noora has to have mark *k*.
Noora got *n* marks in register this year. However, she is afraid that her marks are not enough to get final mark *k*. Noora decided to ask for help in the internet, where hacker Leha immediately responded to her request. He is ready to hack class register for Noora and to add Noora any number of additional marks from 1 to *k*. At the same time, Leha want his hack be unseen to everyone, so he decided to add as less as possible additional marks. Please help Leha to calculate the minimal number of marks he has to add, so that final Noora's mark will become equal to *k*.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100) denoting the number of marks, received by Noora and the value of highest possible mark.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*k*) denoting marks received by Noora before Leha's hack.
|
Print a single integer — minimal number of additional marks, that Leha has to add in order to change Noora's final mark to *k*.
|
[
"2 10\n8 9\n",
"3 5\n4 4 4\n"
] |
[
"4",
"3"
] |
Consider the first example testcase.
Maximal mark is 10, Noora received two marks — 8 and 9, so current final mark is 9. To fix it, Leha can add marks [10, 10, 10, 10] (4 marks in total) to the registry, achieving Noora having average mark equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1b961585522f76271546da990a6228e7c666277f.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Consequently, new final mark is 10. Less number of marks won't fix the situation.
In the second example Leha can add [5, 5, 5] to the registry, so that making average mark equal to 4.5, which is enough to have 5 in the certificate.
| 500
|
[
{
"input": "2 10\n8 9",
"output": "4"
},
{
"input": "3 5\n4 4 4",
"output": "3"
},
{
"input": "3 10\n10 8 9",
"output": "3"
},
{
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"output": "2"
},
{
"input": "5 10\n5 10 10 9 10",
"output": "7"
},
{
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{
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{
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{
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"output": "6650"
},
{
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"output": "11786"
},
{
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},
{
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"output": "0"
},
{
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"output": "1807"
},
{
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},
{
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},
{
"input": "100 27\n16 20 21 10 16 17 18 25 19 18 20 12 11 21 21 23 20 26 20 21 27 16 25 18 25 21 27 12 20 27 18 17 27 13 21 26 12 22 15 21 25 21 18 27 24 15 16 18 23 21 24 27 19 17 24 14 21 16 24 26 13 14 25 18 27 26 22 16 27 27 17 25 17 12 22 10 19 27 19 20 23 22 25 23 17 25 14 20 22 10 22 27 21 20 15 26 24 27 12 16",
"output": "1262"
},
{
"input": "100 29\n20 18 23 24 14 14 16 23 22 17 18 22 21 21 19 19 14 11 18 19 16 22 25 20 14 13 21 24 18 16 18 29 17 25 12 10 18 28 11 16 17 14 15 20 17 20 18 22 10 16 16 20 18 19 29 18 25 27 17 19 24 15 24 25 16 23 19 16 16 20 19 15 12 21 20 13 21 15 15 23 16 23 17 13 17 21 13 18 17 18 18 20 16 12 19 15 27 14 11 18",
"output": "2024"
},
{
"input": "100 30\n16 10 20 11 14 27 15 17 22 26 24 17 15 18 19 22 22 15 21 22 14 21 22 22 21 22 15 17 17 22 18 19 26 18 22 20 22 25 18 18 17 23 18 18 20 13 19 30 17 24 22 19 29 20 20 21 17 18 26 25 22 19 15 18 18 20 19 19 18 18 24 16 19 17 12 21 20 16 23 21 16 17 26 23 25 28 22 20 9 21 17 24 15 19 17 21 29 13 18 15",
"output": "1984"
},
{
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{
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{
"input": "100 100\n30 35 23 43 28 49 31 32 30 44 32 37 33 34 38 28 43 32 33 32 50 32 41 38 33 20 40 36 29 21 42 25 23 34 43 32 37 31 30 27 36 32 45 37 33 29 38 34 35 33 28 19 37 33 28 41 31 29 41 27 32 39 30 34 37 40 33 38 35 32 32 34 35 34 28 39 28 34 40 45 31 25 42 28 29 31 33 21 36 33 34 37 40 42 39 30 36 34 34 40",
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{
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"output": "19700"
},
{
"input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
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"output": "19696"
},
{
"input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "0"
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{
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{
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{
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"output": "16"
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"output": "77"
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{
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{
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{
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{
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"output": "0"
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{
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{
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{
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"output": "0"
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{
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{
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{
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"output": "0"
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{
"input": "1 100\n100",
"output": "0"
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{
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"output": "10"
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{
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"output": "35"
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{
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"output": "33"
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{
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{
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"output": "11"
},
{
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"output": "66"
},
{
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{
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"output": "8"
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{
"input": "1 5\n1",
"output": "7"
},
{
"input": "10 10\n10 10 10 10 10 10 10 10 10 10",
"output": "0"
},
{
"input": "2 3\n2 3",
"output": "0"
},
{
"input": "1 9\n9",
"output": "0"
},
{
"input": "74 2\n2 2 2 2 1 2 2 1 1 1 2 2 1 2 2 2 2 1 2 1 1 1 2 1 1 2 2 1 2 1 1 2 1 1 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 1 1 2 1 1 1 1 1 1 2 2 2 1 1 1 1 1 2 2 2 2 2 2 1 2",
"output": "0"
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{
"input": "5 5\n5 5 5 5 4",
"output": "0"
}
] | 1,541,995,628
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 106
| 124
| 0
|
n, k = [int(c) for c in input().split()]
_sum = sum(int(c) for c in input().split())
ans = 0
while int(_sum / n + 0.5) < k:
_sum += k
n += 1
ans += 1
print(ans)
|
Title: Straight <<A>>
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year. However, she has to get an «A» graduation certificate in order to apply to a prestigious one.
In school, where Noora is studying, teachers are putting down marks to the online class register, which are integers from 1 to *k*. The worst mark is 1, the best is *k*. Mark that is going to the certificate, is calculated as an average of all the marks, rounded to the closest integer. If several answers are possible, rounding up is produced. For example, 7.3 is rounded to 7, but 7.5 and 7.8784 — to 8.
For instance, if Noora has marks [8,<=9], then the mark to the certificate is 9, because the average is equal to 8.5 and rounded to 9, but if the marks are [8,<=8,<=9], Noora will have graduation certificate with 8.
To graduate with «A» certificate, Noora has to have mark *k*.
Noora got *n* marks in register this year. However, she is afraid that her marks are not enough to get final mark *k*. Noora decided to ask for help in the internet, where hacker Leha immediately responded to her request. He is ready to hack class register for Noora and to add Noora any number of additional marks from 1 to *k*. At the same time, Leha want his hack be unseen to everyone, so he decided to add as less as possible additional marks. Please help Leha to calculate the minimal number of marks he has to add, so that final Noora's mark will become equal to *k*.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100) denoting the number of marks, received by Noora and the value of highest possible mark.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*k*) denoting marks received by Noora before Leha's hack.
Output Specification:
Print a single integer — minimal number of additional marks, that Leha has to add in order to change Noora's final mark to *k*.
Demo Input:
['2 10\n8 9\n', '3 5\n4 4 4\n']
Demo Output:
['4', '3']
Note:
Consider the first example testcase.
Maximal mark is 10, Noora received two marks — 8 and 9, so current final mark is 9. To fix it, Leha can add marks [10, 10, 10, 10] (4 marks in total) to the registry, achieving Noora having average mark equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1b961585522f76271546da990a6228e7c666277f.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Consequently, new final mark is 10. Less number of marks won't fix the situation.
In the second example Leha can add [5, 5, 5] to the registry, so that making average mark equal to 4.5, which is enough to have 5 in the certificate.
|
```python
n, k = [int(c) for c in input().split()]
_sum = sum(int(c) for c in input().split())
ans = 0
while int(_sum / n + 0.5) < k:
_sum += k
n += 1
ans += 1
print(ans)
```
| 3
|
|
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\nckcccbj\nccbcbcc\ncccbccc\nccbcbcc\ncbcccbc\nbccccdt",
"output": "NO"
},
{
"input": "13\nsooooooooooos\nosoooooooooso\noosooooooosoo\nooosooooosooo\noooosooosoooo\nooooososooooo\noooooosoooooo\nooooososooooo\noooosooosoooo\nooosooooosooo\noosooooooosoo\nosoooooooooso\nsooooooooooos",
"output": "YES"
},
{
"input": "3\naaa\naaa\naaa",
"output": "NO"
},
{
"input": "3\naca\noec\nzba",
"output": "NO"
},
{
"input": "15\nrxeeeeeeeeeeeer\nereeeeeeeeeeere\needeeeeeeeeeoee\neeereeeeeeeewee\neeeereeeeebeeee\nqeeeereeejedyee\neeeeeerereeeeee\neeeeeeereeeeeee\neeeeeerereeeeze\neeeeereeereeeee\neeeereeeeegeeee\neeereeeeeeereee\neereeeeeeqeeved\ncreeeeeeceeeere\nreeerneeeeeeeer",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxxxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxoxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxxxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxoox\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxaxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\noxoxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "3\nxxx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxx\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxxx",
"output": "NO"
},
{
"input": "3\nxax\nxxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naaa\nxax",
"output": "NO"
},
{
"input": "3\naax\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxaa\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\naax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaa",
"output": "NO"
},
{
"input": "3\nxfx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\nafa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaf",
"output": "NO"
},
{
"input": "3\nxox\nxxx\nxxx",
"output": "NO"
},
{
"input": "3\naxa\naax\nxxa",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxox",
"output": "NO"
},
{
"input": "3\nxox\nooo\nxox",
"output": "NO"
},
{
"input": "3\naaa\naab\nbbb",
"output": "NO"
},
{
"input": "3\nxxx\nsxs\nxsx",
"output": "NO"
},
{
"input": "5\nabbba\nbabab\nbbbbb\nbaaab\nabbba",
"output": "NO"
},
{
"input": "5\nabaaa\nbbbbb\nbbabb\nbabab\nabbba",
"output": "NO"
},
{
"input": "5\nxoxox\noxoxo\nooxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxxx",
"output": "NO"
},
{
"input": "5\nxoooo\noxooo\nooxoo\noooxo\noooox",
"output": "NO"
},
{
"input": "5\nxoooo\noxoxx\nooxoo\noxoxo\noxoox",
"output": "NO"
},
{
"input": "3\naaa\nbab\naba",
"output": "NO"
}
] | 1,632,683,755
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 61
| 6,758,400
|
n = int(input())
s = ''
for i in range(n):
s+=input()
print("Yes") if (s == s[::-1] and s.count(s[0]) == 2*n-1) 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 = ''
for i in range(n):
s+=input()
print("Yes") if (s == s[::-1] and s.count(s[0]) == 2*n-1) else print("NO")
```
| 0
|
|
21
|
C
|
Stripe 2
|
PROGRAMMING
| 2,000
|
[
"binary search",
"dp",
"sortings"
] |
C. Stripe 2
|
1
|
64
|
Once Bob took a paper stripe of n squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into three pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?
|
The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains n space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.
|
Output the amount of ways to cut the stripe into three non-empty pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.
|
[
"4\n1 2 3 3\n",
"5\n1 2 3 4 5\n"
] |
[
"1\n",
"0\n"
] |
none
| 1,500
|
[
{
"input": "1\n-3",
"output": "0"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "3\n0 0 0",
"output": "1"
},
{
"input": "4\n-2 3 3 2",
"output": "0"
},
{
"input": "5\n-6 3 -1 2 -7",
"output": "0"
},
{
"input": "6\n2 3 -3 0 -3 1",
"output": "0"
},
{
"input": "7\n-1 1 -3 4 3 0 2",
"output": "0"
},
{
"input": "8\n2 0 0 2 -1 3 4 5",
"output": "0"
},
{
"input": "9\n-5 -2 1 1 5 0 -4 4 0",
"output": "3"
},
{
"input": "10\n-1 5 2 3 1 5 0 2 2 5",
"output": "0"
},
{
"input": "100\n3 0 -5 2 -3 -1 -1 0 -2 -5 -4 2 1 2 -2 -1 -1 -4 3 -1 -3 -1 5 0 -4 -4 -1 0 -2 -2 0 1 -1 -2 -1 -5 -4 -2 3 1 -3 0 -1 1 0 -1 2 0 -2 -1 -3 1 -2 2 3 2 -3 -5 2 2 -2 -2 1 2 -2 -1 3 0 -4 7 -2 2 1 4 -9 -1 -2 -1 0 -1 0 -2 -2 -1 1 1 -4 2 -3 -3 7 1 1 -3 -7 0 -2 0 5 -2",
"output": "5"
}
] | 1,692,524,230
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 31
| 0
|
n = int(input())
pref = list(map(int, input().split()))
s = sum(pref)
ans = 0
sett = set()
if s % 3 != 0 or n < 3:
print(0)
else:
for i in range(1, n):
pref[i] += pref[i-1]
for i in range(0, n):
if pref[i] % s/3 == 0 and pref[i] not in sett:
sett.add(pref[i])
print(len(sett)-2)
|
Title: Stripe 2
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Once Bob took a paper stripe of n squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into three pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains n space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.
Output Specification:
Output the amount of ways to cut the stripe into three non-empty pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.
Demo Input:
['4\n1 2 3 3\n', '5\n1 2 3 4 5\n']
Demo Output:
['1\n', '0\n']
Note:
none
|
```python
n = int(input())
pref = list(map(int, input().split()))
s = sum(pref)
ans = 0
sett = set()
if s % 3 != 0 or n < 3:
print(0)
else:
for i in range(1, n):
pref[i] += pref[i-1]
for i in range(0, n):
if pref[i] % s/3 == 0 and pref[i] not in sett:
sett.add(pref[i])
print(len(sett)-2)
```
| -1
|
11
|
A
|
Increasing Sequence
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"implementation",
"math"
] |
A. Increasing Sequence
|
1
|
64
|
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
|
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
|
Output the minimal number of moves needed to make the sequence increasing.
|
[
"4 2\n1 3 3 2\n"
] |
[
"3\n"
] |
none
| 0
|
[
{
"input": "4 2\n1 3 3 2",
"output": "3"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 1\n2 5",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 7\n10 20",
"output": "0"
},
{
"input": "2 7\n1 1",
"output": "1"
},
{
"input": "3 3\n18 1 9",
"output": "10"
},
{
"input": "3 3\n15 17 9",
"output": "3"
},
{
"input": "3 3\n10 9 12",
"output": "2"
},
{
"input": "10 3\n2 1 17 10 5 16 8 4 15 17",
"output": "31"
},
{
"input": "10 3\n6 11 4 12 22 15 23 26 24 26",
"output": "13"
},
{
"input": "10 3\n10 24 13 15 18 14 15 26 33 35",
"output": "29"
},
{
"input": "100 3\n529 178 280 403 326 531 671 427 188 866 669 646 421 804 494 609 53 1012 211 243 887 833 900 543 226 42 859 718 454 372 971 692 846 770 511 395 499 479 641 756 115 269 206 45 1039 727 400 779 859 614 146 214 196 919 702 959 380 830 535 878 859 784 316 305 782 924 536 243 236 978 564 150 291 877 808 983 537 839 490 120 168 838 267 650 900 170 211 504 326 771 895 984 994 483 776 100 471 1078 317 580",
"output": "15717"
},
{
"input": "100 3\n329 226 331 909 962 112 837 1005 194 818 506 416 125 648 367 459 400 582 989 547 329 438 234 121 272 226 821 376 834 427 718 164 834 113 654 177 737 212 169 696 744 180 89 944 233 147 667 990 809 1072 1085 1093 814 265 1067 312 833 572 303 901 1032 504 185 817 389 158 613 723 239 269 911 352 769 404 225 822 897 606 947 323 913 804 923 1084 552 901 486 249 209 898 847 610 728 1122 986 669 1116 1076 367 327",
"output": "16133"
}
] | 1,669,728,957
| 2,147,483,647
|
PyPy 3-64
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
d=gets.split[1].to_i;s=gets.split;x=s.shift.to_i;a=0;s.each{|b|b=b.to_i;a+=z=b>x ?0:(x-b)/d+1;x=b+z*d};puts a
|
Title: Increasing Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
Input Specification:
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Output the minimal number of moves needed to make the sequence increasing.
Demo Input:
['4 2\n1 3 3 2\n']
Demo Output:
['3\n']
Note:
none
|
```python
d=gets.split[1].to_i;s=gets.split;x=s.shift.to_i;a=0;s.each{|b|b=b.to_i;a+=z=b>x ?0:(x-b)/d+1;x=b+z*d};puts a
```
| -1
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,623,379,423
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 81
| 218
| 0
|
forces = int(input())
sx, sy, sz = (0, 0 ,0)
for i in range(forces):
force = input().split(' ')
sx += int(force[0])
sy += int(force[1])
sz += int(force[2])
if sx == 0 and sy == 0 and sz == 0:
print('YES')
else:
print('NO')
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
forces = int(input())
sx, sy, sz = (0, 0 ,0)
for i in range(forces):
force = input().split(' ')
sx += int(force[0])
sy += int(force[1])
sz += int(force[2])
if sx == 0 and sy == 0 and sz == 0:
print('YES')
else:
print('NO')
```
| 3.9455
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,593,842,402
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 32
| 312
| 0
|
n = int(input())
lis = [int(_) for _ in input().split()]
dic = {0:[], 1:[]}
for i,x in enumerate(lis):
dic[x&1].append(i+1)
if len(dic[0]) == 1:
print(dic[0][0])
else:
print(dic[1][0])
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n = int(input())
lis = [int(_) for _ in input().split()]
dic = {0:[], 1:[]}
for i,x in enumerate(lis):
dic[x&1].append(i+1)
if len(dic[0]) == 1:
print(dic[0][0])
else:
print(dic[1][0])
```
| 3.922
|
858
|
C
|
Did you mean...
|
PROGRAMMING
| 1,500
|
[
"dp",
"greedy",
"implementation"
] | null | null |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
- the following words have typos: "hellno", "hackcerrs" and "backtothefutttture"; - the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
|
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
|
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
|
[
"hellno\n",
"abacaba\n",
"asdfasdf\n"
] |
[
"hell no \n",
"abacaba \n",
"asd fasd f \n"
] |
none
| 1,500
|
[
{
"input": "hellno",
"output": "hell no "
},
{
"input": "abacaba",
"output": "abacaba "
},
{
"input": "asdfasdf",
"output": "asd fasd f "
},
{
"input": "ooo",
"output": "ooo "
},
{
"input": "moyaoborona",
"output": "moyaoborona "
},
{
"input": "jxegxxx",
"output": "jxegx xx "
},
{
"input": "orfyaenanabckumulsboloyhljhacdgcmnooxvxrtuhcslxgslfpnfnyejbxqisxjyoyvcvuddboxkqgbogkfz",
"output": "orf yaenanabc kumuls boloyh lj hacd gc mnooxv xr tuhc sl xg sl fp nf nyejb xqisx jyoyv cvudd boxk qg bogk fz "
},
{
"input": "zxdgmhsjotvajkwshjpvzcuwehpeyfhakhtlvuoftkgdmvpafmxcliqvrztloocziqdkexhzcbdgxaoyvte",
"output": "zx dg mh sjotvajk ws hj pv zcuwehpeyf hakh tl vuoft kg dm vpafm xc liqv rz tloocziqd kexh zc bd gxaoyv te "
},
{
"input": "niblehmwtycadhbfuginpyafszjbucaszihijndzjtuyuaxkrovotshtsajmdcflnfdmahzbvpymiczqqleedpofcnvhieknlz",
"output": "niblehm wt ycadh bfuginp yafs zj bucaszihijn dz jtuyuaxk rovots ht sajm dc fl nf dmahz bv py micz qq leedpofc nv hiekn lz "
},
{
"input": "pqvtgtctpkgjgxnposjqedofficoyznxlerxyqypyzpoehejtjvyafjxjppywwgeakf",
"output": "pq vt gt ct pk gj gx nposj qedofficoyz nx lerx yq yp yz poehejt jv yafj xj pp yw wgeakf "
},
{
"input": "mvjajoyeg",
"output": "mv jajoyeg "
},
{
"input": "dipxocwjosvdaillxolmthjhzhsxskzqslebpixpuhpgeesrkedhohisdsjsrkiktbjzlhectrfcathvewzficirqbdvzq",
"output": "dipxocw josv daill xolm th jh zh sx sk zq slebpixpuhp geesr kedhohisd sj sr kikt bj zl hect rf cath vewz ficirq bd vz q "
},
{
"input": "ibbtvelwjirxqermucqrgmoauonisgmarjxxybllktccdykvef",
"output": "ibb tvelw jirx qermucq rg moauonisg marj xx yb ll kt cc dy kvef "
},
{
"input": "jxevkmrwlomaaahaubvjzqtyfqhqbhpqhomxqpiuersltohinvfyeykmlooujymldjqhgqjkvqknlyj",
"output": "jxevk mr wlomaaahaubv jz qt yf qh qb hp qhomx qpiuers ltohinv fyeyk mlooujy ml dj qh gq jk vq kn ly j "
},
{
"input": "hzxkuwqxonsulnndlhygvmallghjerwp",
"output": "hz xkuwq xonsuln nd lh yg vmall gh jerw p "
},
{
"input": "jbvcsjdyzlzmxwcvmixunfzxidzvwzaqqdhguvelwbdosbd",
"output": "jb vc sj dy zl zm xw cv mixunf zxidz vw zaqq dh guvelw bdosb d "
},
{
"input": "uyrsxaqmtibbxpfabprvnvbinjoxubupvfyjlqnfrfdeptipketwghr",
"output": "uyr sxaqm tibb xp fabp rv nv binjoxubupv fy jl qn fr fdeptipketw gh r "
},
{
"input": "xfcftysljytybkkzkpqdzralahgvbkxdtheqrhfxpecdjqofnyiahggnkiuusalu",
"output": "xf cf ty sl jy ty bk kz kp qd zralahg vb kx dt heqr hf xpecd jqofn yiahg gn kiuusalu "
},
{
"input": "a",
"output": "a "
},
{
"input": "b",
"output": "b "
},
{
"input": "aa",
"output": "aa "
},
{
"input": "ab",
"output": "ab "
},
{
"input": "ba",
"output": "ba "
},
{
"input": "bb",
"output": "bb "
},
{
"input": "aaa",
"output": "aaa "
},
{
"input": "aab",
"output": "aab "
},
{
"input": "aba",
"output": "aba "
},
{
"input": "abb",
"output": "abb "
},
{
"input": "baa",
"output": "baa "
},
{
"input": "bab",
"output": "bab "
},
{
"input": "bba",
"output": "bba "
},
{
"input": "bbb",
"output": "bbb "
},
{
"input": "bbc",
"output": "bb c "
},
{
"input": "bcb",
"output": "bc b "
},
{
"input": "cbb",
"output": "cb b "
},
{
"input": "bababcdfabbcabcdfacbbabcdfacacabcdfacbcabcdfaccbabcdfacaaabcdfabacabcdfabcbabcdfacbaabcdfabaaabcdfabbaabcdfacababcdfabbbabcdfabcaabcdfaaababcdfabccabcdfacccabcdfaacbabcdfaabaabcdfaabcabcdfaaacabcdfaccaabcdfaabbabcdfaaaaabcdfaacaabcdfaacc",
"output": "bababc dfabb cabc dfacb babc dfacacabc dfacb cabc dfacc babc dfacaaabc dfabacabc dfabc babc dfacbaabc dfabaaabc dfabbaabc dfacababc dfabbbabc dfabcaabc dfaaababc dfabc cabc dfacccabc dfaacbabc dfaabaabc dfaabcabc dfaaacabc dfaccaabc dfaabbabc dfaaaaabc dfaacaabc dfaacc "
},
{
"input": "bddabcdfaccdabcdfadddabcdfabbdabcdfacddabcdfacdbabcdfacbbabcdfacbcabcdfacbdabcdfadbbabcdfabdbabcdfabdcabcdfabbcabcdfabccabcdfabbbabcdfaddcabcdfaccbabcdfadbdabcdfacccabcdfadcdabcdfadcbabcdfabcbabcdfadbcabcdfacdcabcdfabcdabcdfadccabcdfaddb",
"output": "bd dabc dfacc dabc dfadddabc dfabb dabc dfacd dabc dfacd babc dfacb babc dfacb cabc dfacb dabc dfadb babc dfabd babc dfabd cabc dfabb cabc dfabc cabc dfabbbabc dfadd cabc dfacc babc dfadb dabc dfacccabc dfadc dabc dfadc babc dfabc babc dfadb cabc dfacd cabc dfabc dabc dfadc cabc dfadd b "
},
{
"input": "helllllooooo",
"output": "helllllooooo "
},
{
"input": "bbbzxxx",
"output": "bbb zx xx "
},
{
"input": "ffff",
"output": "ffff "
},
{
"input": "cdddddddddddddddddd",
"output": "cd ddddddddddddddddd "
},
{
"input": "bbbc",
"output": "bbb c "
},
{
"input": "lll",
"output": "lll "
},
{
"input": "bbbbb",
"output": "bbbbb "
},
{
"input": "llll",
"output": "llll "
},
{
"input": "bbbbbbccc",
"output": "bbbbbb ccc "
},
{
"input": "lllllb",
"output": "lllll b "
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz "
},
{
"input": "lllll",
"output": "lllll "
},
{
"input": "bbbbbbbbbc",
"output": "bbbbbbbbb c "
},
{
"input": "helllllno",
"output": "helllll no "
},
{
"input": "nnnnnnnnnnnn",
"output": "nnnnnnnnnnnn "
},
{
"input": "bbbbbccc",
"output": "bbbbb ccc "
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzz "
},
{
"input": "nnnnnnnnnnnnnnnnnn",
"output": "nnnnnnnnnnnnnnnnnn "
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzz",
"output": "zzzzzzzzzzzzzzzzzzzzzzz "
},
{
"input": "hhhh",
"output": "hhhh "
},
{
"input": "nnnnnnnnnnnnnnnnnnnnnnnnn",
"output": "nnnnnnnnnnnnnnnnnnnnnnnnn "
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz "
},
{
"input": "dddd",
"output": "dddd "
},
{
"input": "heffffffgggggghhhhhh",
"output": "heffffff gggggg hhhhhh "
},
{
"input": "bcddd",
"output": "bc ddd "
},
{
"input": "x",
"output": "x "
},
{
"input": "nnn",
"output": "nnn "
},
{
"input": "xxxxxxxx",
"output": "xxxxxxxx "
},
{
"input": "cclcc",
"output": "cc lc c "
},
{
"input": "tttttttttttttt",
"output": "tttttttttttttt "
},
{
"input": "xxxxxxx",
"output": "xxxxxxx "
},
{
"input": "ccccb",
"output": "cccc b "
},
{
"input": "bcecccc",
"output": "bcecccc "
},
{
"input": "jjja",
"output": "jjja "
},
{
"input": "zzz",
"output": "zzz "
},
{
"input": "xxxxxxxxxzzzzzzzzzzzz",
"output": "xxxxxxxxx zzzzzzzzzzzz "
},
{
"input": "alllewww",
"output": "alllewww "
},
{
"input": "bbbbbbbbb",
"output": "bbbbbbbbb "
},
{
"input": "jjj",
"output": "jjj "
},
{
"input": "bbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbb "
},
{
"input": "kkkkkkkkkklllllllllllllll",
"output": "kkkkkkkkkk lllllllllllllll "
},
{
"input": "helllllllllllo",
"output": "helllllllllllo "
},
{
"input": "ttttsttttt",
"output": "tttt st tttt "
},
{
"input": "tttttttsssssss",
"output": "ttttttt sssssss "
},
{
"input": "assstttttatsfatsfdjfdhtsjdsaatttssssststsss",
"output": "asss tttttats fats fd jf dh ts jd saattt sssss ts ts ss "
},
{
"input": "xxxxx",
"output": "xxxxx "
},
{
"input": "bbbhbbb",
"output": "bbb hb bb "
},
{
"input": "bbbbbbbb",
"output": "bbbbbbbb "
}
] | 1,505,657,108
| 3,608
|
Python 3
|
RUNTIME_ERROR
|
PRETESTS
| 5
| 61
| 0
|
s = input()
d = ['a', 'e', 'i', 'o', 'u']
st = []
for i in range(len(s)):
if s[i] in d:
st = []
print(s[i], end="")
continue
st.append(s[i])
if st == [s[i], s[i], s[i]]:
st = st[1:]
print(s[i], end='')
elif len(st) == 3:
st = s[i]
print(" " + s[i], end="")
else:
print(s[i], end='')
|
Title: Did you mean...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
- the following words have typos: "hellno", "hackcerrs" and "backtothefutttture"; - the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input Specification:
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output Specification:
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Demo Input:
['hellno\n', 'abacaba\n', 'asdfasdf\n']
Demo Output:
['hell no \n', 'abacaba \n', 'asd fasd f \n']
Note:
none
|
```python
s = input()
d = ['a', 'e', 'i', 'o', 'u']
st = []
for i in range(len(s)):
if s[i] in d:
st = []
print(s[i], end="")
continue
st.append(s[i])
if st == [s[i], s[i], s[i]]:
st = st[1:]
print(s[i], end='')
elif len(st) == 3:
st = s[i]
print(" " + s[i], end="")
else:
print(s[i], end='')
```
| -1
|
|
961
|
B
|
Lecture Sleep
|
PROGRAMMING
| 1,200
|
[
"data structures",
"dp",
"implementation",
"two pointers"
] | null | null |
Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
|
The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture.
|
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
|
[
"6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n"
] |
[
"16\n"
] |
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
| 0
|
[
{
"input": "6 3\n1 3 5 2 5 4\n1 1 0 1 0 0",
"output": "16"
},
{
"input": "5 3\n1 9999 10000 10000 10000\n0 0 0 0 0",
"output": "30000"
},
{
"input": "3 3\n10 10 10\n1 1 0",
"output": "30"
},
{
"input": "1 1\n423\n0",
"output": "423"
},
{
"input": "6 6\n1 3 5 2 5 4\n1 1 0 1 0 0",
"output": "20"
},
{
"input": "5 2\n1 2 3 4 20\n0 0 0 1 0",
"output": "24"
},
{
"input": "3 1\n1 2 3\n0 0 1",
"output": "5"
},
{
"input": "4 2\n4 5 6 8\n1 0 1 0",
"output": "18"
},
{
"input": "6 3\n1 3 5 2 1 15\n1 1 0 1 0 0",
"output": "22"
},
{
"input": "5 5\n1 2 3 4 5\n1 1 1 0 1",
"output": "15"
},
{
"input": "3 3\n3 3 3\n1 0 1",
"output": "9"
},
{
"input": "5 5\n500 44 3 4 50\n1 0 0 0 0",
"output": "601"
},
{
"input": "2 2\n3 2\n1 0",
"output": "5"
},
{
"input": "7 6\n4 9 1 7 1 8 4\n0 0 0 1 0 1 0",
"output": "30"
},
{
"input": "4 3\n6 5 9 6\n1 1 0 1",
"output": "26"
},
{
"input": "2 1\n3 2\n0 0",
"output": "3"
},
{
"input": "1 1\n10\n0",
"output": "10"
},
{
"input": "2 1\n3 2\n1 0",
"output": "5"
},
{
"input": "4 2\n3 6 7 2\n0 0 1 1",
"output": "18"
},
{
"input": "10 5\n3 5 9 2 5 9 3 8 8 1\n0 1 1 1 0 1 0 0 0 0",
"output": "49"
},
{
"input": "10 4\n9 5 6 4 3 9 5 1 10 7\n0 0 0 0 0 0 1 0 0 1",
"output": "36"
},
{
"input": "9 8\n3 3 7 7 1 9 10 7 1\n1 1 1 1 1 1 1 1 1",
"output": "48"
},
{
"input": "2 1\n3 4\n0 0",
"output": "4"
},
{
"input": "2 1\n3 2\n0 1",
"output": "5"
},
{
"input": "10 1\n6 6 8 7 6 6 3 2 5 6\n0 0 1 0 0 1 0 1 1 1",
"output": "34"
},
{
"input": "3 2\n10 10 6\n0 0 0",
"output": "20"
},
{
"input": "6 3\n1 3 5 2 5 4\n1 1 1 1 1 1",
"output": "20"
},
{
"input": "10 5\n1 1 1 1 1 1 1 1 10000 1\n1 1 1 1 1 1 1 1 0 1",
"output": "10009"
}
] | 1,648,039,540
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 186
| 6,144,000
|
def solve():
n, k = list(map(int, input().split(" ")))
a = list(map(int, input().split(" ")))
b = list(map(int, input().split(" ")))
prefix = [0]*(n+1)
res = float('-inf')
for i in range(n):
prefix[i+1] += prefix[i]
if(b[i] == 1):
prefix[i+1] += a[i]
sum = 0
i = 0
j = 0
while j<n:
sum += a[j]
res = max(prefix[i] + sum + (prefix[n] - prefix[j]), res)
if j-i+1 == k:
sum -= a[i]
i+=1
j+=1
print(res)
solve()
|
Title: Lecture Sleep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input Specification:
The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture.
Output Specification:
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Demo Input:
['6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n']
Demo Output:
['16\n']
Note:
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
|
```python
def solve():
n, k = list(map(int, input().split(" ")))
a = list(map(int, input().split(" ")))
b = list(map(int, input().split(" ")))
prefix = [0]*(n+1)
res = float('-inf')
for i in range(n):
prefix[i+1] += prefix[i]
if(b[i] == 1):
prefix[i+1] += a[i]
sum = 0
i = 0
j = 0
while j<n:
sum += a[j]
res = max(prefix[i] + sum + (prefix[n] - prefix[j]), res)
if j-i+1 == k:
sum -= a[i]
i+=1
j+=1
print(res)
solve()
```
| 0
|
|
137
|
A
|
Postcards and photos
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Polycarpus has postcards and photos hung in a row on the wall. He decided to put them away to the closet and hang on the wall a famous painter's picture. Polycarpus does it like that: he goes from the left to the right and removes the objects consecutively. As Polycarpus doesn't want any mix-ups to happen, he will not carry in his hands objects of two different types. In other words, Polycarpus can't carry both postcards and photos simultaneously. Sometimes he goes to the closet and puts the objects there, thus leaving his hands free. Polycarpus must put all the postcards and photos to the closet. He cannot skip objects. What minimum number of times he should visit the closet if he cannot carry more than 5 items?
|
The only line of the input data contains a non-empty string consisting of letters "С" and "P" whose length does not exceed 100 characters. If the *i*-th character in the string is the letter "С", that means that the *i*-th object (the numbering goes from the left to the right) on Polycarpus' wall is a postcard. And if the *i*-th character is the letter "P", than the *i*-th object on the wall is a photo.
|
Print the only number — the minimum number of times Polycarpus has to visit the closet.
|
[
"CPCPCPC\n",
"CCCCCCPPPPPP\n",
"CCCCCCPPCPPPPPPPPPP\n",
"CCCCCCCCCC\n"
] |
[
"7\n",
"4\n",
"6\n",
"2\n"
] |
In the first sample Polycarpus needs to take one item to the closet 7 times.
In the second sample Polycarpus can first take 3 postcards to the closet; then 3 more. He can take the 6 photos that are left in the similar way, going to the closet twice.
In the third sample Polycarpus can visit the closet twice, both times carrying 3 postcards. Then he can take there 2 photos at once, then one postcard and finally, he can carry the last 10 photos if he visits the closet twice.
In the fourth sample Polycarpus can visit the closet twice and take there all 10 postcards (5 items during each go).
| 500
|
[
{
"input": "CPCPCPC",
"output": "7"
},
{
"input": "CCCCCCPPPPPP",
"output": "4"
},
{
"input": "CCCCCCPPCPPPPPPPPPP",
"output": "6"
},
{
"input": "CCCCCCCCCC",
"output": "2"
},
{
"input": "CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC",
"output": "20"
},
{
"input": "CPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCP",
"output": "100"
},
{
"input": "CCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPP",
"output": "28"
},
{
"input": "P",
"output": "1"
},
{
"input": "C",
"output": "1"
},
{
"input": "PC",
"output": "2"
},
{
"input": "PPPPP",
"output": "1"
},
{
"input": "PPPP",
"output": "1"
},
{
"input": "CCCCCCCCCC",
"output": "2"
},
{
"input": "CP",
"output": "2"
},
{
"input": "CPCCPCPPPC",
"output": "7"
},
{
"input": "PPCPCCPCPPCCPPPPPPCP",
"output": "12"
},
{
"input": "PCPCCPCPPCCPCPCCPPPPPCPCPCPCCC",
"output": "20"
},
{
"input": "CCPPPPPCPCCPPPCCPPCPCCPCPPCPPCCCPPCPPPCC",
"output": "21"
},
{
"input": "CPPCCCCCCPCCCCPCCPCPPPCPCCCCCCCPCCPPCCCPCCCCCPPCCC",
"output": "23"
},
{
"input": "PPCCCCPPCCPPPCCCCPPPPPCPPPCPPPCCCPCCCPCPPPCPCCCPCCPPCCPPPPPC",
"output": "26"
},
{
"input": "PPCPPCCCCCPCCCPCCPCCCCPPPCCCCPCPCCPCPCPCPPPPCCPPPPPPPCPCPPPCPCPCPCPPPC",
"output": "39"
},
{
"input": "CCPCPPPPCPPPPCCCCPCCPCPCCPPCPCCCPPCCCCPCCCPCPCCPPPCPPPCPCPPPPPCPCCPCCPPCCCPCPPPC",
"output": "43"
},
{
"input": "CCPPCPCPCPPCCCPCPPPCCCCCPCPPCCCPPCPCPPPPCPPCPPPPCCCPCCPCPPPCPCPPCCCPCCCCCCPCCCCPCCPPPPCCPP",
"output": "47"
},
{
"input": "PPCPPPPCCCCPPPPCPPPPPPPPCPCPPCCPPPPPPPPCPPPPCCCCPPPPCPPCPCPPPCCPPCPPCCCPCPPCCCCCCPCPCPCPPCPCPCPPPCCC",
"output": "49"
},
{
"input": "CCPCCCPPCPPCPCCCPCPPCPPCPPCCCCCCCPCPPCPCCPCCPCPCPCCCPCCCPPPCCPCCPPCCCCCPPPPCPCPPCPCPCCPCPPP",
"output": "53"
},
{
"input": "PCPCPPPPCPCPPPCPPCCCPCPCPCPPCPPPPCCPPPCPPPCPPPPCCPPCCCPCCPCCCCPCCPCPPCPCCCPCPPCP",
"output": "47"
},
{
"input": "PCCPPCCCPPCPPCC",
"output": "8"
},
{
"input": "CCCPPPPPPCCCCPCCPCCCCCCPCCCPPPCPC",
"output": "15"
},
{
"input": "CPPCCPPCCPPPCCCPPPPCPPPPPPPCCPCPCCPPPPCCCPPCCPCCPPCCCPCCPCPPPPCCPP",
"output": "31"
},
{
"input": "CCCCCPPPCCPCPCCPPPPCPCCCPCPPCPCPPPPPCCPCPCPC",
"output": "25"
},
{
"input": "PPPPPPPPPCPCP",
"output": "6"
},
{
"input": "PPPCPCPCCCPPCPCCPPPPCCCPCCP",
"output": "15"
},
{
"input": "PCPCCPCPPPPPPCPCCPCPCPCCPPPCPCPCPPCPPCCPCPCCCPCCCPPCPCPCCPCPPPPCCCCCCPPCCPCCCCCPCCCCPPPCPCCCCCPCPCP",
"output": "59"
},
{
"input": "PCCPCPPCCCCCPCCCPCCCPPCCCCCPPPCCPPPPPPPPCPPPCCPPCPPCPCP",
"output": "26"
},
{
"input": "CPCPCCPPPPCCPPCPPCPPCCCCCCPCCPPPCPPCPCCCCCCPCPCCCCCPCCCCCCPCCPPCCP",
"output": "35"
},
{
"input": "PPCCCCCCPP",
"output": "4"
},
{
"input": "CCCCCCCCCCCCPPCCCCPP",
"output": "6"
},
{
"input": "PPPPPPPPPPPCCCCCCCCCCCCCCCCCCP",
"output": "8"
},
{
"input": "PPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCPPPPCCCC",
"output": "10"
},
{
"input": "PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPCCCCCCCCCPPPC",
"output": "12"
},
{
"input": "CCCCCCCCCPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCCCCCCCCCCCCCCPPPPPCC",
"output": "13"
},
{
"input": "CCCCCCCCCCCCCCCCCCCCCCCCCPPPCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC",
"output": "15"
},
{
"input": "CCCCCPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCCCPPPPPPPCCPPP",
"output": "18"
},
{
"input": "PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCCCCCCCCCCCCCCCCCPPPPPPPPPPPPPPPPPPP",
"output": "19"
},
{
"input": "PPPPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCPPPPPCCCPPPPCCCCCPCC",
"output": "23"
}
] | 1,589,541,017
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 45
| 312
| 20,172,800
|
s = str(input())
i=0
count = 0
while len(s)>0:
if s[0]=='C':
temp = s.lstrip('C')
if len(s)-len(temp)>5:
s = s[5:]
else:
s = temp
count = count+1
else:
temp = s.lstrip('P')
if len(s)-len(temp)>5:
s = s[5:]
else:
s = temp
count = count+1
print(count)
|
Title: Postcards and photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has postcards and photos hung in a row on the wall. He decided to put them away to the closet and hang on the wall a famous painter's picture. Polycarpus does it like that: he goes from the left to the right and removes the objects consecutively. As Polycarpus doesn't want any mix-ups to happen, he will not carry in his hands objects of two different types. In other words, Polycarpus can't carry both postcards and photos simultaneously. Sometimes he goes to the closet and puts the objects there, thus leaving his hands free. Polycarpus must put all the postcards and photos to the closet. He cannot skip objects. What minimum number of times he should visit the closet if he cannot carry more than 5 items?
Input Specification:
The only line of the input data contains a non-empty string consisting of letters "С" and "P" whose length does not exceed 100 characters. If the *i*-th character in the string is the letter "С", that means that the *i*-th object (the numbering goes from the left to the right) on Polycarpus' wall is a postcard. And if the *i*-th character is the letter "P", than the *i*-th object on the wall is a photo.
Output Specification:
Print the only number — the minimum number of times Polycarpus has to visit the closet.
Demo Input:
['CPCPCPC\n', 'CCCCCCPPPPPP\n', 'CCCCCCPPCPPPPPPPPPP\n', 'CCCCCCCCCC\n']
Demo Output:
['7\n', '4\n', '6\n', '2\n']
Note:
In the first sample Polycarpus needs to take one item to the closet 7 times.
In the second sample Polycarpus can first take 3 postcards to the closet; then 3 more. He can take the 6 photos that are left in the similar way, going to the closet twice.
In the third sample Polycarpus can visit the closet twice, both times carrying 3 postcards. Then he can take there 2 photos at once, then one postcard and finally, he can carry the last 10 photos if he visits the closet twice.
In the fourth sample Polycarpus can visit the closet twice and take there all 10 postcards (5 items during each go).
|
```python
s = str(input())
i=0
count = 0
while len(s)>0:
if s[0]=='C':
temp = s.lstrip('C')
if len(s)-len(temp)>5:
s = s[5:]
else:
s = temp
count = count+1
else:
temp = s.lstrip('P')
if len(s)-len(temp)>5:
s = s[5:]
else:
s = temp
count = count+1
print(count)
```
| 3
|
|
75
|
A
|
Life Without Zeros
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Life Without Zeros
|
2
|
256
|
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
|
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
|
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
|
[
"101\n102\n",
"105\n106\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "101\n102",
"output": "YES"
},
{
"input": "105\n106",
"output": "NO"
},
{
"input": "544\n397",
"output": "YES"
},
{
"input": "822\n280",
"output": "NO"
},
{
"input": "101\n413",
"output": "NO"
},
{
"input": "309\n139",
"output": "NO"
},
{
"input": "693\n970",
"output": "NO"
},
{
"input": "981\n1",
"output": "YES"
},
{
"input": "352\n276",
"output": "YES"
},
{
"input": "164\n691",
"output": "YES"
},
{
"input": "110036\n43",
"output": "YES"
},
{
"input": "100\n1000",
"output": "NO"
},
{
"input": "1000000000\n1000000000",
"output": "YES"
},
{
"input": "999999999\n999999999",
"output": "YES"
},
{
"input": "6\n4",
"output": "NO"
},
{
"input": "123456\n876543",
"output": "YES"
},
{
"input": "1234567\n9876543",
"output": "NO"
},
{
"input": "1111111\n1119111",
"output": "NO"
},
{
"input": "10502405\n40403403",
"output": "YES"
},
{
"input": "1\n1",
"output": "YES"
},
{
"input": "5\n4",
"output": "YES"
},
{
"input": "1\n6",
"output": "YES"
},
{
"input": "40\n17",
"output": "NO"
},
{
"input": "35\n81",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "2081\n7450",
"output": "NO"
},
{
"input": "84355694\n373629503",
"output": "NO"
},
{
"input": "885992048\n510468672",
"output": "NO"
},
{
"input": "540144752\n647307841",
"output": "NO"
},
{
"input": "341781108\n784147010",
"output": "NO"
},
{
"input": "960440938\n572334653",
"output": "NO"
},
{
"input": "657414510\n768771903",
"output": "NO"
},
{
"input": "501871728\n725074574",
"output": "NO"
},
{
"input": "968\n851",
"output": "YES"
},
{
"input": "943\n939",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "611\n41",
"output": "YES"
},
{
"input": "737\n289",
"output": "NO"
},
{
"input": "168\n571",
"output": "YES"
},
{
"input": "868\n463",
"output": "YES"
},
{
"input": "815732007\n913882008",
"output": "NO"
},
{
"input": "459039901\n453033850",
"output": "NO"
},
{
"input": "975305404\n484601256",
"output": "NO"
}
] | 1,617,687,505
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 154
| 0
|
a = input()
b = input()
c = int(a) + int(b)
c1 = int(str(c).translate({ord('0'): None}))
a1 = int(a.translate({ord('0'): None}))
b1 = int(b.translate({ord('0'): None}))
if a1 + b1 == c1:
print("YES")
else:
print("NO")
|
Title: Life Without Zeros
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
Input Specification:
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
Output Specification:
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Demo Input:
['101\n102\n', '105\n106\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
a = input()
b = input()
c = int(a) + int(b)
c1 = int(str(c).translate({ord('0'): None}))
a1 = int(a.translate({ord('0'): None}))
b1 = int(b.translate({ord('0'): None}))
if a1 + b1 == c1:
print("YES")
else:
print("NO")
```
| 3.9615
|
190
|
A
|
Vasya and the Bus
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
One day Vasya heard a story: "In the city of High Bertown a bus number 62 left from the bus station. It had *n* grown-ups and *m* kids..."
The latter events happen to be of no importance to us. Vasya is an accountant and he loves counting money. So he wondered what maximum and minimum sum of money these passengers could have paid for the ride.
The bus fare equals one berland ruble in High Bertown. However, not everything is that easy — no more than one child can ride for free with each grown-up passenger. That means that a grown-up passenger who rides with his *k* (*k*<=><=0) children, pays overall *k* rubles: a ticket for himself and (*k*<=-<=1) tickets for his children. Also, a grown-up can ride without children, in this case he only pays one ruble.
We know that in High Bertown children can't ride in a bus unaccompanied by grown-ups.
Help Vasya count the minimum and the maximum sum in Berland rubles, that all passengers of this bus could have paid in total.
|
The input file consists of a single line containing two space-separated numbers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=105) — the number of the grown-ups and the number of the children in the bus, correspondingly.
|
If *n* grown-ups and *m* children could have ridden in the bus, then print on a single line two space-separated integers — the minimum and the maximum possible total bus fare, correspondingly.
Otherwise, print "Impossible" (without the quotes).
|
[
"1 2\n",
"0 5\n",
"2 2\n"
] |
[
"2 2",
"Impossible",
"2 3"
] |
In the first sample a grown-up rides with two children and pays two rubles.
In the second sample there are only children in the bus, so the situation is impossible.
In the third sample there are two cases: - Each of the two grown-ups rides with one children and pays one ruble for the tickets. In this case the passengers pay two rubles in total. - One of the grown-ups ride with two children's and pays two rubles, the another one rides alone and pays one ruble for himself. So, they pay three rubles in total.
| 500
|
[
{
"input": "1 2",
"output": "2 2"
},
{
"input": "0 5",
"output": "Impossible"
},
{
"input": "2 2",
"output": "2 3"
},
{
"input": "2 7",
"output": "7 8"
},
{
"input": "4 10",
"output": "10 13"
},
{
"input": "6 0",
"output": "6 6"
},
{
"input": "7 1",
"output": "7 7"
},
{
"input": "0 0",
"output": "0 0"
},
{
"input": "71 24",
"output": "71 94"
},
{
"input": "16 70",
"output": "70 85"
},
{
"input": "0 1",
"output": "Impossible"
},
{
"input": "1 0",
"output": "1 1"
},
{
"input": "1 1",
"output": "1 1"
},
{
"input": "63 82",
"output": "82 144"
},
{
"input": "8 26",
"output": "26 33"
},
{
"input": "21 27",
"output": "27 47"
},
{
"input": "0 38",
"output": "Impossible"
},
{
"input": "46 84",
"output": "84 129"
},
{
"input": "59 96",
"output": "96 154"
},
{
"input": "63028 0",
"output": "63028 63028"
},
{
"input": "9458 0",
"output": "9458 9458"
},
{
"input": "80236 0",
"output": "80236 80236"
},
{
"input": "26666 0",
"output": "26666 26666"
},
{
"input": "59617 0",
"output": "59617 59617"
},
{
"input": "0 6048",
"output": "Impossible"
},
{
"input": "63028 28217",
"output": "63028 91244"
},
{
"input": "9458 39163",
"output": "39163 48620"
},
{
"input": "80236 14868",
"output": "80236 95103"
},
{
"input": "26666 52747",
"output": "52747 79412"
},
{
"input": "59617 28452",
"output": "59617 88068"
},
{
"input": "6048 4158",
"output": "6048 10205"
},
{
"input": "76826 4210",
"output": "76826 81035"
},
{
"input": "23256 15156",
"output": "23256 38411"
},
{
"input": "56207 53035",
"output": "56207 109241"
},
{
"input": "2637 28740",
"output": "28740 31376"
},
{
"input": "73415 4445",
"output": "73415 77859"
},
{
"input": "82019 4498",
"output": "82019 86516"
},
{
"input": "28449 80204",
"output": "80204 108652"
},
{
"input": "99227 53323",
"output": "99227 152549"
},
{
"input": "45657 29028",
"output": "45657 74684"
},
{
"input": "78608 4733",
"output": "78608 83340"
},
{
"input": "25038 4786",
"output": "25038 29823"
},
{
"input": "95816 80492",
"output": "95816 176307"
},
{
"input": "42246 94024",
"output": "94024 136269"
},
{
"input": "0 100000",
"output": "Impossible"
},
{
"input": "100000 0",
"output": "100000 100000"
},
{
"input": "1 100000",
"output": "100000 100000"
},
{
"input": "100000 1",
"output": "100000 100000"
},
{
"input": "63028 63028",
"output": "63028 126055"
},
{
"input": "9458 9458",
"output": "9458 18915"
},
{
"input": "80236 80236",
"output": "80236 160471"
},
{
"input": "26666 26666",
"output": "26666 53331"
},
{
"input": "59617 59617",
"output": "59617 119233"
},
{
"input": "100000 100000",
"output": "100000 199999"
}
] | 1,618,418,815
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 122
| 0
|
n, m = map(int, input().split())
if n == 0:
print("Impossible")
else:
if m != 0:
if n <= m:
print(n + (m - n), n + (m - 1))
else:
print(n, n + (m - 1))
elif m == 0:
print(n, n)
|
Title: Vasya and the Bus
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya heard a story: "In the city of High Bertown a bus number 62 left from the bus station. It had *n* grown-ups and *m* kids..."
The latter events happen to be of no importance to us. Vasya is an accountant and he loves counting money. So he wondered what maximum and minimum sum of money these passengers could have paid for the ride.
The bus fare equals one berland ruble in High Bertown. However, not everything is that easy — no more than one child can ride for free with each grown-up passenger. That means that a grown-up passenger who rides with his *k* (*k*<=><=0) children, pays overall *k* rubles: a ticket for himself and (*k*<=-<=1) tickets for his children. Also, a grown-up can ride without children, in this case he only pays one ruble.
We know that in High Bertown children can't ride in a bus unaccompanied by grown-ups.
Help Vasya count the minimum and the maximum sum in Berland rubles, that all passengers of this bus could have paid in total.
Input Specification:
The input file consists of a single line containing two space-separated numbers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=105) — the number of the grown-ups and the number of the children in the bus, correspondingly.
Output Specification:
If *n* grown-ups and *m* children could have ridden in the bus, then print on a single line two space-separated integers — the minimum and the maximum possible total bus fare, correspondingly.
Otherwise, print "Impossible" (without the quotes).
Demo Input:
['1 2\n', '0 5\n', '2 2\n']
Demo Output:
['2 2', 'Impossible', '2 3']
Note:
In the first sample a grown-up rides with two children and pays two rubles.
In the second sample there are only children in the bus, so the situation is impossible.
In the third sample there are two cases: - Each of the two grown-ups rides with one children and pays one ruble for the tickets. In this case the passengers pay two rubles in total. - One of the grown-ups ride with two children's and pays two rubles, the another one rides alone and pays one ruble for himself. So, they pay three rubles in total.
|
```python
n, m = map(int, input().split())
if n == 0:
print("Impossible")
else:
if m != 0:
if n <= m:
print(n + (m - n), n + (m - 1))
else:
print(n, n + (m - 1))
elif m == 0:
print(n, n)
```
| 0
|
|
32
|
B
|
Borze
|
PROGRAMMING
| 800
|
[
"expression parsing",
"implementation"
] |
B. Borze
|
2
|
256
|
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
|
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
|
Output the decoded ternary number. It can have leading zeroes.
|
[
".-.--\n",
"--.\n",
"-..-.--\n"
] |
[
"012",
"20",
"1012"
] |
none
| 1,000
|
[
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "10000220"
},
{
"input": ".",
"output": "0"
},
{
"input": "-.",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "..",
"output": "00"
},
{
"input": "--.",
"output": "20"
},
{
"input": ".--.",
"output": "020"
},
{
"input": ".-.-..",
"output": "0110"
},
{
"input": "----.-.",
"output": "2201"
},
{
"input": "-..--.-.",
"output": "10201"
},
{
"input": "..--..--.",
"output": "0020020"
},
{
"input": "-.-.---.--..-..-.-.-..-..-.--.",
"output": "112120010111010120"
},
{
"input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.",
"output": "21112220010101011012011011221011"
},
{
"input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------",
"output": "11020111110111222212021020002022111100201121222020012022110010222"
},
{
"input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..",
"output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110"
},
{
"input": ".-..-.-.---.-----.--.---...-.--.-.-....-..",
"output": "01011212212021001201100010"
},
{
"input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..",
"output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210"
},
{
"input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.",
"output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220"
},
{
"input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----",
"output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122"
},
{
"input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..",
"output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110"
},
{
"input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.",
"output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020"
},
{
"input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..",
"output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200"
},
{
"input": "-.---...----...--.--...-.--.----",
"output": "121002200020200012022"
},
{
"input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.",
"output": "202020212012111010201021100001021221200011222112012201012220"
}
] | 1,649,258,173
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
Cad=input()
Res=""
Digi=""
K=0
while(K<len(Cad)):
if(Cad[K]=="."):
Digi="0"
K+=1
else:
if((Cad[K]=="-")and(Cad[K+1]==".")):
Digi="1"
else:
Digi="2"
K+=2
Res+=Digi
print(Res)
|
Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none
|
```python
Cad=input()
Res=""
Digi=""
K=0
while(K<len(Cad)):
if(Cad[K]=="."):
Digi="0"
K+=1
else:
if((Cad[K]=="-")and(Cad[K+1]==".")):
Digi="1"
else:
Digi="2"
K+=2
Res+=Digi
print(Res)
```
| 3.977
|
914
|
A
|
Perfect Squares
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Given an array *a*1,<=*a*2,<=...,<=*a**n* of *n* integers, find the largest number in the array that is not a perfect square.
A number *x* is said to be a perfect square if there exists an integer *y* such that *x*<==<=*y*2.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
|
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
|
[
"2\n4 2\n",
"8\n1 2 4 8 16 32 64 576\n"
] |
[
"2\n",
"32\n"
] |
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
| 500
|
[
{
"input": "2\n4 2",
"output": "2"
},
{
"input": "8\n1 2 4 8 16 32 64 576",
"output": "32"
},
{
"input": "3\n-1 -4 -9",
"output": "-1"
},
{
"input": "5\n918375 169764 598796 76602 538757",
"output": "918375"
},
{
"input": "5\n804610 765625 2916 381050 93025",
"output": "804610"
},
{
"input": "5\n984065 842724 127449 525625 573049",
"output": "984065"
},
{
"input": "2\n226505 477482",
"output": "477482"
},
{
"input": "2\n370881 659345",
"output": "659345"
},
{
"input": "2\n4 5",
"output": "5"
},
{
"input": "2\n3 4",
"output": "3"
},
{
"input": "2\n999999 1000000",
"output": "999999"
},
{
"input": "3\n-1 -2 -3",
"output": "-1"
},
{
"input": "2\n-1000000 1000000",
"output": "-1000000"
},
{
"input": "2\n-1 0",
"output": "-1"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "35\n-871271 -169147 -590893 -400197 -476793 0 -15745 -890852 -124052 -631140 -238569 -597194 -147909 -928925 -587628 -569656 -581425 -963116 -665954 -506797 -196044 -309770 -701921 -926257 -152426 -991371 -624235 -557143 -689886 -59804 -549134 -107407 -182016 -24153 -607462",
"output": "-15745"
},
{
"input": "16\n-882343 -791322 0 -986738 -415891 -823354 -840236 -552554 -760908 -331993 -549078 -863759 -913261 -937429 -257875 -602322",
"output": "-257875"
},
{
"input": "71\n908209 289 44521 240100 680625 274576 212521 91809 506944 499849 3844 15376 592900 58081 240100 984064 732736 257049 600625 180625 130321 580644 261121 75625 46225 853776 485809 700569 817216 268324 293764 528529 25921 399424 175561 99856 295936 20736 611524 13924 470596 574564 5329 15376 676 431649 145161 697225 41616 550564 514089 9409 227529 1681 839056 3721 552049 465124 38809 197136 659344 214369 998001 44944 3844 186624 362404 -766506 739600 10816 299209",
"output": "-766506"
},
{
"input": "30\n192721 -950059 -734656 625 247009 -423468 318096 622521 678976 777924 1444 748303 27556 62001 795664 89401 221841 -483208 467856 477109 196 -461813 831744 772641 574564 -519370 861184 67600 -717966 -259259",
"output": "748303"
},
{
"input": "35\n628849 962361 436921 944784 444889 29241 -514806 171396 685584 -823202 -929730 6982 198025 783225 552049 -957165 782287 -659167 -414846 695556 -336330 41616 963781 71289 119639 952576 -346713 178929 232324 121802 393266 841 649636 179555 998001",
"output": "963781"
},
{
"input": "53\n280988 756430 -515570 -248578 170649 -21608 642677 216770 827291 589500 940901 216097 -118956 -919104 -319264 -761585 289479 499613 588276 883036 480518 -323196 -274570 -406556 -381484 -956025 702135 -445274 -783543 136593 153664 897473 352651 737974 -21123 -284944 501734 898033 604429 624138 40804 248782 -786059 -304592 -209210 -312904 419820 -328648 -47331 -919227 -280955 104827 877304",
"output": "940901"
},
{
"input": "15\n256 -227055 427717 827239 462070 66049 987533 -175306 -552810 -867915 -408251 -693957 -972981 -245827 896904",
"output": "987533"
},
{
"input": "3\n-1 1 0",
"output": "-1"
},
{
"input": "2\n0 -5",
"output": "-5"
},
{
"input": "3\n-1 -2 0",
"output": "-1"
},
{
"input": "2\n-5 0",
"output": "-5"
},
{
"input": "1\n-439",
"output": "-439"
},
{
"input": "1\n-1000000",
"output": "-1000000"
},
{
"input": "1\n-917455",
"output": "-917455"
},
{
"input": "3\n1 1 -1",
"output": "-1"
},
{
"input": "2\n131073 1",
"output": "131073"
},
{
"input": "2\n99999 3",
"output": "99999"
},
{
"input": "2\n-524272 -1000000",
"output": "-524272"
},
{
"input": "2\n15 131073",
"output": "131073"
}
] | 1,619,153,989
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 124
| 2,150,400
|
n = int(input())
daf = list(map(int, input().split()))
maxP = 0
for i in range(n):
x = daf[i]**0.5
y = int(x)
if abs(x-y) <= 1e-3:
continue
else:
maxP = max(maxP, daf[i])
print(maxP)
|
Title: Perfect Squares
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Given an array *a*1,<=*a*2,<=...,<=*a**n* of *n* integers, find the largest number in the array that is not a perfect square.
A number *x* is said to be a perfect square if there exists an integer *y* such that *x*<==<=*y*2.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output Specification:
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Demo Input:
['2\n4 2\n', '8\n1 2 4 8 16 32 64 576\n']
Demo Output:
['2\n', '32\n']
Note:
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
|
```python
n = int(input())
daf = list(map(int, input().split()))
maxP = 0
for i in range(n):
x = daf[i]**0.5
y = int(x)
if abs(x-y) <= 1e-3:
continue
else:
maxP = max(maxP, daf[i])
print(maxP)
```
| -1
|
|
633
|
B
|
A Trivial Problem
|
PROGRAMMING
| 1,300
|
[
"brute force",
"constructive algorithms",
"math",
"number theory"
] | null | null |
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer *m* and asks for the number of positive integers *n*, such that the factorial of *n* ends with exactly *m* zeroes. Are you among those great programmers who can solve this problem?
|
The only line of input contains an integer *m* (1<=≤<=*m*<=≤<=100<=000) — the required number of trailing zeroes in factorial.
|
First print *k* — the number of values of *n* such that the factorial of *n* ends with *m* zeroes. Then print these *k* integers in increasing order.
|
[
"1\n",
"5\n"
] |
[
"5\n5 6 7 8 9 ",
"0"
] |
The factorial of *n* is equal to the product of all integers from 1 to *n* inclusive, that is *n*! = 1·2·3·...·*n*.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
| 500
|
[
{
"input": "1",
"output": "5\n5 6 7 8 9 "
},
{
"input": "5",
"output": "0"
},
{
"input": "2",
"output": "5\n10 11 12 13 14 "
},
{
"input": "3",
"output": "5\n15 16 17 18 19 "
},
{
"input": "7",
"output": "5\n30 31 32 33 34 "
},
{
"input": "12",
"output": "5\n50 51 52 53 54 "
},
{
"input": "15",
"output": "5\n65 66 67 68 69 "
},
{
"input": "18",
"output": "5\n75 76 77 78 79 "
},
{
"input": "38",
"output": "5\n155 156 157 158 159 "
},
{
"input": "47",
"output": "5\n195 196 197 198 199 "
},
{
"input": "58",
"output": "5\n240 241 242 243 244 "
},
{
"input": "66",
"output": "5\n270 271 272 273 274 "
},
{
"input": "70",
"output": "5\n285 286 287 288 289 "
},
{
"input": "89",
"output": "5\n365 366 367 368 369 "
},
{
"input": "417",
"output": "5\n1675 1676 1677 1678 1679 "
},
{
"input": "815",
"output": "5\n3265 3266 3267 3268 3269 "
},
{
"input": "394",
"output": "5\n1585 1586 1587 1588 1589 "
},
{
"input": "798",
"output": "0"
},
{
"input": "507",
"output": "5\n2035 2036 2037 2038 2039 "
},
{
"input": "406",
"output": "5\n1630 1631 1632 1633 1634 "
},
{
"input": "570",
"output": "5\n2290 2291 2292 2293 2294 "
},
{
"input": "185",
"output": "0"
},
{
"input": "765",
"output": "0"
},
{
"input": "967",
"output": "0"
},
{
"input": "112",
"output": "5\n455 456 457 458 459 "
},
{
"input": "729",
"output": "5\n2925 2926 2927 2928 2929 "
},
{
"input": "4604",
"output": "5\n18425 18426 18427 18428 18429 "
},
{
"input": "8783",
"output": "5\n35140 35141 35142 35143 35144 "
},
{
"input": "1059",
"output": "0"
},
{
"input": "6641",
"output": "5\n26575 26576 26577 26578 26579 "
},
{
"input": "9353",
"output": "5\n37425 37426 37427 37428 37429 "
},
{
"input": "1811",
"output": "5\n7250 7251 7252 7253 7254 "
},
{
"input": "2528",
"output": "0"
},
{
"input": "8158",
"output": "5\n32640 32641 32642 32643 32644 "
},
{
"input": "3014",
"output": "5\n12070 12071 12072 12073 12074 "
},
{
"input": "7657",
"output": "5\n30640 30641 30642 30643 30644 "
},
{
"input": "4934",
"output": "0"
},
{
"input": "9282",
"output": "5\n37140 37141 37142 37143 37144 "
},
{
"input": "2610",
"output": "5\n10450 10451 10452 10453 10454 "
},
{
"input": "2083",
"output": "5\n8345 8346 8347 8348 8349 "
},
{
"input": "26151",
"output": "5\n104620 104621 104622 104623 104624 "
},
{
"input": "64656",
"output": "5\n258640 258641 258642 258643 258644 "
},
{
"input": "46668",
"output": "5\n186690 186691 186692 186693 186694 "
},
{
"input": "95554",
"output": "5\n382235 382236 382237 382238 382239 "
},
{
"input": "37320",
"output": "0"
},
{
"input": "52032",
"output": "5\n208140 208141 208142 208143 208144 "
},
{
"input": "11024",
"output": "5\n44110 44111 44112 44113 44114 "
},
{
"input": "63218",
"output": "5\n252885 252886 252887 252888 252889 "
},
{
"input": "40095",
"output": "5\n160390 160391 160392 160393 160394 "
},
{
"input": "42724",
"output": "5\n170910 170911 170912 170913 170914 "
},
{
"input": "24381",
"output": "5\n97530 97531 97532 97533 97534 "
},
{
"input": "73138",
"output": "5\n292570 292571 292572 292573 292574 "
},
{
"input": "93346",
"output": "5\n373400 373401 373402 373403 373404 "
},
{
"input": "18338",
"output": "5\n73370 73371 73372 73373 73374 "
},
{
"input": "42662",
"output": "5\n170660 170661 170662 170663 170664 "
},
{
"input": "81221",
"output": "5\n324900 324901 324902 324903 324904 "
},
{
"input": "100000",
"output": "5\n400005 400006 400007 400008 400009 "
},
{
"input": "100000",
"output": "5\n400005 400006 400007 400008 400009 "
},
{
"input": "99998",
"output": "0"
},
{
"input": "30",
"output": "0"
},
{
"input": "11",
"output": "0"
},
{
"input": "780",
"output": "0"
},
{
"input": "97656",
"output": "5\n390625 390626 390627 390628 390629 "
},
{
"input": "12499",
"output": "5\n50000 50001 50002 50003 50004 "
},
{
"input": "65",
"output": "5\n265 266 267 268 269 "
},
{
"input": "41",
"output": "5\n170 171 172 173 174 "
},
{
"input": "31",
"output": "5\n125 126 127 128 129 "
},
{
"input": "86577",
"output": "0"
}
] | 1,625,060,122
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 68
| 186
| 0
|
n=int(input())
count=i=0
while count<n:
i+=1
k=i
while k%5==0:
count+=1
k=(k//5)
if count==n:
print(5)
print(i,i+1,i+2,i+3,i+4)
else:
print(0)
|
Title: A Trivial Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer *m* and asks for the number of positive integers *n*, such that the factorial of *n* ends with exactly *m* zeroes. Are you among those great programmers who can solve this problem?
Input Specification:
The only line of input contains an integer *m* (1<=≤<=*m*<=≤<=100<=000) — the required number of trailing zeroes in factorial.
Output Specification:
First print *k* — the number of values of *n* such that the factorial of *n* ends with *m* zeroes. Then print these *k* integers in increasing order.
Demo Input:
['1\n', '5\n']
Demo Output:
['5\n5 6 7 8 9 ', '0']
Note:
The factorial of *n* is equal to the product of all integers from 1 to *n* inclusive, that is *n*! = 1·2·3·...·*n*.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
|
```python
n=int(input())
count=i=0
while count<n:
i+=1
k=i
while k%5==0:
count+=1
k=(k//5)
if count==n:
print(5)
print(i,i+1,i+2,i+3,i+4)
else:
print(0)
```
| 3
|
|
31
|
E
|
TV Game
|
PROGRAMMING
| 2,400
|
[
"dp"
] |
E. TV Game
|
2
|
256
|
There is a new TV game on BerTV. In this game two players get a number *A* consisting of 2*n* digits. Before each turn players determine who will make the next move. Each player should make exactly *n* moves. On it's turn *i*-th player takes the leftmost digit of *A* and appends it to his or her number *S**i*. After that this leftmost digit is erased from *A*. Initially the numbers of both players (*S*1 and *S*2) are «empty». Leading zeroes in numbers *A*,<=*S*1,<=*S*2 are allowed. In the end of the game the first player gets *S*1 dollars, and the second gets *S*2 dollars.
One day Homer and Marge came to play the game. They managed to know the number *A* beforehand. They want to find such sequence of their moves that both of them makes exactly *n* moves and which maximizes their total prize. Help them.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=18). The second line contains integer *A* consisting of exactly 2*n* digits. This number can have leading zeroes.
|
Output the line of 2*n* characters «H» and «M» — the sequence of moves of Homer and Marge, which gives them maximum possible total prize. Each player must make exactly *n* moves. If there are several solutions, output any of them.
|
[
"2\n1234\n",
"2\n9911\n"
] |
[
"HHMM",
"HMHM"
] |
none
| 2,500
|
[
{
"input": "2\n1234",
"output": "HHMM"
},
{
"input": "2\n9911",
"output": "HMHM"
},
{
"input": "2\n0153",
"output": "HHMM"
},
{
"input": "3\n614615",
"output": "HHHMMM"
},
{
"input": "4\n21305374",
"output": "HHHHMMMM"
},
{
"input": "4\n00013213",
"output": "HHHHMMMM"
},
{
"input": "1\n01",
"output": "HM"
},
{
"input": "1\n21",
"output": "HM"
},
{
"input": "1\n99",
"output": "HM"
},
{
"input": "18\n999999999999999999999999999999999999",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "10\n89959999998998796989",
"output": "HHHHHHMMMMHMMHHMHMMM"
},
{
"input": "10\n99999988899989998889",
"output": "HHHHHHHHHMMMHMMMMMMM"
},
{
"input": "7\n10210320200120",
"output": "HHHHHMMHMHMMMM"
},
{
"input": "18\n949787136121830145537930861689777414",
"output": "HHMHMHHHHHHHMHHHHHHHHMMMMMMMMMMMMMMM"
},
{
"input": "18\n956859579789834858167218778893796384",
"output": "HHHHHMHHMHHMMHHMHMHHHHHMHMMMMMMMMMMM"
},
{
"input": "18\n789998768896689887879979878577696879",
"output": "HHHHMHHHHHMHHHMHHHHHMMHMMMMMMMMMMMMM"
},
{
"input": "18\n899898999999899789998999898998699998",
"output": "HHHHHHHHHHHMHMMHHMMMHMMMHMHMMMHMMMMM"
},
{
"input": "18\n998999899889999999999999999999998999",
"output": "HHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMHMMM"
},
{
"input": "18\n999999999999999999999999999999999999",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "18\n520301003123441003000011410650200262",
"output": "HHHHHHHHHHHHMMHHHMHHHMMMMMMMMMMMMMMM"
},
{
"input": "18\n003003010010211000120021200200013010",
"output": "HHMHHMHHHHHHMHHHHHHMHHMMMMMMMMMMMMMM"
},
{
"input": "18\n101011411002041200101000000000001000",
"output": "HHHHHHMHHHHHHMHMHHMHMHHMMMMMMMMMMMMM"
},
{
"input": "18\n010000000000010000000000000101001000",
"output": "HHHHHHHHHHHHHMHHHHHMMMMMMMMMMMMMMMMM"
},
{
"input": "18\n000000000000000000000000000000001000",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "18\n999999999999999999999999999999999999",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "18\n000000000000000000000000000000000000",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "18\n999999999999999999999999999999999899",
"output": "HHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMHMM"
},
{
"input": "18\n000000000000000000000000000000000000",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "18\n000000000000000000000000000000000000",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "18\n998877665544332211998877665544332211",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "9\n998877665544332211",
"output": "HMHMHMHMHMHMHMHMHM"
},
{
"input": "18\n999988887777666655554444333322221111",
"output": "HHMMHHMMHHMMHHMMHHMMHHMMHHMMHHMMHHMM"
},
{
"input": "18\n111111111111111111111111111111111111",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "9\n112233445566778899",
"output": "HHHHHHHHHMMMMMMMMM"
},
{
"input": "18\n112233445566778899112233445566778899",
"output": "HHHHHHHHHHHHHHHHMMHHMMMMMMMMMMMMMMMM"
},
{
"input": "18\n111122223333444455556666777788889999",
"output": "HHHHHHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMM"
},
{
"input": "7\n98887870656634",
"output": "HHHMHMMHMHMMHM"
},
{
"input": "6\n389531391995",
"output": "HHMHHHHMMMMM"
},
{
"input": "7\n98887870656634",
"output": "HHHMHMMHMHMMHM"
},
{
"input": "8\n6770379299275795",
"output": "HHHHHHMHMMHMMMMM"
},
{
"input": "9\n256276354396913650",
"output": "HHHHMHHHHHMMMMMMMM"
},
{
"input": "17\n0505221537264872903260121660510735",
"output": "HHHHHHHHHHHHHHHHMHMMMMMMMMMMMMMMMM"
},
{
"input": "16\n94037617932892444411587756822347",
"output": "HHHHHHHHMHHHMHHHMMHHMMMMMMMMMMMM"
},
{
"input": "15\n051911296923361789993851872792",
"output": "HHHHHHHMHMHHHHHHHMMMMMMMMMMMMM"
},
{
"input": "14\n1637752215267288357827332779",
"output": "HHHHHHHHHHHHHHMMMMMMMMMMMMMM"
},
{
"input": "13\n04432024633126608067252737",
"output": "HHHHHHHHMHHHHMMHMMMMMMMMMM"
},
{
"input": "12\n254163163934611135461589",
"output": "HHHHHHHHHMHHMHMMMMMMMMMM"
},
{
"input": "11\n3657071886390583694504",
"output": "HHHHHHHHHHHMMMMMMMMMMM"
},
{
"input": "10\n27555210523159542624",
"output": "HMHHHHHHMHHHMMMMMMMM"
},
{
"input": "6\n389531391995",
"output": "HHMHHHHMMMMM"
},
{
"input": "5\n9823921785",
"output": "HHHHMMHMMM"
}
] | 1,551,368,320
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 12
| 248
| 0
|
def compute():
from sys import stdin
[n] = list(map(int, stdin.readline().split()))
i = list(map(int, stdin.readline().strip()))
dp = {}
INF = (0,"")
def f(x,h,m):
if x >= 2*n:
return INF
args = (h,m)
if args in dp:
return dp[args]
res0, res1 = INF, INF
if h>=0:
ff = f(x+1, h-1,m)
res0 = (ff[0] + pow(10,h)*i[x], "H"+ff[1])
if m>=0:
ff = f(x+1, h,m-1)
res1 = (ff[0] + pow(10,m)*i[x], "M"+ff[1])
dp[args] = res0 if res0[0]>res1[0] else res1
return dp[args]
ans = f(0, n-1, n-1)[1]
print(ans)
if __name__ == "__main__":
compute()
|
Title: TV Game
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a new TV game on BerTV. In this game two players get a number *A* consisting of 2*n* digits. Before each turn players determine who will make the next move. Each player should make exactly *n* moves. On it's turn *i*-th player takes the leftmost digit of *A* and appends it to his or her number *S**i*. After that this leftmost digit is erased from *A*. Initially the numbers of both players (*S*1 and *S*2) are «empty». Leading zeroes in numbers *A*,<=*S*1,<=*S*2 are allowed. In the end of the game the first player gets *S*1 dollars, and the second gets *S*2 dollars.
One day Homer and Marge came to play the game. They managed to know the number *A* beforehand. They want to find such sequence of their moves that both of them makes exactly *n* moves and which maximizes their total prize. Help them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=18). The second line contains integer *A* consisting of exactly 2*n* digits. This number can have leading zeroes.
Output Specification:
Output the line of 2*n* characters «H» and «M» — the sequence of moves of Homer and Marge, which gives them maximum possible total prize. Each player must make exactly *n* moves. If there are several solutions, output any of them.
Demo Input:
['2\n1234\n', '2\n9911\n']
Demo Output:
['HHMM', 'HMHM']
Note:
none
|
```python
def compute():
from sys import stdin
[n] = list(map(int, stdin.readline().split()))
i = list(map(int, stdin.readline().strip()))
dp = {}
INF = (0,"")
def f(x,h,m):
if x >= 2*n:
return INF
args = (h,m)
if args in dp:
return dp[args]
res0, res1 = INF, INF
if h>=0:
ff = f(x+1, h-1,m)
res0 = (ff[0] + pow(10,h)*i[x], "H"+ff[1])
if m>=0:
ff = f(x+1, h,m-1)
res1 = (ff[0] + pow(10,m)*i[x], "M"+ff[1])
dp[args] = res0 if res0[0]>res1[0] else res1
return dp[args]
ans = f(0, n-1, n-1)[1]
print(ans)
if __name__ == "__main__":
compute()
```
| 0
|
149
|
A
|
Business trip
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
|
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
|
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
|
[
"5\n1 1 1 1 2 2 3 2 2 1 1 1\n",
"0\n0 0 0 0 0 0 0 1 1 2 3 0\n",
"11\n1 1 4 1 1 5 1 1 4 1 1 1\n"
] |
[
"2\n",
"0\n",
"3\n"
] |
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all.
| 500
|
[
{
"input": "5\n1 1 1 1 2 2 3 2 2 1 1 1",
"output": "2"
},
{
"input": "0\n0 0 0 0 0 0 0 1 1 2 3 0",
"output": "0"
},
{
"input": "11\n1 1 4 1 1 5 1 1 4 1 1 1",
"output": "3"
},
{
"input": "15\n20 1 1 1 1 2 2 1 2 2 1 1",
"output": "1"
},
{
"input": "7\n8 9 100 12 14 17 21 10 11 100 23 10",
"output": "1"
},
{
"input": "52\n1 12 3 11 4 5 10 6 9 7 8 2",
"output": "6"
},
{
"input": "50\n2 2 3 4 5 4 4 5 7 3 2 7",
"output": "-1"
},
{
"input": "0\n55 81 28 48 99 20 67 95 6 19 10 93",
"output": "0"
},
{
"input": "93\n85 40 93 66 92 43 61 3 64 51 90 21",
"output": "1"
},
{
"input": "99\n36 34 22 0 0 0 52 12 0 0 33 47",
"output": "2"
},
{
"input": "99\n28 32 31 0 10 35 11 18 0 0 32 28",
"output": "3"
},
{
"input": "99\n19 17 0 1 18 11 29 9 29 22 0 8",
"output": "4"
},
{
"input": "76\n2 16 11 10 12 0 20 4 4 14 11 14",
"output": "5"
},
{
"input": "41\n2 1 7 7 4 2 4 4 9 3 10 0",
"output": "6"
},
{
"input": "47\n8 2 2 4 3 1 9 4 2 7 7 8",
"output": "7"
},
{
"input": "58\n6 11 7 0 5 6 3 9 4 9 5 1",
"output": "8"
},
{
"input": "32\n5 2 4 1 5 0 5 1 4 3 0 3",
"output": "9"
},
{
"input": "31\n6 1 0 4 4 5 1 0 5 3 2 0",
"output": "9"
},
{
"input": "35\n2 3 0 0 6 3 3 4 3 5 0 6",
"output": "9"
},
{
"input": "41\n3 1 3 4 3 6 6 1 4 4 0 6",
"output": "11"
},
{
"input": "97\n0 5 3 12 10 16 22 8 21 17 21 10",
"output": "5"
},
{
"input": "100\n21 21 0 0 4 13 0 26 0 0 0 15",
"output": "6"
},
{
"input": "100\n0 0 16 5 22 0 5 0 25 0 14 13",
"output": "7"
},
{
"input": "97\n17 0 10 0 0 0 18 0 14 23 15 0",
"output": "6"
},
{
"input": "100\n0 9 0 18 7 0 0 14 33 3 0 16",
"output": "7"
},
{
"input": "95\n5 2 13 0 15 18 17 0 6 11 0 8",
"output": "9"
},
{
"input": "94\n11 13 0 9 15 8 8 16 3 7 1 3",
"output": "11"
},
{
"input": "96\n8 4 12 15 8 0 4 10 6 6 12 11",
"output": "11"
},
{
"input": "100\n5 5 3 8 6 5 0 3 3 8 1 3",
"output": "-1"
},
{
"input": "100\n1 0 0 1 1 0 1 1 1 1 2 1",
"output": "-1"
},
{
"input": "100\n6 3 2 0 4 1 2 2 2 2 1 1",
"output": "-1"
},
{
"input": "0\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
},
{
"input": "0\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "1"
},
{
"input": "12\n1 1 1 1 1 1 1 1 1 1 1 1",
"output": "12"
},
{
"input": "13\n1 1 1 1 1 1 1 1 1 1 1 2",
"output": "12"
},
{
"input": "15\n10 1 1 1 1 1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
}
] | 1,680,088,924
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
import java.util.Scanner;
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] li = new int[n];
for (int i = 0; i < n; i++) {
li[i] = sc.nextInt();
}
int res = 0;
if (n == 0) {
System.out.println(0);
} else {
Arrays.sort(li);
int i = n - 1;
while (i >= 0 && n > 0) {
n -= li[i];
res++;
i--;
}
if (n <= 0) {
System.out.println(res);
} else {
System.out.println(-1);
}
}
sc.close();
}
}
|
Title: Business trip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
Input Specification:
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
Output Specification:
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
Demo Input:
['5\n1 1 1 1 2 2 3 2 2 1 1 1\n', '0\n0 0 0 0 0 0 0 1 1 2 3 0\n', '11\n1 1 4 1 1 5 1 1 4 1 1 1\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all.
|
```python
import java.util.Scanner;
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] li = new int[n];
for (int i = 0; i < n; i++) {
li[i] = sc.nextInt();
}
int res = 0;
if (n == 0) {
System.out.println(0);
} else {
Arrays.sort(li);
int i = n - 1;
while (i >= 0 && n > 0) {
n -= li[i];
res++;
i--;
}
if (n <= 0) {
System.out.println(res);
} else {
System.out.println(-1);
}
}
sc.close();
}
}
```
| -1
|
|
57
|
A
|
Square Earth?
|
PROGRAMMING
| 1,300
|
[
"dfs and similar",
"greedy",
"implementation"
] |
A. Square Earth?
|
2
|
256
|
Meg the Rabbit decided to do something nice, specifically — to determine the shortest distance between two points on the surface of our planet. But Meg... what can you say, she wants everything simple. So, she already regards our planet as a two-dimensional circle. No, wait, it's even worse — as a square of side *n*. Thus, the task has been reduced to finding the shortest path between two dots on a square (the path should go through the square sides). To simplify the task let us consider the vertices of the square to lie at points whose coordinates are: (0,<=0), (*n*,<=0), (0,<=*n*) and (*n*,<=*n*).
|
The single line contains 5 space-separated integers: *n*,<=*x*1,<=*y*1,<=*x*2,<=*y*2 (1<=≤<=*n*<=≤<=1000,<=0<=≤<=*x*1,<=*y*1,<=*x*2,<=*y*2<=≤<=*n*) which correspondingly represent a side of the square, the coordinates of the first point and the coordinates of the second point. It is guaranteed that the points lie on the sides of the square.
|
You must print on a single line the shortest distance between the points.
|
[
"2 0 0 1 0\n",
"2 0 1 2 1\n",
"100 0 0 100 100\n"
] |
[
"1\n",
"4\n",
"200\n"
] |
none
| 500
|
[
{
"input": "2 0 0 1 0",
"output": "1"
},
{
"input": "2 0 1 2 1",
"output": "4"
},
{
"input": "100 0 0 100 100",
"output": "200"
},
{
"input": "4 0 3 1 4",
"output": "2"
},
{
"input": "10 8 10 10 0",
"output": "12"
},
{
"input": "26 21 0 26 14",
"output": "19"
},
{
"input": "15 0 1 11 0",
"output": "12"
},
{
"input": "26 26 7 26 12",
"output": "5"
},
{
"input": "6 6 0 2 6",
"output": "10"
},
{
"input": "5 1 5 2 5",
"output": "1"
},
{
"input": "99 12 0 35 99",
"output": "146"
},
{
"input": "44 44 31 28 0",
"output": "47"
},
{
"input": "42 42 36 5 0",
"output": "73"
},
{
"input": "87 87 66 0 5",
"output": "158"
},
{
"input": "85 0 32 0 31",
"output": "1"
},
{
"input": "30 20 30 3 0",
"output": "53"
},
{
"input": "5 4 0 5 1",
"output": "2"
},
{
"input": "40 24 40 4 0",
"output": "68"
},
{
"input": "11 0 2 11 4",
"output": "17"
},
{
"input": "82 0 11 35 0",
"output": "46"
},
{
"input": "32 19 32 0 1",
"output": "50"
},
{
"input": "54 12 0 0 44",
"output": "56"
},
{
"input": "75 42 75 28 0",
"output": "145"
},
{
"input": "48 31 48 0 4",
"output": "75"
},
{
"input": "69 4 69 69 59",
"output": "75"
},
{
"input": "561 0 295 233 0",
"output": "528"
},
{
"input": "341 158 0 0 190",
"output": "348"
},
{
"input": "887 887 461 39 887",
"output": "1274"
},
{
"input": "700 0 288 700 368",
"output": "1356"
},
{
"input": "512 70 512 512 99",
"output": "855"
},
{
"input": "826 188 826 592 0",
"output": "1606"
},
{
"input": "953 0 773 0 903",
"output": "130"
},
{
"input": "80 80 4 0 54",
"output": "138"
},
{
"input": "208 73 0 208 123",
"output": "258"
},
{
"input": "983 0 894 199 0",
"output": "1093"
},
{
"input": "686 615 686 470 686",
"output": "145"
},
{
"input": "869 869 833 0 578",
"output": "1196"
},
{
"input": "169 0 94 0 132",
"output": "38"
},
{
"input": "68 42 68 68 28",
"output": "66"
},
{
"input": "967 967 607 279 0",
"output": "1295"
},
{
"input": "489 489 139 455 489",
"output": "384"
},
{
"input": "964 205 964 604 964",
"output": "399"
},
{
"input": "86 0 34 86 21",
"output": "141"
},
{
"input": "209 166 209 131 0",
"output": "330"
},
{
"input": "684 684 113 314 684",
"output": "941"
},
{
"input": "16 0 6 0 8",
"output": "2"
},
{
"input": "862 154 862 297 862",
"output": "143"
},
{
"input": "418 222 0 254 418",
"output": "778"
},
{
"input": "571 504 571 143 571",
"output": "361"
},
{
"input": "371 371 210 81 371",
"output": "451"
},
{
"input": "1000 0 0 1000 1000",
"output": "2000"
},
{
"input": "1000 564 0 436 1000",
"output": "2000"
},
{
"input": "1000 0 573 12 1000",
"output": "439"
},
{
"input": "1000 984 0 1000 999",
"output": "1015"
},
{
"input": "100 10 0 10 0",
"output": "0"
}
] | 1,638,199,739
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 55
| 156
| 716,800
|
### Codeforces handle - "@titanalpha"
#Email id - [email protected]
from collections import Counter, deque
import math
from random import randint
from bisect import bisect
import sys
import re
import copy
from decimal import Decimal,getcontext
getcontext().prec = 32
MOD = 1000000007
MAX = sys.maxsize
MIN = -sys.maxsize
prime = []
class get:
def int():
return list(map(int,input().split()))
def str():
return input().split()
def float():
return list(map(float,input().split()))
def map():
return map(int,input().split())
def sqrt(x):
return int(math.sqrt(x))+1
def gcd(*args):
result = args[0]
for item in args:
result = math.gcd(result,item)
return result
def LCM(a,b):
return (a*b)//(math.gcd(a,b))
def isPrime(n):
if n == 1:
return False
for i in range(2,sqrt(n)):
if n%i == 0:
return False
return True
def solve():
n,x1,y1,x2,y2 = get.int()
if x1 == x2 and x1 in [0,n]:
print(abs(y1-y2))
return
if y1 == y2 and y1 in [0,n]:
print(abs(x1-x2))
return
if (x1 == 0 and y2 == 0) or (x2 == 0 and y1 == 0):
print(y1+x2+x1+y2)
return
if x1 == 0 and y2 == n:
print(n-y1+x2)
return
if x2 == 0 and y1 == n:
print(n-y2+x1)
return
if y1 == n and x2 == n:
print(n-x1+n-y2)
return
if y2 == n and x1 == n:
print(n-x2+n-y1)
return
if y1 == 0 and x2 == n:
print(y2 + n-x1)
return
if y2 == 0 and x1 == n:
print(y1+n-x2)
return
if (x1 == 0 and x2 == n) or (x2 == 0 and x1 == n):
k1 = n + n - y1 + n - y2
k2 = n + y1+y2
print(min(k1,k2))
return
if (y1 == 0 and y2 == n) or (y2 == 0 and y1 == n):
k1 = n + n - x1+n-x2
k2 = n+x1+x2
print(min(k1,k2))
return
#for _ in range(int(input())):
solve()
|
Title: Square Earth?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Meg the Rabbit decided to do something nice, specifically — to determine the shortest distance between two points on the surface of our planet. But Meg... what can you say, she wants everything simple. So, she already regards our planet as a two-dimensional circle. No, wait, it's even worse — as a square of side *n*. Thus, the task has been reduced to finding the shortest path between two dots on a square (the path should go through the square sides). To simplify the task let us consider the vertices of the square to lie at points whose coordinates are: (0,<=0), (*n*,<=0), (0,<=*n*) and (*n*,<=*n*).
Input Specification:
The single line contains 5 space-separated integers: *n*,<=*x*1,<=*y*1,<=*x*2,<=*y*2 (1<=≤<=*n*<=≤<=1000,<=0<=≤<=*x*1,<=*y*1,<=*x*2,<=*y*2<=≤<=*n*) which correspondingly represent a side of the square, the coordinates of the first point and the coordinates of the second point. It is guaranteed that the points lie on the sides of the square.
Output Specification:
You must print on a single line the shortest distance between the points.
Demo Input:
['2 0 0 1 0\n', '2 0 1 2 1\n', '100 0 0 100 100\n']
Demo Output:
['1\n', '4\n', '200\n']
Note:
none
|
```python
### Codeforces handle - "@titanalpha"
#Email id - [email protected]
from collections import Counter, deque
import math
from random import randint
from bisect import bisect
import sys
import re
import copy
from decimal import Decimal,getcontext
getcontext().prec = 32
MOD = 1000000007
MAX = sys.maxsize
MIN = -sys.maxsize
prime = []
class get:
def int():
return list(map(int,input().split()))
def str():
return input().split()
def float():
return list(map(float,input().split()))
def map():
return map(int,input().split())
def sqrt(x):
return int(math.sqrt(x))+1
def gcd(*args):
result = args[0]
for item in args:
result = math.gcd(result,item)
return result
def LCM(a,b):
return (a*b)//(math.gcd(a,b))
def isPrime(n):
if n == 1:
return False
for i in range(2,sqrt(n)):
if n%i == 0:
return False
return True
def solve():
n,x1,y1,x2,y2 = get.int()
if x1 == x2 and x1 in [0,n]:
print(abs(y1-y2))
return
if y1 == y2 and y1 in [0,n]:
print(abs(x1-x2))
return
if (x1 == 0 and y2 == 0) or (x2 == 0 and y1 == 0):
print(y1+x2+x1+y2)
return
if x1 == 0 and y2 == n:
print(n-y1+x2)
return
if x2 == 0 and y1 == n:
print(n-y2+x1)
return
if y1 == n and x2 == n:
print(n-x1+n-y2)
return
if y2 == n and x1 == n:
print(n-x2+n-y1)
return
if y1 == 0 and x2 == n:
print(y2 + n-x1)
return
if y2 == 0 and x1 == n:
print(y1+n-x2)
return
if (x1 == 0 and x2 == n) or (x2 == 0 and x1 == n):
k1 = n + n - y1 + n - y2
k2 = n + y1+y2
print(min(k1,k2))
return
if (y1 == 0 and y2 == n) or (y2 == 0 and y1 == n):
k1 = n + n - x1+n-x2
k2 = n+x1+x2
print(min(k1,k2))
return
#for _ in range(int(input())):
solve()
```
| 3.959665
|
691
|
A
|
Fashion in Berland
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with *n* buttons. Determine if it is fastened in a right way.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of buttons on the jacket.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1). The number *a**i*<==<=0 if the *i*-th button is not fastened. Otherwise *a**i*<==<=1.
|
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
|
[
"3\n1 0 1\n",
"3\n1 0 0\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "3\n1 0 1",
"output": "YES"
},
{
"input": "3\n1 0 0",
"output": "NO"
},
{
"input": "3\n1 1 0",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "NO"
},
{
"input": "3\n0 0 1",
"output": "NO"
},
{
"input": "3\n0 0 0",
"output": "NO"
},
{
"input": "3\n0 1 1",
"output": "YES"
},
{
"input": "3\n0 1 0",
"output": "NO"
},
{
"input": "2\n0 1",
"output": "YES"
},
{
"input": "2\n1 0",
"output": "YES"
},
{
"input": "2\n0 0",
"output": "NO"
},
{
"input": "2\n1 1",
"output": "NO"
},
{
"input": "1\n1",
"output": "YES"
},
{
"input": "1\n0",
"output": "NO"
},
{
"input": "10\n1 0 1 1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "5\n1 1 1 1 0",
"output": "YES"
},
{
"input": "4\n1 0 1 0",
"output": "NO"
},
{
"input": "4\n1 1 0 0",
"output": "NO"
},
{
"input": "10\n0 0 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "123\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "124\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "94\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "64\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "34\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "35\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1",
"output": "NO"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "NO"
},
{
"input": "231\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "NO"
},
{
"input": "10\n1 1 0 1 1 1 0 1 1 1",
"output": "NO"
}
] | 1,554,298,782
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 93
| 0
|
l = []
n = int(input())
l = map(int, raw_input().split())
if n == 1:
if l[0] == 1:
print("YES")
else:
print("NO")
else:
cnt = 0
for i in range(n):
if l[i] == 0:
cnt = cnt + 1
if cnt == 1:
print("YES")
else:
print("NO")
|
Title: Fashion in Berland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with *n* buttons. Determine if it is fastened in a right way.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of buttons on the jacket.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1). The number *a**i*<==<=0 if the *i*-th button is not fastened. Otherwise *a**i*<==<=1.
Output Specification:
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Demo Input:
['3\n1 0 1\n', '3\n1 0 0\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
l = []
n = int(input())
l = map(int, raw_input().split())
if n == 1:
if l[0] == 1:
print("YES")
else:
print("NO")
else:
cnt = 0
for i in range(n):
if l[i] == 0:
cnt = cnt + 1
if cnt == 1:
print("YES")
else:
print("NO")
```
| -1
|
|
911
|
A
|
Nearest Minimums
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
You are given an array of *n* integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.
|
The first line contains positive integer *n* (2<=≤<=*n*<=≤<=105) — size of the given array. The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=109) — elements of the array. It is guaranteed that in the array a minimum occurs at least two times.
|
Print the only number — distance between two nearest minimums in the array.
|
[
"2\n3 3\n",
"3\n5 6 5\n",
"9\n2 1 3 5 4 1 2 3 1\n"
] |
[
"1\n",
"2\n",
"3\n"
] |
none
| 0
|
[
{
"input": "2\n3 3",
"output": "1"
},
{
"input": "3\n5 6 5",
"output": "2"
},
{
"input": "9\n2 1 3 5 4 1 2 3 1",
"output": "3"
},
{
"input": "6\n4 6 7 8 6 4",
"output": "5"
},
{
"input": "2\n1000000000 1000000000",
"output": "1"
},
{
"input": "42\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "2\n10000000 10000000",
"output": "1"
},
{
"input": "5\n100000000 100000001 100000000 100000001 100000000",
"output": "2"
},
{
"input": "9\n4 3 4 3 4 1 3 3 1",
"output": "3"
},
{
"input": "3\n10000000 1000000000 10000000",
"output": "2"
},
{
"input": "12\n5 6 6 5 6 1 9 9 9 9 9 1",
"output": "6"
},
{
"input": "5\n5 5 1 2 1",
"output": "2"
},
{
"input": "5\n2 2 1 3 1",
"output": "2"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "3\n100000005 1000000000 100000005",
"output": "2"
},
{
"input": "5\n1 2 2 2 1",
"output": "4"
},
{
"input": "3\n10000 1000000 10000",
"output": "2"
},
{
"input": "3\n999999999 999999998 999999998",
"output": "1"
},
{
"input": "6\n2 1 1 2 3 4",
"output": "1"
},
{
"input": "4\n1000000000 900000000 900000000 1000000000",
"output": "1"
},
{
"input": "5\n7 7 2 7 2",
"output": "2"
},
{
"input": "6\n10 10 1 20 20 1",
"output": "3"
},
{
"input": "2\n999999999 999999999",
"output": "1"
},
{
"input": "10\n100000 100000 1 2 3 4 5 6 7 1",
"output": "7"
},
{
"input": "10\n3 3 1 2 2 1 10 10 10 10",
"output": "3"
},
{
"input": "5\n900000000 900000001 900000000 900000001 900000001",
"output": "2"
},
{
"input": "5\n3 3 2 5 2",
"output": "2"
},
{
"input": "2\n100000000 100000000",
"output": "1"
},
{
"input": "10\n10 15 10 2 54 54 54 54 2 10",
"output": "5"
},
{
"input": "2\n999999 999999",
"output": "1"
},
{
"input": "6\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "5\n1000000000 100000000 1000000000 1000000000 100000000",
"output": "3"
},
{
"input": "4\n10 9 10 9",
"output": "2"
},
{
"input": "5\n1 3 2 3 1",
"output": "4"
},
{
"input": "5\n2 2 1 4 1",
"output": "2"
},
{
"input": "6\n1 2 2 2 2 1",
"output": "5"
},
{
"input": "7\n3 7 6 7 6 7 3",
"output": "6"
},
{
"input": "8\n1 2 2 2 2 1 2 2",
"output": "5"
},
{
"input": "10\n2 2 2 3 3 1 3 3 3 1",
"output": "4"
},
{
"input": "2\n88888888 88888888",
"output": "1"
},
{
"input": "3\n100000000 100000000 100000000",
"output": "1"
},
{
"input": "10\n1 3 2 4 5 5 4 3 2 1",
"output": "9"
},
{
"input": "5\n2 2 1 2 1",
"output": "2"
},
{
"input": "6\n900000005 900000000 900000001 900000000 900000001 900000001",
"output": "2"
},
{
"input": "5\n41 41 1 41 1",
"output": "2"
},
{
"input": "6\n5 5 1 3 3 1",
"output": "3"
},
{
"input": "8\n1 2 2 2 1 2 2 2",
"output": "4"
},
{
"input": "7\n6 6 6 6 1 8 1",
"output": "2"
},
{
"input": "3\n999999999 1000000000 999999999",
"output": "2"
},
{
"input": "5\n5 5 4 10 4",
"output": "2"
},
{
"input": "11\n2 2 3 4 1 5 3 4 2 5 1",
"output": "6"
},
{
"input": "5\n3 5 4 5 3",
"output": "4"
},
{
"input": "6\n6 6 6 6 1 1",
"output": "1"
},
{
"input": "7\n11 1 3 2 3 1 11",
"output": "4"
},
{
"input": "5\n3 3 1 2 1",
"output": "2"
},
{
"input": "5\n4 4 2 5 2",
"output": "2"
},
{
"input": "4\n10000099 10000567 10000099 10000234",
"output": "2"
},
{
"input": "4\n100000009 100000011 100000012 100000009",
"output": "3"
},
{
"input": "2\n1000000 1000000",
"output": "1"
},
{
"input": "2\n10000010 10000010",
"output": "1"
},
{
"input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "8\n2 6 2 8 1 9 8 1",
"output": "3"
},
{
"input": "5\n7 7 1 8 1",
"output": "2"
},
{
"input": "7\n1 3 2 3 2 3 1",
"output": "6"
},
{
"input": "7\n2 3 2 1 3 4 1",
"output": "3"
},
{
"input": "5\n1000000000 999999999 1000000000 1000000000 999999999",
"output": "3"
},
{
"input": "4\n1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "5\n5 5 3 5 3",
"output": "2"
},
{
"input": "6\n2 3 3 3 3 2",
"output": "5"
},
{
"input": "4\n1 1 2 2",
"output": "1"
},
{
"input": "5\n1 1 2 2 2",
"output": "1"
},
{
"input": "6\n2 1 1 2 2 2",
"output": "1"
},
{
"input": "5\n1000000000 1000000000 100000000 1000000000 100000000",
"output": "2"
},
{
"input": "7\n2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "8\n2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "10\n2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "11\n2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "12\n2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "13\n2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "14\n2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "15\n2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "16\n2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "17\n2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "18\n2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "19\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "20\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "4\n1000000000 100000000 100000000 1000000000",
"output": "1"
},
{
"input": "21\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "4\n1 2 3 1",
"output": "3"
},
{
"input": "8\n5 5 5 5 3 5 5 3",
"output": "3"
},
{
"input": "7\n2 3 2 1 4 4 1",
"output": "3"
},
{
"input": "6\n3 3 1 2 4 1",
"output": "3"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "5\n3 3 2 8 2",
"output": "2"
},
{
"input": "5\n1 2 1 2 2",
"output": "2"
},
{
"input": "4\n1 2 1 2",
"output": "2"
},
{
"input": "5\n3 1 1 3 2",
"output": "1"
},
{
"input": "4\n1 1 2 1",
"output": "1"
},
{
"input": "4\n2 2 1 1",
"output": "1"
},
{
"input": "5\n1 2 2 1 2",
"output": "3"
},
{
"input": "7\n2 1 2 1 1 2 1",
"output": "1"
},
{
"input": "9\n200000 500000 500000 500000 200000 500000 500000 500000 500000",
"output": "4"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "85\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 1",
"output": "84"
},
{
"input": "5\n1000000000 1000000000 999999999 1000000000 999999999",
"output": "2"
},
{
"input": "5\n2 1 2 2 1",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n1 2 1 1",
"output": "1"
},
{
"input": "6\n1 3 4 2 4 1",
"output": "5"
},
{
"input": "9\n2 2 5 1 6 8 7 9 1",
"output": "5"
},
{
"input": "10\n1000000000 1000000000 1000000000 999999999 1000000000 1000000000 1000000000 1000000000 1000000000 999999999",
"output": "6"
},
{
"input": "7\n3 3 1 2 4 1 2",
"output": "3"
},
{
"input": "7\n3 3 1 2 3 4 1",
"output": "4"
},
{
"input": "8\n10 5 10 1 10 10 10 1",
"output": "4"
}
] | 1,516,622,681
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 140
| 124
| 14,438,400
|
n = int(input())
l = list(map(int,input().split()))
x = min(l)
k = l.index(x)
dmin = n+1
for i in range(k+1,n):
if l[i] == x:
d = i - k
k = i
if d < dmin:
dmin = d
print(dmin)
|
Title: Nearest Minimums
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of *n* integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.
Input Specification:
The first line contains positive integer *n* (2<=≤<=*n*<=≤<=105) — size of the given array. The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=109) — elements of the array. It is guaranteed that in the array a minimum occurs at least two times.
Output Specification:
Print the only number — distance between two nearest minimums in the array.
Demo Input:
['2\n3 3\n', '3\n5 6 5\n', '9\n2 1 3 5 4 1 2 3 1\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
none
|
```python
n = int(input())
l = list(map(int,input().split()))
x = min(l)
k = l.index(x)
dmin = n+1
for i in range(k+1,n):
if l[i] == x:
d = i - k
k = i
if d < dmin:
dmin = d
print(dmin)
```
| 3
|
|
935
|
A
|
Fafa and his Company
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
|
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
|
Print a single integer representing the answer to the problem.
|
[
"2\n",
"10\n"
] |
[
"1\n",
"3\n"
] |
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
| 500
|
[
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
},
{
"input": "1024",
"output": "10"
},
{
"input": "99999",
"output": "11"
},
{
"input": "10007",
"output": "1"
},
{
"input": "4096",
"output": "12"
},
{
"input": "65536",
"output": "16"
},
{
"input": "40320",
"output": "95"
},
{
"input": "30030",
"output": "63"
},
{
"input": "161",
"output": "3"
},
{
"input": "1000",
"output": "15"
},
{
"input": "10000",
"output": "24"
},
{
"input": "777",
"output": "7"
},
{
"input": "121",
"output": "2"
},
{
"input": "25",
"output": "2"
},
{
"input": "40000",
"output": "34"
},
{
"input": "99990",
"output": "47"
},
{
"input": "98765",
"output": "3"
},
{
"input": "56789",
"output": "3"
},
{
"input": "13579",
"output": "3"
},
{
"input": "97531",
"output": "3"
},
{
"input": "12345",
"output": "7"
},
{
"input": "54321",
"output": "7"
},
{
"input": "83160",
"output": "127"
},
{
"input": "9",
"output": "2"
},
{
"input": "21",
"output": "3"
},
{
"input": "11",
"output": "1"
},
{
"input": "15",
"output": "3"
}
] | 1,682,811,035
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 61
| 0
|
n = int(input())
c_work = 0
if n % 2 == 1:
m = n / 2 - 0.5
m = int(m)
else:
m = n / 2
m = int(m)
for i in range(1, m + 1):
x = i
n = n - 1
if n % x == 0:
c_work += 1
print(str(c_work))
|
Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
|
```python
n = int(input())
c_work = 0
if n % 2 == 1:
m = n / 2 - 0.5
m = int(m)
else:
m = n / 2
m = int(m)
for i in range(1, m + 1):
x = i
n = n - 1
if n % x == 0:
c_work += 1
print(str(c_work))
```
| 3
|
|
261
|
B
|
Maxim and Restaurant
|
PROGRAMMING
| 1,900
|
[
"dp",
"math",
"probabilities"
] | null | null |
Maxim has opened his own restaurant! The restaurant has got a huge table, the table's length is *p* meters.
Maxim has got a dinner party tonight, *n* guests will come to him. Let's index the guests of Maxim's restaurant from 1 to *n*. Maxim knows the sizes of all guests that are going to come to him. The *i*-th guest's size (*a**i*) represents the number of meters the guest is going to take up if he sits at the restaurant table.
Long before the dinner, the guests line up in a queue in front of the restaurant in some order. Then Maxim lets the guests in, one by one. Maxim stops letting the guests in when there is no place at the restaurant table for another guest in the queue. There is no place at the restaurant table for another guest in the queue, if the sum of sizes of all guests in the restaurant plus the size of this guest from the queue is larger than *p*. In this case, not to offend the guest who has no place at the table, Maxim doesn't let any other guest in the restaurant, even if one of the following guests in the queue would have fit in at the table.
Maxim is now wondering, what is the average number of visitors who have come to the restaurant for all possible *n*! orders of guests in the queue. Help Maxim, calculate this number.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of guests in the restaurant. The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=50) — the guests' sizes in meters. The third line contains integer *p* (1<=≤<=*p*<=≤<=50) — the table's length in meters.
The numbers in the lines are separated by single spaces.
|
In a single line print a real number — the answer to the problem. The answer will be considered correct, if the absolute or relative error doesn't exceed 10<=-<=4.
|
[
"3\n1 2 3\n3\n"
] |
[
"1.3333333333\n"
] |
In the first sample the people will come in the following orders:
- (1, 2, 3) — there will be two people in the restaurant; - (1, 3, 2) — there will be one person in the restaurant; - (2, 1, 3) — there will be two people in the restaurant; - (2, 3, 1) — there will be one person in the restaurant; - (3, 1, 2) — there will be one person in the restaurant; - (3, 2, 1) — there will be one person in the restaurant.
In total we get (2 + 1 + 2 + 1 + 1 + 1) / 6 = 8 / 6 = 1.(3).
| 1,000
|
[
{
"input": "3\n1 2 3\n3",
"output": "1.3333333333"
},
{
"input": "9\n2 2 2 2 2 2 2 1 2\n9",
"output": "4.5555555556"
},
{
"input": "7\n2 1 1 2 1 1 2\n2",
"output": "1.2857142857"
},
{
"input": "8\n3 1 5 6 1 5 4 4\n7",
"output": "1.6250000000"
},
{
"input": "2\n1 3\n3",
"output": "1.0000000000"
},
{
"input": "2\n1 2\n2",
"output": "1.0000000000"
},
{
"input": "5\n1 2 3 1 2\n3",
"output": "1.5000000000"
},
{
"input": "9\n1 2 2 7 4 4 1 4 7\n7",
"output": "1.7341269841"
},
{
"input": "6\n1 1 1 1 1 1\n1",
"output": "1.0000000000"
},
{
"input": "10\n35 5 7 41 17 27 32 9 45 40\n30",
"output": "0.6500000000"
},
{
"input": "27\n38 39 19 33 30 42 34 16 40 9 5 31 28 7 24 37 22 46 25 23 21 30 28 24 48 13 37\n24",
"output": "0.4333903134"
},
{
"input": "41\n37 6 18 6 25 32 3 1 1 42 25 17 31 8 42 8 38 8 38 4 34 46 10 10 9 22 39 23 47 7 31 14 19 1 42 13 6 11 10 25 38\n12",
"output": "0.5001534565"
},
{
"input": "49\n46 42 3 1 42 37 25 21 47 22 49 50 19 35 32 35 4 50 19 39 1 39 28 18 29 44 49 34 8 22 11 18 14 15 10 17 36 2 1 50 20 7 49 4 25 9 45 10 40\n34",
"output": "0.9154259295"
},
{
"input": "3\n36 44 44\n46",
"output": "1.0000000000"
},
{
"input": "24\n15 4 49 1 9 19 31 47 49 32 40 49 10 8 23 23 39 43 39 30 41 8 9 42\n38",
"output": "0.8604837192"
},
{
"input": "16\n3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4\n39",
"output": "12.3395604396"
},
{
"input": "23\n2 2 1 1 2 2 1 1 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1\n2",
"output": "1.1778656126"
},
{
"input": "18\n2 1 2 2 1 2 1 1 2 2 1 2 2 1 2 1 1 2\n8",
"output": "4.9849398967"
},
{
"input": "40\n5 8 2 14 14 19 14 2 12 10 17 15 2 9 11 24 7 19 14 21 8 3 24 18 20 10 14 1 9 9 18 18 13 2 23 7 23 2 17 13\n24",
"output": "1.6715713966"
},
{
"input": "23\n16 21 14 27 15 30 13 10 4 15 25 21 6 10 17 4 5 3 9 9 8 6 19\n30",
"output": "1.9401705846"
},
{
"input": "42\n3 2 3 1 1 3 1 3 2 3 3 2 1 3 2 3 3 2 3 3 1 3 3 2 3 2 3 1 2 2 2 3 3 1 2 1 1 3 1 3 3 2\n3",
"output": "1.2020905923"
},
{
"input": "23\n2 1 2 1 1 1 2 2 2 1 1 2 2 1 1 1 2 1 2 2 1 1 1\n37",
"output": "23.0000000000"
},
{
"input": "3\n2 2 1\n22",
"output": "3.0000000000"
},
{
"input": "19\n8 11 13 11 7 11 3 11 6 7 3 14 4 10 10 8 2 1 8\n15",
"output": "1.6128310974"
},
{
"input": "28\n3 5 4 24 21 3 13 24 22 13 12 21 1 15 11 3 17 6 2 12 22 12 23 4 21 16 25 14\n25",
"output": "1.6461894466"
},
{
"input": "14\n8 13 28 2 17 31 2 11 39 12 24 3 8 10\n41",
"output": "2.4931734932"
},
{
"input": "8\n9 14 13 2 1 11 4 19\n25",
"output": "2.3500000000"
},
{
"input": "35\n5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3\n6",
"output": "1.9851721334"
},
{
"input": "35\n2 2 1 2 2 2 2 2 2 1 2 1 2 1 1 2 2 2 2 2 2 1 1 1 2 2 1 1 2 2 2 1 2 1 1\n35",
"output": "21.2873098934"
},
{
"input": "44\n24 19 6 4 23 10 11 16 21 15 18 17 13 9 25 3 1 11 24 26 12 12 21 17 19 2 6 24 21 18 7 2 12 2 4 25 17 26 22 10 22 11 13 27\n27",
"output": "1.5513891043"
},
{
"input": "36\n5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1\n10",
"output": "2.9649127243"
},
{
"input": "38\n2 4 2 4 1 2 5 1 5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1\n48",
"output": "15.0079078318"
},
{
"input": "40\n1 26 39 14 16 17 19 28 38 18 23 41 19 22 4 24 18 36 15 21 31 29 34 13 19 19 38 45 4 10 2 14 3 24 21 27 4 30 9 17\n45",
"output": "1.8507376624"
},
{
"input": "41\n31 21 49 18 37 34 36 27 36 39 4 30 25 49 24 10 8 17 45 6 19 27 12 26 6 2 50 47 35 16 15 43 26 14 43 47 49 23 27 7 24\n50",
"output": "1.5535424434"
},
{
"input": "30\n2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2 3 2 1 1 3 5 4 4\n5",
"output": "1.8614767098"
},
{
"input": "50\n2 1 2 1 2 1 1 1 2 2 2 1 1 1 1 1 1 2 2 1 2 1 2 2 1 2 2 1 1 2 1 1 1 2 2 2 1 2 1 2 2 2 2 2 1 1 2 2 1 2\n3",
"output": "1.8379591837"
},
{
"input": "50\n15 28 34 29 17 21 20 34 37 17 10 20 37 10 18 25 31 25 16 1 37 27 39 3 5 18 2 32 10 35 20 17 29 20 3 29 3 25 9 32 37 5 25 23 25 33 35 8 31 29\n39",
"output": "1.4997987526"
},
{
"input": "50\n1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2\n29",
"output": "9.8873093486"
},
{
"input": "50\n3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5 4 4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4\n19",
"output": "5.5762635183"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n25",
"output": "25.0000000000"
},
{
"input": "10\n42 18 35 1 20 25 29 9 50 36\n50",
"output": "1.5269841270"
},
{
"input": "7\n42 35 1 20 29 50 36\n50",
"output": "1.3142857143"
},
{
"input": "50\n1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 5 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7\n50",
"output": "12.0011471293"
},
{
"input": "1\n1\n1",
"output": "1.0000000000"
},
{
"input": "1\n2\n1",
"output": "0.0000000000"
},
{
"input": "5\n2 3 2 3 6\n30",
"output": "5.0000000000"
},
{
"input": "3\n1 2 3\n7",
"output": "3.0000000000"
},
{
"input": "3\n1 1 1\n50",
"output": "3.0000000000"
},
{
"input": "4\n1 2 3 4\n11",
"output": "4.0000000000"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n50",
"output": "50.0000000000"
},
{
"input": "50\n1 2 3 4 4 4 4 4 4 4 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43\n50",
"output": "2.3167627104"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n20",
"output": "20.0000000000"
},
{
"input": "2\n1 2\n3",
"output": "2.0000000000"
},
{
"input": "3\n1 2 3\n10",
"output": "3.0000000000"
},
{
"input": "5\n1 2 3 4 5\n20",
"output": "5.0000000000"
}
] | 1,531,925,913
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <bits/stdc++.h>
const int N = 55;
using namespace std;
using ll = long long;
int n, a[N], p;
ll dp[N][N];
double fac[N];
int main() {
scanf("%d", &n);fac[1] = fac[0] = 1;
for (int i = 2; i < N; i++) fac[i] = (fac[i-1])*i;
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
} scanf("%d", &p); dp[0][0] = 1;
for (int i = 0; i < n; i++) {
for (int j = p; j >= a[i]; j--) {
for (int k = n; k >= 1; k--) {
dp[j][k] += dp[j-a[i]][k-1];
}
}
} double ans = 0;
for (int i = 1; i <= p; i++) {
for (int j = 1; j <= n; j++) {
ans += fac[j]*fac[n-j]*dp[i][j];
}
} return printf("%.12f\n", ans/fac[n]), 0;
}
|
Title: Maxim and Restaurant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Maxim has opened his own restaurant! The restaurant has got a huge table, the table's length is *p* meters.
Maxim has got a dinner party tonight, *n* guests will come to him. Let's index the guests of Maxim's restaurant from 1 to *n*. Maxim knows the sizes of all guests that are going to come to him. The *i*-th guest's size (*a**i*) represents the number of meters the guest is going to take up if he sits at the restaurant table.
Long before the dinner, the guests line up in a queue in front of the restaurant in some order. Then Maxim lets the guests in, one by one. Maxim stops letting the guests in when there is no place at the restaurant table for another guest in the queue. There is no place at the restaurant table for another guest in the queue, if the sum of sizes of all guests in the restaurant plus the size of this guest from the queue is larger than *p*. In this case, not to offend the guest who has no place at the table, Maxim doesn't let any other guest in the restaurant, even if one of the following guests in the queue would have fit in at the table.
Maxim is now wondering, what is the average number of visitors who have come to the restaurant for all possible *n*! orders of guests in the queue. Help Maxim, calculate this number.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of guests in the restaurant. The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=50) — the guests' sizes in meters. The third line contains integer *p* (1<=≤<=*p*<=≤<=50) — the table's length in meters.
The numbers in the lines are separated by single spaces.
Output Specification:
In a single line print a real number — the answer to the problem. The answer will be considered correct, if the absolute or relative error doesn't exceed 10<=-<=4.
Demo Input:
['3\n1 2 3\n3\n']
Demo Output:
['1.3333333333\n']
Note:
In the first sample the people will come in the following orders:
- (1, 2, 3) — there will be two people in the restaurant; - (1, 3, 2) — there will be one person in the restaurant; - (2, 1, 3) — there will be two people in the restaurant; - (2, 3, 1) — there will be one person in the restaurant; - (3, 1, 2) — there will be one person in the restaurant; - (3, 2, 1) — there will be one person in the restaurant.
In total we get (2 + 1 + 2 + 1 + 1 + 1) / 6 = 8 / 6 = 1.(3).
|
```python
#include <bits/stdc++.h>
const int N = 55;
using namespace std;
using ll = long long;
int n, a[N], p;
ll dp[N][N];
double fac[N];
int main() {
scanf("%d", &n);fac[1] = fac[0] = 1;
for (int i = 2; i < N; i++) fac[i] = (fac[i-1])*i;
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
} scanf("%d", &p); dp[0][0] = 1;
for (int i = 0; i < n; i++) {
for (int j = p; j >= a[i]; j--) {
for (int k = n; k >= 1; k--) {
dp[j][k] += dp[j-a[i]][k-1];
}
}
} double ans = 0;
for (int i = 1; i <= p; i++) {
for (int j = 1; j <= n; j++) {
ans += fac[j]*fac[n-j]*dp[i][j];
}
} return printf("%.12f\n", ans/fac[n]), 0;
}
```
| -1
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,670,616,279
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
n = int(input()) # number of numbers in the input list
numbers = list(map(int, input().split())) # input list of numbers
# initialize a variable to track whether all the numbers seen so far are even or odd
evenness = None
# iterate over the numbers in the input list
for i in range(n):
# check if the current number is even or odd
if numbers[i] % 2 == 0:
# if the current number is even, check if all the previous numbers were odd
if evenness == False:
# if all the previous numbers were odd, print the index of the current number and stop the loop
print(i+1)
break
else:
# if not all the previous numbers were odd, update the evenness tracker to reflect that the current number is even
evenness = True
else:
# if the current number is odd, check if all the previous numbers were even
if evenness == True:
# if all the previous numbers were even, print the index of the current number and stop the loop
print(i+1)
break
else:
# if not all the previous numbers were even, update the evenness tracker to reflect that the current number is odd
evenness = False
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n = int(input()) # number of numbers in the input list
numbers = list(map(int, input().split())) # input list of numbers
# initialize a variable to track whether all the numbers seen so far are even or odd
evenness = None
# iterate over the numbers in the input list
for i in range(n):
# check if the current number is even or odd
if numbers[i] % 2 == 0:
# if the current number is even, check if all the previous numbers were odd
if evenness == False:
# if all the previous numbers were odd, print the index of the current number and stop the loop
print(i+1)
break
else:
# if not all the previous numbers were odd, update the evenness tracker to reflect that the current number is even
evenness = True
else:
# if the current number is odd, check if all the previous numbers were even
if evenness == True:
# if all the previous numbers were even, print the index of the current number and stop the loop
print(i+1)
break
else:
# if not all the previous numbers were even, update the evenness tracker to reflect that the current number is odd
evenness = False
```
| 0
|
491
|
A
|
Up the hill
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation"
] | null | null |
Hiking club "Up the hill" just returned from a walk. Now they are trying to remember which hills they've just walked through.
It is known that there were *N* stops, all on different integer heights between 1 and *N* kilometers (inclusive) above the sea level. On the first day they've traveled from the first stop to the second stop, on the second day they've traveled from the second to the third and so on, and on the last day they've traveled from the stop *N*<=-<=1 to the stop *N* and successfully finished their expedition.
They are trying to find out which heights were their stops located at. They have an entry in a travel journal specifying how many days did they travel up the hill, and how many days did they walk down the hill.
Help them by suggesting some possible stop heights satisfying numbers from the travel journal.
|
In the first line there is an integer non-negative number *A* denoting the number of days of climbing up the hill. Second line contains an integer non-negative number *B* — the number of days of walking down the hill (*A*<=+<=*B*<=+<=1<==<=*N*, 1<=≤<=*N*<=≤<=100<=000).
|
Output *N* space-separated distinct integers from 1 to *N* inclusive, denoting possible heights of the stops in order of visiting.
|
[
"0\n1\n",
"2\n1"
] |
[
"2 1 \n",
"1 3 4 2"
] |
none
| 500
|
[
{
"input": "0\n1",
"output": "2 1 "
},
{
"input": "2\n1",
"output": "2 3 4 1 "
},
{
"input": "0\n3",
"output": "4 3 2 1 "
},
{
"input": "1\n1",
"output": "2 3 1 "
},
{
"input": "3\n7",
"output": "8 9 10 11 7 6 5 4 3 2 1 "
},
{
"input": "700\n300",
"output": "301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428..."
},
{
"input": "37\n29",
"output": "30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "177\n191",
"output": "192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319..."
},
{
"input": "50000\n3",
"output": "4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 1..."
},
{
"input": "99999\n0",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "0\n99999",
"output": "100000 99999 99998 99997 99996 99995 99994 99993 99992 99991 99990 99989 99988 99987 99986 99985 99984 99983 99982 99981 99980 99979 99978 99977 99976 99975 99974 99973 99972 99971 99970 99969 99968 99967 99966 99965 99964 99963 99962 99961 99960 99959 99958 99957 99956 99955 99954 99953 99952 99951 99950 99949 99948 99947 99946 99945 99944 99943 99942 99941 99940 99939 99938 99937 99936 99935 99934 99933 99932 99931 99930 99929 99928 99927 99926 99925 99924 99923 99922 99921 99920 99919 99918 99917 99916 ..."
},
{
"input": "24999\n74997",
"output": "74998 74999 75000 75001 75002 75003 75004 75005 75006 75007 75008 75009 75010 75011 75012 75013 75014 75015 75016 75017 75018 75019 75020 75021 75022 75023 75024 75025 75026 75027 75028 75029 75030 75031 75032 75033 75034 75035 75036 75037 75038 75039 75040 75041 75042 75043 75044 75045 75046 75047 75048 75049 75050 75051 75052 75053 75054 75055 75056 75057 75058 75059 75060 75061 75062 75063 75064 75065 75066 75067 75068 75069 75070 75071 75072 75073 75074 75075 75076 75077 75078 75079 75080 75081 75082 7..."
},
{
"input": "17\n61111",
"output": "61112 61113 61114 61115 61116 61117 61118 61119 61120 61121 61122 61123 61124 61125 61126 61127 61128 61129 61111 61110 61109 61108 61107 61106 61105 61104 61103 61102 61101 61100 61099 61098 61097 61096 61095 61094 61093 61092 61091 61090 61089 61088 61087 61086 61085 61084 61083 61082 61081 61080 61079 61078 61077 61076 61075 61074 61073 61072 61071 61070 61069 61068 61067 61066 61065 61064 61063 61062 61061 61060 61059 61058 61057 61056 61055 61054 61053 61052 61051 61050 61049 61048 61047 61046 61045 6..."
},
{
"input": "50021\n40009",
"output": "40010 40011 40012 40013 40014 40015 40016 40017 40018 40019 40020 40021 40022 40023 40024 40025 40026 40027 40028 40029 40030 40031 40032 40033 40034 40035 40036 40037 40038 40039 40040 40041 40042 40043 40044 40045 40046 40047 40048 40049 40050 40051 40052 40053 40054 40055 40056 40057 40058 40059 40060 40061 40062 40063 40064 40065 40066 40067 40068 40069 40070 40071 40072 40073 40074 40075 40076 40077 40078 40079 40080 40081 40082 40083 40084 40085 40086 40087 40088 40089 40090 40091 40092 40093 40094 4..."
},
{
"input": "49999\n49997",
"output": "49998 49999 50000 50001 50002 50003 50004 50005 50006 50007 50008 50009 50010 50011 50012 50013 50014 50015 50016 50017 50018 50019 50020 50021 50022 50023 50024 50025 50026 50027 50028 50029 50030 50031 50032 50033 50034 50035 50036 50037 50038 50039 50040 50041 50042 50043 50044 50045 50046 50047 50048 50049 50050 50051 50052 50053 50054 50055 50056 50057 50058 50059 50060 50061 50062 50063 50064 50065 50066 50067 50068 50069 50070 50071 50072 50073 50074 50075 50076 50077 50078 50079 50080 50081 50082 5..."
},
{
"input": "6777\n57897",
"output": "57898 57899 57900 57901 57902 57903 57904 57905 57906 57907 57908 57909 57910 57911 57912 57913 57914 57915 57916 57917 57918 57919 57920 57921 57922 57923 57924 57925 57926 57927 57928 57929 57930 57931 57932 57933 57934 57935 57936 57937 57938 57939 57940 57941 57942 57943 57944 57945 57946 57947 57948 57949 57950 57951 57952 57953 57954 57955 57956 57957 57958 57959 57960 57961 57962 57963 57964 57965 57966 57967 57968 57969 57970 57971 57972 57973 57974 57975 57976 57977 57978 57979 57980 57981 57982 5..."
},
{
"input": "49999\n49999",
"output": "50000 50001 50002 50003 50004 50005 50006 50007 50008 50009 50010 50011 50012 50013 50014 50015 50016 50017 50018 50019 50020 50021 50022 50023 50024 50025 50026 50027 50028 50029 50030 50031 50032 50033 50034 50035 50036 50037 50038 50039 50040 50041 50042 50043 50044 50045 50046 50047 50048 50049 50050 50051 50052 50053 50054 50055 50056 50057 50058 50059 50060 50061 50062 50063 50064 50065 50066 50067 50068 50069 50070 50071 50072 50073 50074 50075 50076 50077 50078 50079 50080 50081 50082 50083 50084 5..."
},
{
"input": "1\n0",
"output": "1 2 "
},
{
"input": "0\n1",
"output": "2 1 "
},
{
"input": "0\n0",
"output": "1 "
},
{
"input": "2\n0",
"output": "1 2 3 "
},
{
"input": "5\n0",
"output": "1 2 3 4 5 6 "
},
{
"input": "90000\n1",
"output": "2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 1..."
},
{
"input": "100\n4",
"output": "5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 4 3 2 1 "
}
] | 1,620,698,886
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 6,963,200
|
u = int(input())
d = int(input())
n=u+d+1
res = []
res2 = []
if(u==0 and d==0):
print('1')
elif(d==0 and u!=0):
for i in range(1,n+1):
res.append(i)
elif(u==0 and d!=0):
for i in range(n,0,-1):
res.append(i)
else:
res.append(1)
f=1
ct=0
for i in range(2,n+1):
if(ct==d):
f=0
if(f):
res2.append(i)
else:
res.append(i)
ct+=1
res.extend(list(reversed(res2)))
print(res)
|
Title: Up the hill
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hiking club "Up the hill" just returned from a walk. Now they are trying to remember which hills they've just walked through.
It is known that there were *N* stops, all on different integer heights between 1 and *N* kilometers (inclusive) above the sea level. On the first day they've traveled from the first stop to the second stop, on the second day they've traveled from the second to the third and so on, and on the last day they've traveled from the stop *N*<=-<=1 to the stop *N* and successfully finished their expedition.
They are trying to find out which heights were their stops located at. They have an entry in a travel journal specifying how many days did they travel up the hill, and how many days did they walk down the hill.
Help them by suggesting some possible stop heights satisfying numbers from the travel journal.
Input Specification:
In the first line there is an integer non-negative number *A* denoting the number of days of climbing up the hill. Second line contains an integer non-negative number *B* — the number of days of walking down the hill (*A*<=+<=*B*<=+<=1<==<=*N*, 1<=≤<=*N*<=≤<=100<=000).
Output Specification:
Output *N* space-separated distinct integers from 1 to *N* inclusive, denoting possible heights of the stops in order of visiting.
Demo Input:
['0\n1\n', '2\n1']
Demo Output:
['2 1 \n', '1 3 4 2']
Note:
none
|
```python
u = int(input())
d = int(input())
n=u+d+1
res = []
res2 = []
if(u==0 and d==0):
print('1')
elif(d==0 and u!=0):
for i in range(1,n+1):
res.append(i)
elif(u==0 and d!=0):
for i in range(n,0,-1):
res.append(i)
else:
res.append(1)
f=1
ct=0
for i in range(2,n+1):
if(ct==d):
f=0
if(f):
res2.append(i)
else:
res.append(i)
ct+=1
res.extend(list(reversed(res2)))
print(res)
```
| 0
|
|
701
|
C
|
They Are Everywhere
|
PROGRAMMING
| 1,500
|
[
"binary search",
"strings",
"two pointers"
] | null | null |
Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
|
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house.
The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*.
|
Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
|
[
"3\nAaA\n",
"7\nbcAAcbc\n",
"6\naaBCCe\n"
] |
[
"2\n",
"3\n",
"5\n"
] |
In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.
| 1,000
|
[
{
"input": "3\nAaA",
"output": "2"
},
{
"input": "7\nbcAAcbc",
"output": "3"
},
{
"input": "6\naaBCCe",
"output": "5"
},
{
"input": "1\nA",
"output": "1"
},
{
"input": "1\ng",
"output": "1"
},
{
"input": "52\nabcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "52"
},
{
"input": "2\nAA",
"output": "1"
},
{
"input": "4\nqqqE",
"output": "2"
},
{
"input": "10\nrrrrroooro",
"output": "2"
},
{
"input": "15\nOCOCCCCiCOCCCOi",
"output": "3"
},
{
"input": "20\nVEVnVVnWnVEVVnEVBEWn",
"output": "5"
},
{
"input": "25\ncpcyPPjPPcPPPPcppPcPpppcP",
"output": "6"
},
{
"input": "30\nsssssAsesssssssssssssessssssss",
"output": "3"
},
{
"input": "35\ngdXdddgddddddddggggXdbgdggdgddddddb",
"output": "4"
},
{
"input": "40\nIgsggIiIggzgigIIiiIIIiIgIggIzgIiiiggggIi",
"output": "9"
},
{
"input": "45\neteeeeeteaattaeetaetteeettoetettteyeteeeotaae",
"output": "9"
},
{
"input": "50\nlUlUllUlUllllUllllUllllUlUlllUlllUlllllUUlllUUlkUl",
"output": "3"
},
{
"input": "55\nAAAAASAAAASAASAAAAAAAAAAAAASAAAAAAAAAAAAAAAASAAAAAAAAAA",
"output": "2"
},
{
"input": "60\nRRRrSRRRRRRRRRSSRRRSRRRRRRRRrRSRRRRRRRRRRRRRRSRRRRRSSRSRrRRR",
"output": "3"
},
{
"input": "65\nhhMhMhhhhhhhhhhhMhhMMMhhhhBhhhhMhhhhMhhhhhMhhhBhhhhhhhhhhBhhhhhhh",
"output": "5"
},
{
"input": "70\nwAwwwAwwwwwwwwwwwwwwAwAAwwAwwwwwwwwAwAAAwAAwwwwwwwwwAwwwwwwwwwwwwAAwww",
"output": "2"
},
{
"input": "75\niiiXXiiyiiiXyXiiyXiiXiiiiiiXXyiiiiXXiiXiiXifiXiXXiifiiiiiiXfXiyiXXiXiiiiXiX",
"output": "4"
},
{
"input": "80\nSrSrrrrrrrrrrrrrrSSSrrrrrrSrrrrSrrrrrrrrrrSSrrrrrrrrrrrSrrrSrrrrSrrrrSrrrrSSrSSr",
"output": "2"
},
{
"input": "85\nwkMMMwMMkMMMMMMMkkkkMMMMzkkMMwMMkkwMkMwkMMkMMwwMzMMMkkMwwMMMMMMkMMkMzMMMkMMkwMkMMMkMM",
"output": "6"
},
{
"input": "90\nZllZZZyZlZlllZlylllZlllZZllllllllZZllllllllllyylZZyvZvZlllZZlZllZlZlllZyllZyZlllZlllllllZl",
"output": "5"
},
{
"input": "95\nEmuBuEBmmEBBElBlElmmBEmmmEmmEuBEEmummmEmBBBBEWBBBmEEBmmummBBmmlluBBmElmEBEmBmBBmBmuLmEBBmlEBmBu",
"output": "39"
},
{
"input": "100\nfAfhfNNNhfffAfNffNANhffNffffffNfffffhNfffNfffNfffNfNNNffNfffNfhNNNffffNfNfffAffNffNffNNfhfffNhNffNNN",
"output": "4"
},
{
"input": "7\nabcbbab",
"output": "3"
},
{
"input": "65\nkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkjjjjkkkkkkkkkkkkkkkkkkkllllllljk",
"output": "3"
},
{
"input": "9\nbcdddbddc",
"output": "3"
},
{
"input": "71\nlqcrdopdpfagcilpcmheqhjbmnciqmeoabflrenbkrsaabaanpehdmpqcdqnljcrsplnlco",
"output": "36"
},
{
"input": "8\nabbbbbbc",
"output": "8"
}
] | 1,678,480,730
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 93
| 3,481,600
|
N = int(input())
flats = input()
lptr, rptr = 0, 1
letters = set(list(flats))
def checkLetters(word):
for letter in letters:
if letter not in word:
return False
return True
while True:
if flats[lptr] in flats[lptr + 1:rptr]:
lptr += 1
min_rptr = rptr
else:
rptr += 1
if rptr == len(flats): break
print(min_rptr - lptr)
|
Title: They Are Everywhere
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house.
The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*.
Output Specification:
Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
Demo Input:
['3\nAaA\n', '7\nbcAAcbc\n', '6\naaBCCe\n']
Demo Output:
['2\n', '3\n', '5\n']
Note:
In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.
|
```python
N = int(input())
flats = input()
lptr, rptr = 0, 1
letters = set(list(flats))
def checkLetters(word):
for letter in letters:
if letter not in word:
return False
return True
while True:
if flats[lptr] in flats[lptr + 1:rptr]:
lptr += 1
min_rptr = rptr
else:
rptr += 1
if rptr == len(flats): break
print(min_rptr - lptr)
```
| -1
|
|
7
|
A
|
Kalevitch and Chess
|
PROGRAMMING
| 1,100
|
[
"brute force",
"constructive algorithms"
] |
A. Kalevitch and Chess
|
2
|
64
|
A famous Berland's painter Kalevitch likes to shock the public. One of his last obsessions is chess. For more than a thousand years people have been playing this old game on uninteresting, monotonous boards. Kalevitch decided to put an end to this tradition and to introduce a new attitude to chessboards.
As before, the chessboard is a square-checkered board with the squares arranged in a 8<=×<=8 grid, each square is painted black or white. Kalevitch suggests that chessboards should be painted in the following manner: there should be chosen a horizontal or a vertical line of 8 squares (i.e. a row or a column), and painted black. Initially the whole chessboard is white, and it can be painted in the above described way one or more times. It is allowed to paint a square many times, but after the first time it does not change its colour any more and remains black. Kalevitch paints chessboards neatly, and it is impossible to judge by an individual square if it was painted with a vertical or a horizontal stroke.
Kalevitch hopes that such chessboards will gain popularity, and he will be commissioned to paint chessboards, which will help him ensure a comfortable old age. The clients will inform him what chessboard they want to have, and the painter will paint a white chessboard meeting the client's requirements.
It goes without saying that in such business one should economize on everything — for each commission he wants to know the minimum amount of strokes that he has to paint to fulfill the client's needs. You are asked to help Kalevitch with this task.
|
The input file contains 8 lines, each of the lines contains 8 characters. The given matrix describes the client's requirements, W character stands for a white square, and B character — for a square painted black.
It is guaranteed that client's requirments can be fulfilled with a sequence of allowed strokes (vertical/column or horizontal/row).
|
Output the only number — the minimum amount of rows and columns that Kalevitch has to paint on the white chessboard to meet the client's requirements.
|
[
"WWWBWWBW\nBBBBBBBB\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\n",
"WWWWWWWW\nBBBBBBBB\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\n"
] |
[
"3\n",
"1\n"
] |
none
| 0
|
[
{
"input": "WWWBWWBW\nBBBBBBBB\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW",
"output": "3"
},
{
"input": "WWWWWWWW\nBBBBBBBB\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW",
"output": "1"
},
{
"input": "WWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW",
"output": "0"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB",
"output": "8"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBW",
"output": "14"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBWB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB",
"output": "14"
},
{
"input": "BBBBBBBB\nWBBBWBBW\nBBBBBBBB\nWBBBWBBW\nWBBBWBBW\nBBBBBBBB\nBBBBBBBB\nWBBBWBBW",
"output": "9"
},
{
"input": "BBBBBBBB\nWBBWWWBB\nBBBBBBBB\nWBBWWWBB\nBBBBBBBB\nBBBBBBBB\nWBBWWWBB\nBBBBBBBB",
"output": "9"
},
{
"input": "BBBBBWWB\nBBBBBBBB\nBBBBBBBB\nBBBBBWWB\nBBBBBWWB\nBBBBBWWB\nBBBBBWWB\nBBBBBWWB",
"output": "8"
},
{
"input": "WWWWBBBB\nWWWWBBBB\nBBBBBBBB\nBBBBBBBB\nWWWWBBBB\nWWWWBBBB\nBBBBBBBB\nBBBBBBBB",
"output": "8"
},
{
"input": "BBBBBBBB\nWBWWBBBW\nBBBBBBBB\nWBWWBBBW\nWBWWBBBW\nWBWWBBBW\nWBWWBBBW\nBBBBBBBB",
"output": "7"
},
{
"input": "WBWWBBBW\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nWBWWBBBW\nWBWWBBBW",
"output": "9"
},
{
"input": "BBWWBBBW\nBBBBBBBB\nBBBBBBBB\nBBWWBBBW\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB",
"output": "11"
},
{
"input": "WWBWBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nWWBWBBBB\nBBBBBBBB\nWWBWBBBB\nBBBBBBBB",
"output": "10"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBBB\nWWBWBBBB\nWWBWBBBB\nBBBBBBBB\nBBBBBBBB\nWWBWBBBB",
"output": "10"
},
{
"input": "WBBWBBBW\nWBBWBBBW\nWBBWBBBW\nWBBWBBBW\nWBBWBBBW\nBBBBBBBB\nWBBWBBBW\nWBBWBBBW",
"output": "6"
},
{
"input": "BBBWBBBW\nBBBWBBBW\nBBBWBBBW\nBBBBBBBB\nBBBBBBBB\nBBBWBBBW\nBBBBBBBB\nBBBBBBBB",
"output": "10"
},
{
"input": "BBBBBBBB\nBBBWBBBB\nBBBWBBBB\nBBBWBBBB\nBBBBBBBB\nBBBWBBBB\nBBBWBBBB\nBBBWBBBB",
"output": "9"
},
{
"input": "BBBBBBBB\nWWWBBBBB\nWWWBBBBB\nBBBBBBBB\nWWWBBBBB\nWWWBBBBB\nBBBBBBBB\nBBBBBBBB",
"output": "9"
},
{
"input": "WBBBBBWB\nBBBBBBBB\nWBBBBBWB\nWBBBBBWB\nWBBBBBWB\nWBBBBBWB\nWBBBBBWB\nBBBBBBBB",
"output": "8"
},
{
"input": "WBBBWWBW\nWBBBWWBW\nBBBBBBBB\nWBBBWWBW\nBBBBBBBB\nWBBBWWBW\nWBBBWWBW\nWBBBWWBW",
"output": "6"
},
{
"input": "WBBBBWBB\nBBBBBBBB\nBBBBBBBB\nWBBBBWBB\nWBBBBWBB\nBBBBBBBB\nWBBBBWBB\nBBBBBBBB",
"output": "10"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBBB\nWBBBWBBW\nBBBBBBBB\nBBBBBBBB\nWBBBWBBW\nBBBBBBBB",
"output": "11"
},
{
"input": "BBBBBBBB\nBWBBBBBW\nBWBBBBBW\nBBBBBBBB\nBWBBBBBW\nBWBBBBBW\nBBBBBBBB\nBWBBBBBW",
"output": "9"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nWBBBWWWW\nBBBBBBBB\nBBBBBBBB\nWBBBWWWW\nBBBBBBBB\nBBBBBBBB",
"output": "9"
},
{
"input": "BWBBBWWB\nBWBBBWWB\nBBBBBBBB\nBBBBBBBB\nBWBBBWWB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB",
"output": "10"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBWBBWBWB",
"output": "12"
},
{
"input": "BWBBBBWW\nBWBBBBWW\nBWBBBBWW\nBWBBBBWW\nBBBBBBBB\nBWBBBBWW\nBWBBBBWW\nBBBBBBBB",
"output": "7"
},
{
"input": "WWBBWWBB\nBBBBBBBB\nWWBBWWBB\nWWBBWWBB\nWWBBWWBB\nBBBBBBBB\nWWBBWWBB\nWWBBWWBB",
"output": "6"
},
{
"input": "BWBBWWWW\nBWBBWWWW\nBWBBWWWW\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBWBBWWWW\nBBBBBBBB",
"output": "7"
}
] | 1,696,005,904
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 0
|
################################################################################
board=[]
for i in range(8):
board.append(input)
# board=['WWWBWWBW',
# 'BBBBBBBB',
# 'WWWBWWBW',
# 'WWWBWWBW',
# 'WWWBWWBW',
# 'WWWBWWBW',
# 'WWWBWWBW',
# 'WWWBWWBW']
################################################################################
row=0
col=0
for i in range(8):
is_all_black=1
for j in range(8):
if board[i][j]=='W':
is_all_black=0
if is_all_black:
row+=1
for j in range(8):
is_all_black=1
for i in range(8):
if board[i][j]=='W':
is_all_black=0
if is_all_black:
col+=1
out=row+col
if col==8 and row==8:
out=8
print(out)
|
Title: Kalevitch and Chess
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
A famous Berland's painter Kalevitch likes to shock the public. One of his last obsessions is chess. For more than a thousand years people have been playing this old game on uninteresting, monotonous boards. Kalevitch decided to put an end to this tradition and to introduce a new attitude to chessboards.
As before, the chessboard is a square-checkered board with the squares arranged in a 8<=×<=8 grid, each square is painted black or white. Kalevitch suggests that chessboards should be painted in the following manner: there should be chosen a horizontal or a vertical line of 8 squares (i.e. a row or a column), and painted black. Initially the whole chessboard is white, and it can be painted in the above described way one or more times. It is allowed to paint a square many times, but after the first time it does not change its colour any more and remains black. Kalevitch paints chessboards neatly, and it is impossible to judge by an individual square if it was painted with a vertical or a horizontal stroke.
Kalevitch hopes that such chessboards will gain popularity, and he will be commissioned to paint chessboards, which will help him ensure a comfortable old age. The clients will inform him what chessboard they want to have, and the painter will paint a white chessboard meeting the client's requirements.
It goes without saying that in such business one should economize on everything — for each commission he wants to know the minimum amount of strokes that he has to paint to fulfill the client's needs. You are asked to help Kalevitch with this task.
Input Specification:
The input file contains 8 lines, each of the lines contains 8 characters. The given matrix describes the client's requirements, W character stands for a white square, and B character — for a square painted black.
It is guaranteed that client's requirments can be fulfilled with a sequence of allowed strokes (vertical/column or horizontal/row).
Output Specification:
Output the only number — the minimum amount of rows and columns that Kalevitch has to paint on the white chessboard to meet the client's requirements.
Demo Input:
['WWWBWWBW\nBBBBBBBB\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\nWWWBWWBW\n', 'WWWWWWWW\nBBBBBBBB\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
################################################################################
board=[]
for i in range(8):
board.append(input)
# board=['WWWBWWBW',
# 'BBBBBBBB',
# 'WWWBWWBW',
# 'WWWBWWBW',
# 'WWWBWWBW',
# 'WWWBWWBW',
# 'WWWBWWBW',
# 'WWWBWWBW']
################################################################################
row=0
col=0
for i in range(8):
is_all_black=1
for j in range(8):
if board[i][j]=='W':
is_all_black=0
if is_all_black:
row+=1
for j in range(8):
is_all_black=1
for i in range(8):
if board[i][j]=='W':
is_all_black=0
if is_all_black:
col+=1
out=row+col
if col==8 and row==8:
out=8
print(out)
```
| -1
|
960
|
A
|
Check the string
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
|
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
|
Print "YES" or "NO", according to the condition.
|
[
"aaabccc\n",
"bbacc\n",
"aabc\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'.
| 500
|
[
{
"input": "aaabccc",
"output": "YES"
},
{
"input": "bbacc",
"output": "NO"
},
{
"input": "aabc",
"output": "YES"
},
{
"input": "aabbcc",
"output": "YES"
},
{
"input": "aaacccbb",
"output": "NO"
},
{
"input": "abc",
"output": "YES"
},
{
"input": "acba",
"output": "NO"
},
{
"input": "bbabbc",
"output": "NO"
},
{
"input": "bbbabacca",
"output": "NO"
},
{
"input": "aabcbcaca",
"output": "NO"
},
{
"input": "aaaaabbbbbb",
"output": "NO"
},
{
"input": "c",
"output": "NO"
},
{
"input": "cc",
"output": "NO"
},
{
"input": "bbb",
"output": "NO"
},
{
"input": "bc",
"output": "NO"
},
{
"input": "ccbcc",
"output": "NO"
},
{
"input": "aaa",
"output": "NO"
},
{
"input": "aaccaa",
"output": "NO"
},
{
"input": "a",
"output": "NO"
},
{
"input": "b",
"output": "NO"
},
{
"input": "abca",
"output": "NO"
},
{
"input": "aabbcccc",
"output": "NO"
},
{
"input": "abac",
"output": "NO"
},
{
"input": "abcc",
"output": "NO"
},
{
"input": "abcb",
"output": "NO"
},
{
"input": "aacc",
"output": "NO"
},
{
"input": "aabbaacccc",
"output": "NO"
},
{
"input": "aabb",
"output": "NO"
},
{
"input": "ac",
"output": "NO"
},
{
"input": "abbacc",
"output": "NO"
},
{
"input": "abacc",
"output": "NO"
},
{
"input": "ababc",
"output": "NO"
},
{
"input": "aa",
"output": "NO"
},
{
"input": "aabaccc",
"output": "NO"
},
{
"input": "bbcc",
"output": "NO"
},
{
"input": "aaabcbc",
"output": "NO"
},
{
"input": "acbbc",
"output": "NO"
},
{
"input": "babc",
"output": "NO"
},
{
"input": "bbbcc",
"output": "NO"
},
{
"input": "bbc",
"output": "NO"
},
{
"input": "abababccc",
"output": "NO"
},
{
"input": "ccbbaa",
"output": "NO"
}
] | 1,523,122,469
| 5,369
|
Python 3
|
CHALLENGED
|
CHALLENGES
| 7
| 93
| 7,065,600
|
text = input()
Cnum=text.count('c')
Anum=text.count('a')
Bnum=text.count('b')
if Cnum==Anum or Cnum==Bnum:
if 'ba' in text or 'ca' in text or 'cb' in text:
print('NO')
else:
print('YES')
else:
print('NO')
|
Title: Check the string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
Input Specification:
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
Output Specification:
Print "YES" or "NO", according to the condition.
Demo Input:
['aaabccc\n', 'bbacc\n', 'aabc\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'.
|
```python
text = input()
Cnum=text.count('c')
Anum=text.count('a')
Bnum=text.count('b')
if Cnum==Anum or Cnum==Bnum:
if 'ba' in text or 'ca' in text or 'cb' in text:
print('NO')
else:
print('YES')
else:
print('NO')
```
| -1
|
|
507
|
B
|
Amr and Pins
|
PROGRAMMING
| 1,400
|
[
"geometry",
"math"
] | null | null |
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius *r* and center in point (*x*,<=*y*). He wants the circle center to be in new position (*x*',<=*y*').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
|
Input consists of 5 space-separated integers *r*, *x*, *y*, *x*' *y*' (1<=≤<=*r*<=≤<=105, <=-<=105<=≤<=*x*,<=*y*,<=*x*',<=*y*'<=≤<=105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
|
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
|
[
"2 0 0 0 4\n",
"1 1 1 4 4\n",
"4 5 6 5 6\n"
] |
[
"1\n",
"3\n",
"0\n"
] |
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<img class="tex-graphics" src="https://espresso.codeforces.com/4e40fd4cc24a2050a0488aa131e6244369328039.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "2 0 0 0 4",
"output": "1"
},
{
"input": "1 1 1 4 4",
"output": "3"
},
{
"input": "4 5 6 5 6",
"output": "0"
},
{
"input": "10 20 0 40 0",
"output": "1"
},
{
"input": "9 20 0 40 0",
"output": "2"
},
{
"input": "5 -1 -6 -5 1",
"output": "1"
},
{
"input": "99125 26876 -21414 14176 17443",
"output": "1"
},
{
"input": "8066 7339 19155 -90534 -60666",
"output": "8"
},
{
"input": "100000 -100000 -100000 100000 100000",
"output": "2"
},
{
"input": "10 20 0 41 0",
"output": "2"
},
{
"input": "25 -64 -6 -56 64",
"output": "2"
},
{
"input": "125 455 450 439 721",
"output": "2"
},
{
"input": "5 6 3 7 2",
"output": "1"
},
{
"input": "24 130 14786 3147 2140",
"output": "271"
},
{
"input": "125 -363 176 93 330",
"output": "2"
},
{
"input": "1 14 30 30 14",
"output": "12"
},
{
"input": "25 96 13 7 2",
"output": "2"
},
{
"input": "4 100000 -100000 100000 -100000",
"output": "0"
},
{
"input": "1 3 4 2 5",
"output": "1"
},
{
"input": "1 -3 3 2 6",
"output": "3"
},
{
"input": "2 7 20 13 -5",
"output": "7"
},
{
"input": "1 1 1 1 4",
"output": "2"
},
{
"input": "249 -54242 -30537 -45023 -89682",
"output": "121"
},
{
"input": "4 100000 -100000 100000 -99999",
"output": "1"
},
{
"input": "97741 23818 78751 97583 26933",
"output": "1"
},
{
"input": "56767 -29030 51625 79823 -56297",
"output": "2"
},
{
"input": "98260 13729 74998 23701 9253",
"output": "1"
},
{
"input": "67377 -80131 -90254 -57320 14102",
"output": "1"
},
{
"input": "1 100000 100000 100000 -100000",
"output": "100000"
},
{
"input": "19312 19470 82059 58064 62231",
"output": "2"
},
{
"input": "67398 -68747 -79056 -34193 29400",
"output": "1"
},
{
"input": "91099 37184 -71137 75650 -3655",
"output": "1"
},
{
"input": "46456 -2621 -23623 -98302 -99305",
"output": "2"
},
{
"input": "100 100000 -100000 100000 -99999",
"output": "1"
},
{
"input": "1 100000 -100000 100000 -100000",
"output": "0"
},
{
"input": "8 0 0 0 32",
"output": "2"
},
{
"input": "100000 100000 1 -100000 0",
"output": "2"
}
] | 1,666,940,586
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 46
| 0
|
import math
r, x, y, a, b = tuple(map(int, input().split()))
dis = math.sqrt(((x-a)*(x-a)) + ((y-b)*(y-b)))
print(math.ceil(dis/(2*r)))
|
Title: Amr and Pins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius *r* and center in point (*x*,<=*y*). He wants the circle center to be in new position (*x*',<=*y*').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input Specification:
Input consists of 5 space-separated integers *r*, *x*, *y*, *x*' *y*' (1<=≤<=*r*<=≤<=105, <=-<=105<=≤<=*x*,<=*y*,<=*x*',<=*y*'<=≤<=105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output Specification:
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Demo Input:
['2 0 0 0 4\n', '1 1 1 4 4\n', '4 5 6 5 6\n']
Demo Output:
['1\n', '3\n', '0\n']
Note:
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<img class="tex-graphics" src="https://espresso.codeforces.com/4e40fd4cc24a2050a0488aa131e6244369328039.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
import math
r, x, y, a, b = tuple(map(int, input().split()))
dis = math.sqrt(((x-a)*(x-a)) + ((y-b)*(y-b)))
print(math.ceil(dis/(2*r)))
```
| 3
|
|
808
|
D
|
Array Division
|
PROGRAMMING
| 1,900
|
[
"binary search",
"data structures",
"implementation"
] | null | null |
Vasya has an array *a* consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).
Inserting an element in the same position he was erased from is also considered moving.
Can Vasya divide the array after choosing the right element to move and its new position?
|
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
|
Print YES if Vasya can divide the array after moving one element. Otherwise print NO.
|
[
"3\n1 3 2\n",
"5\n1 2 3 4 5\n",
"5\n2 2 3 4 5\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
In the first example Vasya can move the second element to the end of the array.
In the second example no move can make the division possible.
In the third example Vasya can move the fourth element by one position to the left.
| 0
|
[
{
"input": "3\n1 3 2",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5",
"output": "NO"
},
{
"input": "5\n2 2 3 4 5",
"output": "YES"
},
{
"input": "5\n72 32 17 46 82",
"output": "NO"
},
{
"input": "6\n26 10 70 11 69 57",
"output": "NO"
},
{
"input": "7\n4 7 10 7 5 5 1",
"output": "NO"
},
{
"input": "8\n9 5 5 10 4 9 5 8",
"output": "NO"
},
{
"input": "10\n9 6 8 5 5 2 8 9 2 2",
"output": "YES"
},
{
"input": "15\n4 8 10 3 1 4 5 9 3 2 1 7 7 3 8",
"output": "NO"
},
{
"input": "20\n71 83 54 6 10 64 91 98 94 49 65 68 14 39 91 60 74 100 17 13",
"output": "NO"
},
{
"input": "20\n2 8 10 4 6 6 4 1 2 2 6 9 5 1 9 1 9 8 10 6",
"output": "NO"
},
{
"input": "100\n9 9 72 55 14 8 55 58 35 67 3 18 73 92 41 49 15 60 18 66 9 26 97 47 43 88 71 97 19 34 48 96 79 53 8 24 69 49 12 23 77 12 21 88 66 9 29 13 61 69 54 77 41 13 4 68 37 74 7 6 29 76 55 72 89 4 78 27 29 82 18 83 12 4 32 69 89 85 66 13 92 54 38 5 26 56 17 55 29 4 17 39 29 94 3 67 85 98 21 14",
"output": "YES"
},
{
"input": "100\n89 38 63 73 77 4 99 74 30 5 69 57 97 37 88 71 36 59 19 63 46 20 33 58 61 98 100 31 33 53 99 96 34 17 44 95 54 52 22 77 67 88 20 88 26 43 12 23 96 94 14 7 57 86 56 54 32 8 3 43 97 56 74 22 5 100 12 60 93 12 44 68 31 63 7 71 21 29 19 38 50 47 97 43 50 59 88 40 51 61 20 68 32 66 70 48 19 55 91 53",
"output": "NO"
},
{
"input": "100\n80 100 88 52 25 87 85 8 92 62 35 66 74 39 58 41 55 53 23 73 90 72 36 44 97 67 16 54 3 8 25 34 84 47 77 39 93 19 49 20 29 44 21 48 21 56 82 59 8 31 94 95 84 54 72 20 95 91 85 1 67 19 76 28 31 63 87 98 55 28 16 20 36 91 93 39 94 69 80 97 100 96 68 26 91 45 22 84 20 36 20 92 53 75 58 51 60 26 76 25",
"output": "NO"
},
{
"input": "100\n27 95 57 29 91 85 83 36 72 86 39 5 79 61 78 93 100 97 73 23 82 66 41 92 38 92 100 96 48 56 66 47 5 32 69 13 95 23 46 62 99 83 57 66 98 82 81 57 37 37 81 64 45 76 72 43 99 76 86 22 37 39 93 80 99 36 53 83 3 32 52 9 78 34 47 100 33 72 19 40 29 56 77 32 79 72 15 88 100 98 56 50 22 81 88 92 58 70 21 19",
"output": "NO"
},
{
"input": "100\n35 31 83 11 7 94 57 58 30 26 2 99 33 58 98 6 3 52 13 66 21 53 26 94 100 5 1 3 91 13 97 49 86 25 63 90 88 98 57 57 34 81 32 16 65 94 59 83 44 14 46 18 28 89 75 95 87 57 52 18 46 80 31 43 38 54 69 75 82 9 64 96 75 40 96 52 67 85 86 38 95 55 16 57 17 20 22 7 63 3 12 16 42 87 46 12 51 95 67 80",
"output": "NO"
},
{
"input": "6\n1 4 3 100 100 6",
"output": "YES"
},
{
"input": "6\n6 100 100 3 4 1",
"output": "YES"
},
{
"input": "6\n4 2 3 7 1 1",
"output": "YES"
},
{
"input": "4\n6 1 4 5",
"output": "NO"
},
{
"input": "3\n228 114 114",
"output": "YES"
},
{
"input": "3\n229 232 444",
"output": "NO"
},
{
"input": "3\n322 324 555",
"output": "NO"
},
{
"input": "3\n69 34 5",
"output": "NO"
},
{
"input": "6\n5 4 1 2 2 2",
"output": "YES"
},
{
"input": "3\n545 237 546",
"output": "NO"
},
{
"input": "5\n2 3 1 1 1",
"output": "YES"
},
{
"input": "6\n2 2 10 2 2 2",
"output": "YES"
},
{
"input": "5\n5 4 6 5 6",
"output": "NO"
},
{
"input": "5\n6 1 1 1 1",
"output": "NO"
},
{
"input": "2\n1 3",
"output": "NO"
},
{
"input": "5\n5 2 2 3 4",
"output": "YES"
},
{
"input": "2\n2 2",
"output": "YES"
},
{
"input": "5\n1 2 6 1 2",
"output": "YES"
},
{
"input": "5\n1 1 8 5 1",
"output": "YES"
},
{
"input": "10\n73 67 16 51 56 71 37 49 90 6",
"output": "NO"
},
{
"input": "1\n10",
"output": "NO"
},
{
"input": "1\n1",
"output": "NO"
},
{
"input": "2\n1 1",
"output": "YES"
},
{
"input": "5\n8 2 7 5 4",
"output": "YES"
},
{
"input": "1\n2",
"output": "NO"
},
{
"input": "16\n9 10 2 1 6 7 6 5 8 3 2 10 8 4 9 2",
"output": "YES"
},
{
"input": "4\n8 2 2 4",
"output": "YES"
},
{
"input": "19\n9 9 3 2 4 5 5 7 8 10 8 10 1 2 2 6 5 3 3",
"output": "NO"
},
{
"input": "11\n7 2 1 8 8 2 4 10 8 7 1",
"output": "YES"
},
{
"input": "6\n10 20 30 40 99 1",
"output": "YES"
},
{
"input": "10\n3 7 9 2 10 1 9 6 4 1",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "2\n9 3",
"output": "NO"
},
{
"input": "7\n1 2 3 12 1 2 3",
"output": "YES"
},
{
"input": "6\n2 4 4 5 8 5",
"output": "YES"
},
{
"input": "18\n2 10 3 6 6 6 10 8 8 1 10 9 9 3 1 9 7 4",
"output": "YES"
},
{
"input": "20\n9 6 6 10 4 4 8 7 4 10 10 2 10 5 9 5 3 10 1 9",
"output": "NO"
},
{
"input": "12\n3 8 10 2 4 4 6 9 5 10 10 3",
"output": "YES"
},
{
"input": "11\n9 2 7 7 7 3 7 5 4 10 7",
"output": "NO"
},
{
"input": "5\n1 1 4 1 1",
"output": "YES"
},
{
"input": "2\n4 4",
"output": "YES"
},
{
"input": "2\n7 1",
"output": "NO"
},
{
"input": "5\n10 5 6 7 6",
"output": "YES"
},
{
"input": "11\n4 3 10 3 7 8 4 9 2 1 1",
"output": "YES"
},
{
"input": "6\n705032704 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "NO"
},
{
"input": "8\n1 5 6 8 3 1 7 3",
"output": "YES"
},
{
"input": "20\n8 6 3 6 3 5 10 2 6 1 7 6 9 10 8 3 5 9 3 8",
"output": "YES"
},
{
"input": "11\n2 4 8 3 4 7 9 10 5 3 3",
"output": "YES"
},
{
"input": "7\n6 4 2 24 6 4 2",
"output": "YES"
},
{
"input": "17\n7 1 1 1 8 9 1 10 8 8 7 9 7 9 1 6 5",
"output": "NO"
},
{
"input": "7\n7 10 1 2 6 2 2",
"output": "NO"
},
{
"input": "5\n10 10 40 10 10",
"output": "YES"
},
{
"input": "3\n4 3 13",
"output": "NO"
},
{
"input": "5\n5 2 10 2 1",
"output": "YES"
},
{
"input": "7\n7 4 5 62 20 20 6",
"output": "YES"
},
{
"input": "6\n1 5 2 20 10 2",
"output": "YES"
},
{
"input": "2\n5 6",
"output": "NO"
},
{
"input": "14\n5 2 9 7 5 8 3 2 2 4 9 1 3 10",
"output": "YES"
},
{
"input": "5\n1 2 3 4 2",
"output": "YES"
},
{
"input": "5\n2 2 2 5 5",
"output": "NO"
},
{
"input": "11\n1 1 1 1 1 10 1 1 1 1 1",
"output": "YES"
},
{
"input": "9\n8 4 13 19 11 1 8 2 8",
"output": "YES"
},
{
"input": "6\n14 16 14 14 15 11",
"output": "YES"
},
{
"input": "9\n14 19 1 13 11 3 1 1 7",
"output": "YES"
},
{
"input": "6\n16 13 3 7 4 15",
"output": "YES"
},
{
"input": "4\n11 7 12 14",
"output": "NO"
},
{
"input": "3\n3 2 1",
"output": "YES"
},
{
"input": "5\n2 1 3 6 4",
"output": "YES"
},
{
"input": "5\n3 4 8 11 2",
"output": "YES"
},
{
"input": "5\n1 2 10 3 4",
"output": "YES"
},
{
"input": "6\n8 15 12 14 15 4",
"output": "YES"
},
{
"input": "5\n1 2 4 4 5",
"output": "YES"
},
{
"input": "3\n2 4 2",
"output": "YES"
},
{
"input": "5\n2 3 1 6 4",
"output": "YES"
},
{
"input": "7\n1 2 3 12 3 2 1",
"output": "YES"
},
{
"input": "3\n3 4 13",
"output": "NO"
},
{
"input": "6\n1 1 1 1 1000000000 1000000000",
"output": "YES"
},
{
"input": "6\n19 6 5 13 6 13",
"output": "YES"
},
{
"input": "8\n2 2 2 5 1 2 3 3",
"output": "YES"
}
] | 1,559,166,842
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 20
| 2,000
| 8,294,400
|
#Array Division
def main():
n = int(input())
array = input().split()
array = [int(x) for x in array]
soma = 0
for num in array:
soma += num
if soma%2 == 0:
part = 0
for i in range(n):
part += array[i]
falta = soma//2 - part
if falta == 0:
print("YES")
return
if abs(falta) in array[:i+1]:
print("YES")
return
print("NO")
return
main()
|
Title: Array Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has an array *a* consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).
Inserting an element in the same position he was erased from is also considered moving.
Can Vasya divide the array after choosing the right element to move and its new position?
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
Output Specification:
Print YES if Vasya can divide the array after moving one element. Otherwise print NO.
Demo Input:
['3\n1 3 2\n', '5\n1 2 3 4 5\n', '5\n2 2 3 4 5\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example Vasya can move the second element to the end of the array.
In the second example no move can make the division possible.
In the third example Vasya can move the fourth element by one position to the left.
|
```python
#Array Division
def main():
n = int(input())
array = input().split()
array = [int(x) for x in array]
soma = 0
for num in array:
soma += num
if soma%2 == 0:
part = 0
for i in range(n):
part += array[i]
falta = soma//2 - part
if falta == 0:
print("YES")
return
if abs(falta) in array[:i+1]:
print("YES")
return
print("NO")
return
main()
```
| 0
|
|
121
|
A
|
Lucky Sum
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let *next*(*x*) be the minimum lucky number which is larger than or equals *x*. Petya is interested what is the value of the expression *next*(*l*)<=+<=*next*(*l*<=+<=1)<=+<=...<=+<=*next*(*r*<=-<=1)<=+<=*next*(*r*). Help him solve this problem.
|
The single line contains two integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=109) — the left and right interval limits.
|
In the single line print the only number — the sum *next*(*l*)<=+<=*next*(*l*<=+<=1)<=+<=...<=+<=*next*(*r*<=-<=1)<=+<=*next*(*r*).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
|
[
"2 7\n",
"7 7\n"
] |
[
"33\n",
"7\n"
] |
In the first sample: *next*(2) + *next*(3) + *next*(4) + *next*(5) + *next*(6) + *next*(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: *next*(7) = 7
| 500
|
[
{
"input": "2 7",
"output": "33"
},
{
"input": "7 7",
"output": "7"
},
{
"input": "1 9",
"output": "125"
},
{
"input": "4 7",
"output": "25"
},
{
"input": "12 47",
"output": "1593"
},
{
"input": "6 77",
"output": "4012"
},
{
"input": "1 100",
"output": "14247"
},
{
"input": "1000000000 1000000000",
"output": "4444444444"
},
{
"input": "77 77",
"output": "77"
},
{
"input": "69 788",
"output": "452195"
},
{
"input": "474 747",
"output": "202794"
},
{
"input": "4 77777",
"output": "4070145675"
},
{
"input": "1 1000000",
"output": "1394675359387"
},
{
"input": "47 744447",
"output": "381286992761"
},
{
"input": "47444 1000000000",
"output": "1394672348253941136"
},
{
"input": "48 854888",
"output": "749733199853"
},
{
"input": "854444 985555",
"output": "582719941728"
},
{
"input": "774744 774747",
"output": "3098985"
},
{
"input": "654 987654",
"output": "1339803940266"
},
{
"input": "477777 1000000000",
"output": "1394672167300009765"
},
{
"input": "77777 777777777",
"output": "407018021649898097"
},
{
"input": "963 85555574",
"output": "7526978888069560"
},
{
"input": "47 7444",
"output": "38125896"
},
{
"input": "1 1000000000",
"output": "1394672350065645019"
},
{
"input": "474 999888555",
"output": "1394177038954402791"
},
{
"input": "777 1000000000",
"output": "1394672350065239125"
},
{
"input": "7 777777774",
"output": "407018023386632646"
},
{
"input": "369 852",
"output": "606732"
},
{
"input": "47 47",
"output": "47"
},
{
"input": "747 748",
"output": "1521"
},
{
"input": "77777440 77777444",
"output": "388887220"
},
{
"input": "987545885 999875584",
"output": "54798666661186800"
},
{
"input": "2 777777776",
"output": "407018024942188226"
},
{
"input": "1 1",
"output": "4"
},
{
"input": "1 2",
"output": "8"
},
{
"input": "999999999 1000000000",
"output": "8888888888"
},
{
"input": "777777777 1000000000",
"output": "987654325123456789"
},
{
"input": "777777778 1000000000",
"output": "987654324345679012"
},
{
"input": "5 6",
"output": "14"
},
{
"input": "49 49",
"output": "74"
},
{
"input": "6 6",
"output": "7"
},
{
"input": "3 4",
"output": "8"
}
] | 1,684,612,714
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 2,048,000
|
n,m=map(int,input().split())
v=[]
def gen(x=0):
global v
if x>=10e9:
return
v+=[x]
gen(x*10+4)
gen(x*10+7)
gen()
v.sort()
q=[]
for i in v:
if i>=m:
q+=[i]
break
q+=[i]
q.pop(0)
k=[]
k+=[m]
k+=[n]
for j in q:
if j<=m:
k+=[j]
k=list(set(k))
k.sort()
c=[]
for o in range(len(k)-1):
c+=[k[o+1]-k[o]]
c[0]+=1
_=0
for p in range(len(c)):
_+=c[p]*q[p]
print(_)
|
Title: Lucky Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let *next*(*x*) be the minimum lucky number which is larger than or equals *x*. Petya is interested what is the value of the expression *next*(*l*)<=+<=*next*(*l*<=+<=1)<=+<=...<=+<=*next*(*r*<=-<=1)<=+<=*next*(*r*). Help him solve this problem.
Input Specification:
The single line contains two integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=109) — the left and right interval limits.
Output Specification:
In the single line print the only number — the sum *next*(*l*)<=+<=*next*(*l*<=+<=1)<=+<=...<=+<=*next*(*r*<=-<=1)<=+<=*next*(*r*).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Demo Input:
['2 7\n', '7 7\n']
Demo Output:
['33\n', '7\n']
Note:
In the first sample: *next*(2) + *next*(3) + *next*(4) + *next*(5) + *next*(6) + *next*(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: *next*(7) = 7
|
```python
n,m=map(int,input().split())
v=[]
def gen(x=0):
global v
if x>=10e9:
return
v+=[x]
gen(x*10+4)
gen(x*10+7)
gen()
v.sort()
q=[]
for i in v:
if i>=m:
q+=[i]
break
q+=[i]
q.pop(0)
k=[]
k+=[m]
k+=[n]
for j in q:
if j<=m:
k+=[j]
k=list(set(k))
k.sort()
c=[]
for o in range(len(k)-1):
c+=[k[o+1]-k[o]]
c[0]+=1
_=0
for p in range(len(c)):
_+=c[p]*q[p]
print(_)
```
| 0
|
|
139
|
A
|
Petr and Book
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
|
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
|
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
|
[
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] |
[
"6\n",
"1\n"
] |
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
| 500
|
[
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "20\n5 3 7 2 1 6 4",
"output": "6"
},
{
"input": "10\n5 1 1 1 1 1 5",
"output": "6"
},
{
"input": "50\n10 1 10 1 10 1 10",
"output": "1"
},
{
"input": "77\n11 11 11 11 11 11 10",
"output": "1"
},
{
"input": "1\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n100 100 100 100 100 100 100",
"output": "3"
},
{
"input": "999\n10 20 10 20 30 20 10",
"output": "3"
},
{
"input": "433\n109 58 77 10 39 125 15",
"output": "7"
},
{
"input": "1\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n1 0 1 0 1 0 1",
"output": "1"
},
{
"input": "997\n1 1 0 0 1 0 1",
"output": "1"
},
{
"input": "1000\n1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1000\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "1000\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "1000\n1 0 0 1 0 0 1",
"output": "1"
},
{
"input": "509\n105 23 98 0 7 0 155",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2\n1 1 0 0 0 0 0",
"output": "2"
},
{
"input": "1\n0 0 0 0 0 1 0",
"output": "6"
},
{
"input": "10\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 6 0",
"output": "6"
},
{
"input": "3\n0 1 0 0 0 0 0",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "28\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "100\n5 5 5 5 5 5 5",
"output": "6"
},
{
"input": "4\n1 0 0 0 0 0 1",
"output": "7"
},
{
"input": "2\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "7\n0 0 0 0 0 0 7",
"output": "7"
},
{
"input": "7\n2 1 1 1 1 1 0",
"output": "6"
},
{
"input": "2\n0 0 1 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1 0",
"output": "6"
},
{
"input": "5\n1 1 1 0 0 1 1",
"output": "7"
},
{
"input": "100\n10 20 30 10 10 10 10",
"output": "7"
},
{
"input": "1\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "70\n10 10 10 10 10 10 10",
"output": "7"
},
{
"input": "22\n1 2 3 4 5 6 10",
"output": "7"
},
{
"input": "5\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "2\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 0 0 0 0 0 2",
"output": "7"
},
{
"input": "10\n1 2 2 1 2 1 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "4\n0 1 1 0 0 0 0",
"output": "3"
},
{
"input": "100\n0 0 0 0 0 1 0",
"output": "6"
}
] | 1,665,682,902
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 12
| 186
| 17,715,200
|
n = int(input())
lst = list(map(int,input().split(' ')))
i,res = 0,0
while (res<n):
res+=lst[i%7]
i+=1
print(i%7)
|
Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
|
```python
n = int(input())
lst = list(map(int,input().split(' ')))
i,res = 0,0
while (res<n):
res+=lst[i%7]
i+=1
print(i%7)
```
| 0
|
|
155
|
A
|
I_love_\%username\%
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
|
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
|
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
|
[
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] |
[
"2\n",
"4\n"
] |
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing.
| 500
|
[
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
},
{
"input": "5\n7 36 53 81 100",
"output": "4"
},
{
"input": "5\n100 81 53 36 7",
"output": "4"
},
{
"input": "10\n8 6 3 4 9 10 7 7 1 3",
"output": "5"
},
{
"input": "10\n1627 1675 1488 1390 1812 1137 1746 1324 1952 1862",
"output": "6"
},
{
"input": "10\n1 3 3 4 6 7 7 8 9 10",
"output": "7"
},
{
"input": "10\n1952 1862 1812 1746 1675 1627 1488 1390 1324 1137",
"output": "9"
},
{
"input": "25\n1448 4549 2310 2725 2091 3509 1565 2475 2232 3989 4231 779 2967 2702 608 3739 721 1552 2767 530 3114 665 1940 48 4198",
"output": "5"
},
{
"input": "33\n1097 1132 1091 1104 1049 1038 1023 1080 1104 1029 1035 1061 1049 1060 1088 1106 1105 1087 1063 1076 1054 1103 1047 1041 1028 1120 1126 1063 1117 1110 1044 1093 1101",
"output": "5"
},
{
"input": "34\n821 5536 2491 6074 7216 9885 764 1603 778 8736 8987 771 617 1587 8943 7922 439 7367 4115 8886 7878 6899 8811 5752 3184 3401 9760 9400 8995 4681 1323 6637 6554 6498",
"output": "7"
},
{
"input": "68\n6764 6877 6950 6768 6839 6755 6726 6778 6699 6805 6777 6985 6821 6801 6791 6805 6940 6761 6677 6999 6911 6699 6959 6933 6903 6843 6972 6717 6997 6756 6789 6668 6735 6852 6735 6880 6723 6834 6810 6694 6780 6679 6698 6857 6826 6896 6979 6968 6957 6988 6960 6700 6919 6892 6984 6685 6813 6678 6715 6857 6976 6902 6780 6686 6777 6686 6842 6679",
"output": "9"
},
{
"input": "60\n9000 9014 9034 9081 9131 9162 9174 9199 9202 9220 9221 9223 9229 9235 9251 9260 9268 9269 9270 9298 9307 9309 9313 9323 9386 9399 9407 9495 9497 9529 9531 9544 9614 9615 9627 9627 9643 9654 9656 9657 9685 9699 9701 9736 9745 9758 9799 9827 9843 9845 9854 9854 9885 9891 9896 9913 9942 9963 9986 9992",
"output": "57"
},
{
"input": "100\n7 61 12 52 41 16 34 99 30 44 48 89 31 54 21 1 48 52 61 15 35 87 21 76 64 92 44 81 16 93 84 92 32 15 68 76 53 39 26 4 11 26 7 4 99 99 61 65 55 85 65 67 47 39 2 74 63 49 98 87 5 94 22 30 25 42 31 84 49 23 89 60 16 26 92 27 9 57 75 61 94 35 83 47 99 100 63 24 91 88 79 10 15 45 22 64 3 11 89 83",
"output": "4"
},
{
"input": "100\n9999 9999 9999 9998 9998 9998 9997 9996 9996 9995 9993 9993 9991 9990 9989 9986 9984 9984 9983 9981 9981 9980 9980 9980 9979 9977 9977 9977 9977 9977 9976 9976 9975 9975 9973 9972 9972 9972 9972 9971 9969 9969 9969 9968 9967 9965 9962 9962 9962 9962 9960 9958 9958 9955 9954 9954 9954 9954 9950 9949 9949 9947 9943 9943 9940 9940 9939 9937 9937 9937 9936 9935 9932 9932 9931 9929 9927 9927 9926 9923 9919 9919 9918 9918 9918 9917 9917 9916 9916 9915 9914 9911 9910 9907 9907 9906 9905 9905 9904 9901",
"output": "57"
},
{
"input": "100\n158 159 157 156 155 154 160 153 161 152 162 163 151 164 150 165 149 148 147 166 146 167 145 144 143 142 168 141 169 170 140 139 138 137 171 136 135 134 172 173 174 133 175 132 177 178 131 179 180 130 129 181 128 127 182 126 183 125 124 184 123 122 121 120 119 118 117 185 186 187 188 189 116 190 115 191 192 193 114 113 112 111 110 109 108 194 195 107 106 105 196 197 104 198 199 103 102 200 101 100",
"output": "99"
},
{
"input": "2\n0 10000",
"output": "1"
},
{
"input": "2\n5 5",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n10 10",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
}
] | 1,692,937,357
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 37
| 154
| 0
|
n = int(input())
arr = list(map(int, input().split()))
base = arr[0]
greater = smaller = base
c = 0
for i in range(1,n):
if arr[i] > greater:
c += 1
greater = arr[i]
elif arr[i] < smaller:
smaller = arr[i]
c += 1
print(c)
|
Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing.
|
```python
n = int(input())
arr = list(map(int, input().split()))
base = arr[0]
greater = smaller = base
c = 0
for i in range(1,n):
if arr[i] > greater:
c += 1
greater = arr[i]
elif arr[i] < smaller:
smaller = arr[i]
c += 1
print(c)
```
| 3
|
|
847
|
J
|
Students Initiation
|
PROGRAMMING
| 2,400
|
[
"binary search",
"flows",
"graphs"
] | null | null |
Soon the first year students will be initiated into students at the University of Berland. The organizers of the initiation come up with a program for this holiday. In their opinion, it would be good if the first-year students presented small souvenirs to each other. When they voiced this idea to the first-year students, they found out the following:
- some pairs of the new students already know each other; - each new student agrees to give souvenirs only to those with whom they are already familiar; - each new student does not want to present too many souvenirs.
The organizers have written down all the pairs of first-year friends who are familiar with each other and now want to determine for each new student, whom they should give souvenirs to. In their opinion, in each pair of familiar students exactly one student must present a souvenir to another student.
First year students already decided to call the unluckiest the one who will have to present the greatest number of souvenirs. The organizers in return promised that the unluckiest will be unlucky to the minimum possible degree: of course, they will have to present the greatest number of souvenirs compared to the other students, but this number will be as small as possible.
Organizers are very busy, and they asked you to determine for each pair of first-year friends who and to whom should present a souvenir.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=5000, 0<=≤<=*m*<=≤<=*min*(5000,<=*n*·(*n*<=-<=1)<=/<=2)) — the number of the first year students and the number of pairs of the students that know each other. The students are numbered from 1 to *n*.
Each of the following *m* lines contains two integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*) — the students in each pair.
It is guaranteed that each pair is present in the list exactly once. It is also guaranteed that if there is a pair (*x**i*,<=*y**i*) in the list, then there is no pair (*y**i*,<=*x**i*).
|
Print a single integer into the first line — the smallest number of souvenirs that the unluckiest student will have to present.
Following should be *m* lines, each containing two integers — the students which are familiar with each other. The first number in the pair must be the student that will present the souvenir to the second student in the pair.
Pairs can be printed in any order. If there are many solutions, print any of them.
|
[
"5 4\n2 1\n1 3\n2 3\n2 5\n",
"4 3\n1 2\n1 3\n1 4\n",
"4 6\n1 2\n4 1\n4 2\n3 2\n4 3\n1 3\n"
] |
[
"1\n1 2\n2 3\n3 1\n5 2\n",
"1\n1 4\n2 1\n3 1\n",
"2\n1 3\n2 1\n2 4\n3 2\n4 1\n4 3\n"
] |
none
| 0
|
[] | 1,693,488,390
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
print("_RANDOM_GUESS_1693488389.9852984")# 1693488389.9853175
|
Title: Students Initiation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Soon the first year students will be initiated into students at the University of Berland. The organizers of the initiation come up with a program for this holiday. In their opinion, it would be good if the first-year students presented small souvenirs to each other. When they voiced this idea to the first-year students, they found out the following:
- some pairs of the new students already know each other; - each new student agrees to give souvenirs only to those with whom they are already familiar; - each new student does not want to present too many souvenirs.
The organizers have written down all the pairs of first-year friends who are familiar with each other and now want to determine for each new student, whom they should give souvenirs to. In their opinion, in each pair of familiar students exactly one student must present a souvenir to another student.
First year students already decided to call the unluckiest the one who will have to present the greatest number of souvenirs. The organizers in return promised that the unluckiest will be unlucky to the minimum possible degree: of course, they will have to present the greatest number of souvenirs compared to the other students, but this number will be as small as possible.
Organizers are very busy, and they asked you to determine for each pair of first-year friends who and to whom should present a souvenir.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=5000, 0<=≤<=*m*<=≤<=*min*(5000,<=*n*·(*n*<=-<=1)<=/<=2)) — the number of the first year students and the number of pairs of the students that know each other. The students are numbered from 1 to *n*.
Each of the following *m* lines contains two integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*) — the students in each pair.
It is guaranteed that each pair is present in the list exactly once. It is also guaranteed that if there is a pair (*x**i*,<=*y**i*) in the list, then there is no pair (*y**i*,<=*x**i*).
Output Specification:
Print a single integer into the first line — the smallest number of souvenirs that the unluckiest student will have to present.
Following should be *m* lines, each containing two integers — the students which are familiar with each other. The first number in the pair must be the student that will present the souvenir to the second student in the pair.
Pairs can be printed in any order. If there are many solutions, print any of them.
Demo Input:
['5 4\n2 1\n1 3\n2 3\n2 5\n', '4 3\n1 2\n1 3\n1 4\n', '4 6\n1 2\n4 1\n4 2\n3 2\n4 3\n1 3\n']
Demo Output:
['1\n1 2\n2 3\n3 1\n5 2\n', '1\n1 4\n2 1\n3 1\n', '2\n1 3\n2 1\n2 4\n3 2\n4 1\n4 3\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1693488389.9852984")# 1693488389.9853175
```
| 0
|
|
560
|
A
|
Currency System in Geraldion
|
PROGRAMMING
| 1,000
|
[
"implementation",
"sortings"
] | null | null |
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
|
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
|
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
|
[
"5\n1 2 3 4 5\n"
] |
[
"-1\n"
] |
none
| 500
|
[
{
"input": "5\n1 2 3 4 5",
"output": "-1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "10\n371054 506438 397130 1 766759 208409 769264 549213 641270 771837",
"output": "-1"
},
{
"input": "10\n635370 154890 909382 220996 276501 716105 538714 140162 171960 271264",
"output": "1"
},
{
"input": "50\n110876 835020 859879 999908 712969 788264 287153 921820 330355 499311 209594 484829 296329 940051 174081 931503 1 780512 390075 97866 124255 950067 697612 244256 782385 789882 37608 82153 399889 598867 416717 377988 535636 511221 792568 683271 131077 290194 496712 330720 587436 563481 645817 942562 654093 980561 382937 48293 582608 116156",
"output": "-1"
},
{
"input": "50\n474421 421097 217233 156339 27075 733996 281778 863492 184707 956857 288561 70997 393786 337382 663642 131184 637 273801 799870 295017 392338 842567 161819 297705 102013 930684 375703 838048 154915 138503 629056 256591 893619 19263 787927 684541 320265 841090 421423 490879 394582 493952 619247 633202 612928 50907 276653 407819 489945 153173",
"output": "1"
},
{
"input": "1\n1",
"output": "-1"
},
{
"input": "1\n1000000",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
}
] | 1,686,153,880
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 31
| 0
|
n=int(input())
b=input()
if '1' in b:
print('-1')
else:
print('1')
|
Title: Currency System in Geraldion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
Output Specification:
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
Demo Input:
['5\n1 2 3 4 5\n']
Demo Output:
['-1\n']
Note:
none
|
```python
n=int(input())
b=input()
if '1' in b:
print('-1')
else:
print('1')
```
| 0
|
|
455
|
A
|
Boredom
|
PROGRAMMING
| 1,500
|
[
"dp"
] | null | null |
Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105).
|
Print a single integer — the maximum number of points that Alex can earn.
|
[
"2\n1 2\n",
"3\n1 2 3\n",
"9\n1 2 1 3 2 2 2 2 3\n"
] |
[
"2\n",
"4\n",
"10\n"
] |
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
| 500
|
[
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 2 3",
"output": "4"
},
{
"input": "9\n1 2 1 3 2 2 2 2 3",
"output": "10"
},
{
"input": "5\n3 3 4 5 4",
"output": "11"
},
{
"input": "5\n5 3 5 3 4",
"output": "16"
},
{
"input": "5\n4 2 3 2 5",
"output": "9"
},
{
"input": "10\n10 5 8 9 5 6 8 7 2 8",
"output": "46"
},
{
"input": "10\n1 1 1 1 1 1 2 3 4 4",
"output": "14"
},
{
"input": "100\n6 6 8 9 7 9 6 9 5 7 7 4 5 3 9 1 10 3 4 5 8 9 6 5 6 4 10 9 1 4 1 7 1 4 9 10 8 2 9 9 10 5 8 9 5 6 8 7 2 8 7 6 2 6 10 8 6 2 5 5 3 2 8 8 5 3 6 2 1 4 7 2 7 3 7 4 10 10 7 5 4 7 5 10 7 1 1 10 7 7 7 2 3 4 2 8 4 7 4 4",
"output": "296"
},
{
"input": "100\n6 1 5 7 10 10 2 7 3 7 2 10 7 6 3 5 5 5 3 7 2 4 2 7 7 4 2 8 2 10 4 7 9 1 1 7 9 7 1 10 10 9 5 6 10 1 7 5 8 1 1 5 3 10 2 4 3 5 2 7 4 9 5 10 1 3 7 6 6 9 3 6 6 10 1 10 6 1 10 3 4 1 7 9 2 7 8 9 3 3 2 4 6 6 1 2 9 4 1 2",
"output": "313"
},
{
"input": "100\n7 6 3 8 8 3 10 5 3 8 6 4 6 9 6 7 3 9 10 7 5 5 9 10 7 2 3 8 9 5 4 7 9 3 6 4 9 10 7 6 8 7 6 6 10 3 7 4 5 7 7 5 1 5 4 8 7 3 3 4 7 8 5 9 2 2 3 1 6 4 6 6 6 1 7 10 7 4 5 3 9 2 4 1 5 10 9 3 9 6 8 5 2 1 10 4 8 5 10 9",
"output": "298"
},
{
"input": "100\n2 10 9 1 2 6 7 2 2 8 9 9 9 5 6 2 5 1 1 10 7 4 5 5 8 1 9 4 10 1 9 3 1 8 4 10 8 8 2 4 6 5 1 4 2 2 1 2 8 5 3 9 4 10 10 7 8 6 1 8 2 6 7 1 6 7 3 10 10 3 7 7 6 9 6 8 8 10 4 6 4 3 3 3 2 3 10 6 8 5 5 10 3 7 3 1 1 1 5 5",
"output": "312"
},
{
"input": "100\n4 9 7 10 4 7 2 6 1 9 1 8 7 5 5 7 6 7 9 8 10 5 3 5 7 10 3 2 1 3 8 9 4 10 4 7 6 4 9 6 7 1 9 4 3 5 8 9 2 7 10 5 7 5 3 8 10 3 8 9 3 4 3 10 6 5 1 8 3 2 5 8 4 7 5 3 3 2 6 9 9 8 2 7 6 3 2 2 8 8 4 5 6 9 2 3 2 2 5 2",
"output": "287"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n10 5 8 4 4 4 1 4 5 8 3 10 2 4 1 10 8 1 1 6 8 4 2 9 1 3 1 7 7 9 3 5 5 8 6 9 9 4 8 1 3 3 2 6 1 5 4 5 3 5 5 6 7 5 7 9 3 5 4 9 2 6 8 1 1 7 7 3 8 9 8 7 3 2 4 1 6 1 3 9 4 2 2 8 5 10 1 8 8 5 1 5 6 9 4 5 6 5 10 2",
"output": "265"
},
{
"input": "100\n7 5 1 8 5 6 6 2 6 2 7 7 3 6 2 4 4 2 10 2 2 2 10 6 6 1 5 10 9 1 5 9 8 9 4 1 10 5 7 5 7 6 4 8 8 1 7 8 3 8 2 1 8 4 10 3 5 6 6 10 9 6 5 1 10 7 6 9 9 2 10 10 9 1 2 1 7 7 4 10 1 10 5 5 3 8 9 8 1 4 10 2 4 5 4 4 1 6 2 9",
"output": "328"
},
{
"input": "100\n5 6 10 7 1 7 10 1 9 1 5 1 4 1 3 3 7 9 1 6 1 6 5 7 1 6 3 1 3 6 3 8 2 4 1 5 2 10 7 3 10 4 10 1 5 4 2 9 7 9 5 7 10 4 1 4 8 9 3 1 3 7 7 4 3 7 7 10 6 9 5 5 6 5 3 9 8 8 5 5 4 10 9 4 10 4 1 8 3 5 4 10 9 3 10 4 10 7 10 9",
"output": "324"
},
{
"input": "10\n7 4 5 3 9 1 10 3 4 5",
"output": "34"
},
{
"input": "10\n8 9 6 5 6 4 10 9 1 4",
"output": "39"
},
{
"input": "10\n1 7 1 4 9 10 8 2 9 9",
"output": "40"
},
{
"input": "1\n100000",
"output": "100000"
}
] | 1,695,411,759
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 31
| 0
|
tests = int(input())
nums = [int(n) for n in input().split()]
points = []
def test(num, nums = nums):
newNums = nums.copy()
for n in nums:
if n == num-1 or n == num+1:
newNums.remove(n)
return sum(newNums)
for num in nums:
points.append(test(num))
print(max(points))
|
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
tests = int(input())
nums = [int(n) for n in input().split()]
points = []
def test(num, nums = nums):
newNums = nums.copy()
for n in nums:
if n == num-1 or n == num+1:
newNums.remove(n)
return sum(newNums)
for num in nums:
points.append(test(num))
print(max(points))
```
| 0
|
|
217
|
A
|
Ice Skating
|
PROGRAMMING
| 1,200
|
[
"brute force",
"dfs and similar",
"dsu",
"graphs"
] | null | null |
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
|
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
|
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
|
[
"2\n2 1\n1 2\n",
"2\n2 1\n4 1\n"
] |
[
"1\n",
"0\n"
] |
none
| 500
|
[
{
"input": "2\n2 1\n1 2",
"output": "1"
},
{
"input": "2\n2 1\n4 1",
"output": "0"
},
{
"input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182 184\n806 663\n468 459",
"output": "21"
},
{
"input": "17\n660 646\n440 442\n689 618\n441 415\n922 865\n950 972\n312 366\n203 229\n873 860\n219 199\n344 308\n169 176\n961 992\n153 84\n201 230\n987 938\n834 815",
"output": "16"
},
{
"input": "11\n798 845\n722 911\n374 270\n629 537\n748 856\n831 885\n486 641\n751 829\n609 492\n98 27\n654 663",
"output": "10"
},
{
"input": "1\n321 88",
"output": "0"
},
{
"input": "9\n811 859\n656 676\n76 141\n945 951\n497 455\n18 55\n335 294\n267 275\n656 689",
"output": "7"
},
{
"input": "7\n948 946\n130 130\n761 758\n941 938\n971 971\n387 385\n509 510",
"output": "6"
},
{
"input": "6\n535 699\n217 337\n508 780\n180 292\n393 112\n732 888",
"output": "5"
},
{
"input": "14\n25 23\n499 406\n193 266\n823 751\n219 227\n101 138\n978 992\n43 74\n997 932\n237 189\n634 538\n774 740\n842 767\n742 802",
"output": "13"
},
{
"input": "12\n548 506\n151 198\n370 380\n655 694\n654 690\n407 370\n518 497\n819 827\n765 751\n802 771\n741 752\n653 662",
"output": "11"
},
{
"input": "40\n685 711\n433 403\n703 710\n491 485\n616 619\n288 282\n884 871\n367 352\n500 511\n977 982\n51 31\n576 564\n508 519\n755 762\n22 20\n368 353\n232 225\n953 955\n452 436\n311 330\n967 988\n369 364\n791 803\n150 149\n651 661\n118 93\n398 387\n748 766\n852 852\n230 228\n555 545\n515 519\n667 678\n867 862\n134 146\n859 863\n96 99\n486 469\n303 296\n780 786",
"output": "38"
},
{
"input": "3\n175 201\n907 909\n388 360",
"output": "2"
},
{
"input": "7\n312 298\n86 78\n73 97\n619 594\n403 451\n538 528\n71 86",
"output": "6"
},
{
"input": "19\n802 820\n368 248\n758 794\n455 378\n876 888\n771 814\n245 177\n586 555\n844 842\n364 360\n820 856\n731 624\n982 975\n825 856\n122 121\n862 896\n42 4\n792 841\n828 820",
"output": "16"
},
{
"input": "32\n643 877\n842 614\n387 176\n99 338\n894 798\n652 728\n611 648\n622 694\n579 781\n243 46\n322 305\n198 438\n708 579\n246 325\n536 459\n874 593\n120 277\n989 907\n223 110\n35 130\n761 692\n690 661\n518 766\n226 93\n678 597\n725 617\n661 574\n775 496\n56 416\n14 189\n358 359\n898 901",
"output": "31"
},
{
"input": "32\n325 327\n20 22\n72 74\n935 933\n664 663\n726 729\n785 784\n170 171\n315 314\n577 580\n984 987\n313 317\n434 435\n962 961\n55 54\n46 44\n743 742\n434 433\n617 612\n332 332\n883 886\n940 936\n793 792\n645 644\n611 607\n418 418\n465 465\n219 218\n167 164\n56 54\n403 405\n210 210",
"output": "29"
},
{
"input": "32\n652 712\n260 241\n27 154\n188 16\n521 351\n518 356\n452 540\n790 827\n339 396\n336 551\n897 930\n828 627\n27 168\n180 113\n134 67\n794 671\n812 711\n100 241\n686 813\n138 289\n384 506\n884 932\n913 959\n470 508\n730 734\n373 478\n788 862\n392 426\n148 68\n113 49\n713 852\n924 894",
"output": "29"
},
{
"input": "14\n685 808\n542 677\n712 747\n832 852\n187 410\n399 338\n626 556\n530 635\n267 145\n215 209\n559 684\n944 949\n753 596\n601 823",
"output": "13"
},
{
"input": "5\n175 158\n16 2\n397 381\n668 686\n957 945",
"output": "4"
},
{
"input": "5\n312 284\n490 509\n730 747\n504 497\n782 793",
"output": "4"
},
{
"input": "2\n802 903\n476 348",
"output": "1"
},
{
"input": "4\n325 343\n425 442\n785 798\n275 270",
"output": "3"
},
{
"input": "28\n462 483\n411 401\n118 94\n111 127\n5 6\n70 52\n893 910\n73 63\n818 818\n182 201\n642 633\n900 886\n893 886\n684 700\n157 173\n953 953\n671 660\n224 225\n832 801\n152 157\n601 585\n115 101\n739 722\n611 606\n659 642\n461 469\n702 689\n649 653",
"output": "25"
},
{
"input": "36\n952 981\n885 900\n803 790\n107 129\n670 654\n143 132\n66 58\n813 819\n849 837\n165 198\n247 228\n15 39\n619 618\n105 138\n868 855\n965 957\n293 298\n613 599\n227 212\n745 754\n723 704\n877 858\n503 487\n678 697\n592 595\n155 135\n962 982\n93 89\n660 673\n225 212\n967 987\n690 680\n804 813\n489 518\n240 221\n111 124",
"output": "34"
},
{
"input": "30\n89 3\n167 156\n784 849\n943 937\n144 95\n24 159\n80 120\n657 683\n585 596\n43 147\n909 964\n131 84\n345 389\n333 321\n91 126\n274 325\n859 723\n866 922\n622 595\n690 752\n902 944\n127 170\n426 383\n905 925\n172 284\n793 810\n414 510\n890 884\n123 24\n267 255",
"output": "29"
},
{
"input": "5\n664 666\n951 941\n739 742\n844 842\n2 2",
"output": "4"
},
{
"input": "3\n939 867\n411 427\n757 708",
"output": "2"
},
{
"input": "36\n429 424\n885 972\n442 386\n512 511\n751 759\n4 115\n461 497\n496 408\n8 23\n542 562\n296 331\n448 492\n412 395\n109 166\n622 640\n379 355\n251 262\n564 586\n66 115\n275 291\n666 611\n629 534\n510 567\n635 666\n738 803\n420 369\n92 17\n101 144\n141 92\n258 258\n184 235\n492 456\n311 210\n394 357\n531 512\n634 636",
"output": "34"
},
{
"input": "29\n462 519\n871 825\n127 335\n156 93\n576 612\n885 830\n634 779\n340 105\n744 795\n716 474\n93 139\n563 805\n137 276\n177 101\n333 14\n391 437\n873 588\n817 518\n460 597\n572 670\n140 303\n392 441\n273 120\n862 578\n670 639\n410 161\n544 577\n193 116\n252 195",
"output": "28"
},
{
"input": "23\n952 907\n345 356\n812 807\n344 328\n242 268\n254 280\n1000 990\n80 78\n424 396\n595 608\n755 813\n383 380\n55 56\n598 633\n203 211\n508 476\n600 593\n206 192\n855 882\n517 462\n967 994\n642 657\n493 488",
"output": "22"
},
{
"input": "10\n579 816\n806 590\n830 787\n120 278\n677 800\n16 67\n188 251\n559 560\n87 67\n104 235",
"output": "8"
},
{
"input": "23\n420 424\n280 303\n515 511\n956 948\n799 803\n441 455\n362 369\n299 289\n823 813\n982 967\n876 878\n185 157\n529 551\n964 989\n655 656\n1 21\n114 112\n45 56\n935 937\n1000 997\n934 942\n360 366\n648 621",
"output": "22"
},
{
"input": "23\n102 84\n562 608\n200 127\n952 999\n465 496\n322 367\n728 690\n143 147\n855 867\n861 866\n26 59\n300 273\n255 351\n192 246\n70 111\n365 277\n32 104\n298 319\n330 354\n241 141\n56 125\n315 298\n412 461",
"output": "22"
},
{
"input": "7\n429 506\n346 307\n99 171\n853 916\n322 263\n115 157\n906 924",
"output": "6"
},
{
"input": "3\n1 1\n2 1\n2 2",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "5\n1 1\n1 2\n2 2\n3 1\n3 3",
"output": "0"
},
{
"input": "6\n1 1\n1 2\n2 2\n3 1\n3 2\n3 3",
"output": "0"
},
{
"input": "20\n1 1\n2 2\n3 3\n3 9\n4 4\n5 2\n5 5\n5 7\n5 8\n6 2\n6 6\n6 9\n7 7\n8 8\n9 4\n9 7\n9 9\n10 2\n10 9\n10 10",
"output": "1"
},
{
"input": "21\n1 1\n1 9\n2 1\n2 2\n2 5\n2 6\n2 9\n3 3\n3 8\n4 1\n4 4\n5 5\n5 8\n6 6\n7 7\n8 8\n9 9\n10 4\n10 10\n11 5\n11 11",
"output": "1"
},
{
"input": "22\n1 1\n1 3\n1 4\n1 8\n1 9\n1 11\n2 2\n3 3\n4 4\n4 5\n5 5\n6 6\n6 8\n7 7\n8 3\n8 4\n8 8\n9 9\n10 10\n11 4\n11 9\n11 11",
"output": "3"
},
{
"input": "50\n1 1\n2 2\n2 9\n3 3\n4 4\n4 9\n4 16\n4 24\n5 5\n6 6\n7 7\n8 8\n8 9\n8 20\n9 9\n10 10\n11 11\n12 12\n13 13\n14 7\n14 14\n14 16\n14 25\n15 4\n15 6\n15 15\n15 22\n16 6\n16 16\n17 17\n18 18\n19 6\n19 19\n20 20\n21 21\n22 6\n22 22\n23 23\n24 6\n24 7\n24 8\n24 9\n24 24\n25 1\n25 3\n25 5\n25 7\n25 23\n25 24\n25 25",
"output": "7"
},
{
"input": "55\n1 1\n1 14\n2 2\n2 19\n3 1\n3 3\n3 8\n3 14\n3 23\n4 1\n4 4\n5 5\n5 8\n5 15\n6 2\n6 3\n6 4\n6 6\n7 7\n8 8\n8 21\n9 9\n10 1\n10 10\n11 9\n11 11\n12 12\n13 13\n14 14\n15 15\n15 24\n16 5\n16 16\n17 5\n17 10\n17 17\n17 18\n17 22\n17 27\n18 18\n19 19\n20 20\n21 20\n21 21\n22 22\n23 23\n24 14\n24 24\n25 25\n26 8\n26 11\n26 26\n27 3\n27 27\n28 28",
"output": "5"
},
{
"input": "3\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "6\n4 4\n3 4\n5 4\n4 5\n4 3\n3 1",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n1 2",
"output": "0"
},
{
"input": "8\n1 3\n1 1\n4 1\n2 2\n2 5\n5 9\n5 1\n5 4",
"output": "1"
},
{
"input": "10\n1 1\n1 2\n1 3\n1 4\n5 5\n6 6\n7 7\n8 8\n9 9\n100 100",
"output": "6"
},
{
"input": "7\n1 1\n2 2\n3 3\n4 4\n1 2\n2 3\n3 4",
"output": "0"
},
{
"input": "6\n1 1\n2 1\n2 2\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "4\n3 1\n2 1\n2 2\n1 2",
"output": "0"
},
{
"input": "6\n1 1\n2 2\n2 1\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "3\n1 2\n1 3\n1 4",
"output": "0"
},
{
"input": "4\n1 1\n2 2\n1 2\n2 1",
"output": "0"
},
{
"input": "4\n1 3\n2 1\n3 2\n3 1",
"output": "1"
},
{
"input": "7\n1 1\n1 2\n2 2\n3 3\n3 4\n4 4\n1 4",
"output": "0"
},
{
"input": "21\n12 12\n13 12\n12 11\n13 13\n10 10\n11 10\n11 11\n501 500\n501 501\n503 502\n500 500\n503 503\n502 501\n502 502\n700 700\n702 702\n703 702\n701 701\n702 701\n703 703\n701 700",
"output": "2"
},
{
"input": "6\n1 11\n6 8\n11 10\n1 10\n11 11\n6 9",
"output": "1"
},
{
"input": "4\n1 1\n2 2\n3 2\n3 1",
"output": "0"
},
{
"input": "3\n1 2\n3 4\n3 2",
"output": "0"
},
{
"input": "3\n1 1\n1 2\n2 2",
"output": "0"
},
{
"input": "4\n5 5\n5 4\n6 3\n6 4",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n2 1",
"output": "0"
}
] | 1,585,943,754
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 76
| 248
| 307,200
|
def search(coord, start, n):
if coord[start]['pass']:
return
coord[start]['pass'] = True
for i in range(n):
if coord[i]['x'] == coord[start]['x'] or coord[i]['y'] == coord[start]['y']:
search(coord, i, n)
snow_drifts_number = input()
snow_drifts_number = int(snow_drifts_number)
drift_coord = []
drifts_needed = 0
for i in range(snow_drifts_number):
x, y = input().split(' ')
x = int(x)
y = int(y)
drift_coord.append({'x': x, 'y': y, 'pass': False})
for i in range(snow_drifts_number):
if drift_coord[i]['pass']:
continue
drifts_needed += 1
search(drift_coord, i, snow_drifts_number)
print(drifts_needed - 1)
|
Title: Ice Skating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output Specification:
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
Demo Input:
['2\n2 1\n1 2\n', '2\n2 1\n4 1\n']
Demo Output:
['1\n', '0\n']
Note:
none
|
```python
def search(coord, start, n):
if coord[start]['pass']:
return
coord[start]['pass'] = True
for i in range(n):
if coord[i]['x'] == coord[start]['x'] or coord[i]['y'] == coord[start]['y']:
search(coord, i, n)
snow_drifts_number = input()
snow_drifts_number = int(snow_drifts_number)
drift_coord = []
drifts_needed = 0
for i in range(snow_drifts_number):
x, y = input().split(' ')
x = int(x)
y = int(y)
drift_coord.append({'x': x, 'y': y, 'pass': False})
for i in range(snow_drifts_number):
if drift_coord[i]['pass']:
continue
drifts_needed += 1
search(drift_coord, i, snow_drifts_number)
print(drifts_needed - 1)
```
| 3
|
|
177
|
B1
|
Rectangular Game
|
PROGRAMMING
| 1,000
|
[
"number theory"
] | null | null |
The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*.
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where:
- *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1
The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game.
|
The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=50
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=109
|
Print a single number — the maximum possible result of the game.
|
[
"10\n",
"8\n"
] |
[
"16\n",
"15\n"
] |
Consider the first example (*c*<sub class="lower-index">1</sub> = 10). The possible options for the game development are:
- Arrange the pebbles in 10 rows, one pebble per row. Then *c*<sub class="lower-index">2</sub> = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*<sub class="lower-index">2</sub> = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*<sub class="lower-index">2</sub> = 5, *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 16 — the maximum possible result.
| 30
|
[
{
"input": "10",
"output": "16"
},
{
"input": "8",
"output": "15"
},
{
"input": "4",
"output": "7"
},
{
"input": "36",
"output": "67"
},
{
"input": "32",
"output": "63"
},
{
"input": "46",
"output": "70"
},
{
"input": "6",
"output": "10"
},
{
"input": "13",
"output": "14"
},
{
"input": "9",
"output": "13"
},
{
"input": "18",
"output": "31"
},
{
"input": "20",
"output": "36"
},
{
"input": "37",
"output": "38"
},
{
"input": "42",
"output": "71"
},
{
"input": "50",
"output": "81"
},
{
"input": "48",
"output": "94"
},
{
"input": "33",
"output": "45"
},
{
"input": "49",
"output": "57"
},
{
"input": "34",
"output": "52"
},
{
"input": "35",
"output": "43"
},
{
"input": "37",
"output": "38"
},
{
"input": "39",
"output": "53"
},
{
"input": "42",
"output": "71"
},
{
"input": "14",
"output": "22"
},
{
"input": "2",
"output": "3"
},
{
"input": "3",
"output": "4"
},
{
"input": "5",
"output": "6"
}
] | 1,591,450,249
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS1
| 2
| 218
| 307,200
|
def hcf(x,y):
if x > y:
smaller = y
else:
smaller=x
for i in range(1, smaller+1):
if((x % i == 0) and (y % i == 0)):
hcf = i
return hcf
c=10
A=[]
p=10
n=int(input())
for i in range(n-1,0,-1):
if n%i==0:
A.append(i)
for i in range(len(A)):
p=hcf(A[i],p)
if p!=1:
c+=p
else:
if i==len(A)-1:
c+=p
else:
continue
print(c)
|
Title: Rectangular Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*.
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where:
- *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1
The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game.
Input Specification:
The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=50
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=109
Output Specification:
Print a single number — the maximum possible result of the game.
Demo Input:
['10\n', '8\n']
Demo Output:
['16\n', '15\n']
Note:
Consider the first example (*c*<sub class="lower-index">1</sub> = 10). The possible options for the game development are:
- Arrange the pebbles in 10 rows, one pebble per row. Then *c*<sub class="lower-index">2</sub> = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*<sub class="lower-index">2</sub> = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*<sub class="lower-index">2</sub> = 5, *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 16 — the maximum possible result.
|
```python
def hcf(x,y):
if x > y:
smaller = y
else:
smaller=x
for i in range(1, smaller+1):
if((x % i == 0) and (y % i == 0)):
hcf = i
return hcf
c=10
A=[]
p=10
n=int(input())
for i in range(n-1,0,-1):
if n%i==0:
A.append(i)
for i in range(len(A)):
p=hcf(A[i],p)
if p!=1:
c+=p
else:
if i==len(A)-1:
c+=p
else:
continue
print(c)
```
| 0
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,603,604,008
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 312
| 0
|
w1 = input()
w2 = input()
size = len(w2)
flag = True
for i in range(size):
if w1[i] != w2[size - i - 1]:
flag = False
break
if flag:
print("YES")
else:
print("NO")
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
w1 = input()
w2 = input()
size = len(w2)
flag = True
for i in range(size):
if w1[i] != w2[size - i - 1]:
flag = False
break
if flag:
print("YES")
else:
print("NO")
```
| 3.922
|
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"
},
{
"input": "2 3",
"output": "2 0 4"
},
{
"input": "5 6",
"output": "5 0 1"
},
{
"input": "4 4",
"output": "0 6 0"
},
{
"input": "1 1",
"output": "0 6 0"
},
{
"input": "6 4",
"output": "1 1 4"
},
{
"input": "1 4",
"output": "2 0 4"
},
{
"input": "5 5",
"output": "0 6 0"
},
{
"input": "4 5",
"output": "4 0 2"
},
{
"input": "4 3",
"output": "3 0 3"
},
{
"input": "1 5",
"output": "2 1 3"
},
{
"input": "6 5",
"output": "1 0 5"
},
{
"input": "2 2",
"output": "0 6 0"
},
{
"input": "1 3",
"output": "1 1 4"
},
{
"input": "3 6",
"output": "4 0 2"
},
{
"input": "3 1",
"output": "4 1 1"
},
{
"input": "3 2",
"output": "4 0 2"
},
{
"input": "3 5",
"output": "3 1 2"
},
{
"input": "3 3",
"output": "0 6 0"
},
{
"input": "6 2",
"output": "2 1 3"
},
{
"input": "4 1",
"output": "4 0 2"
},
{
"input": "5 2",
"output": "3 0 3"
},
{
"input": "4 2",
"output": "3 1 2"
},
{
"input": "2 1",
"output": "5 0 1"
},
{
"input": "6 1",
"output": "3 0 3"
},
{
"input": "4 6",
"output": "4 1 1"
},
{
"input": "2 6",
"output": "3 1 2"
},
{
"input": "3 4",
"output": "3 0 3"
},
{
"input": "1 2",
"output": "1 0 5"
},
{
"input": "6 6",
"output": "0 6 0"
},
{
"input": "5 4",
"output": "2 0 4"
},
{
"input": "3 3",
"output": "0 6 0"
},
{
"input": "1 1",
"output": "0 6 0"
}
] | 1,594,240,331
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 109
| 6,656,000
|
a, b = map(int, input().split())
x=y=z=0
for i in range(1, 7):
if abs(a-i)<abs(b-i):x+=1
elif abs(a-i)>abs(b-i):y+=1
else:z+=1
print(x,z,y)
|
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 = map(int, input().split())
x=y=z=0
for i in range(1, 7):
if abs(a-i)<abs(b-i):x+=1
elif abs(a-i)>abs(b-i):y+=1
else:z+=1
print(x,z,y)
```
| 3
|
|
465
|
B
|
Inbox (100500)
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Over time, Alexey's mail box got littered with too many letters. Some of them are read, while others are unread.
Alexey's mail program can either show a list of all letters or show the content of a single letter. As soon as the program shows the content of an unread letter, it becomes read letter (if the program shows the content of a read letter nothing happens). In one click he can do any of the following operations:
- Move from the list of letters to the content of any single letter.- Return to the list of letters from single letter viewing mode.- In single letter viewing mode, move to the next or to the previous letter in the list. You cannot move from the first letter to the previous one or from the last letter to the next one.
The program cannot delete the letters from the list or rearrange them.
Alexey wants to read all the unread letters and go watch football. Now he is viewing the list of all letters and for each letter he can see if it is read or unread. What minimum number of operations does Alexey need to perform to read all unread letters?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of letters in the mailbox.
The second line contains *n* space-separated integers (zeros and ones) — the state of the letter list. The *i*-th number equals either 1, if the *i*-th number is unread, or 0, if the *i*-th letter is read.
|
Print a single number — the minimum number of operations needed to make all the letters read.
|
[
"5\n0 1 0 1 0\n",
"5\n1 1 0 0 1\n",
"2\n0 0\n"
] |
[
"3\n",
"4\n",
"0\n"
] |
In the first sample Alexey needs three operations to cope with the task: open the second letter, move to the third one, move to the fourth one.
In the second sample the action plan: open the first letter, move to the second letter, return to the list, open the fifth letter.
In the third sample all letters are already read.
| 1,000
|
[
{
"input": "5\n0 1 0 1 0",
"output": "3"
},
{
"input": "5\n1 1 0 0 1",
"output": "4"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "9\n1 0 1 0 1 0 1 0 1",
"output": "9"
},
{
"input": "5\n1 1 1 1 1",
"output": "5"
},
{
"input": "14\n0 0 1 1 1 0 1 1 1 0 1 1 1 0",
"output": "11"
},
{
"input": "23\n1 1 1 0 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 0 1 1 1",
"output": "23"
},
{
"input": "27\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0",
"output": "25"
},
{
"input": "10\n1 0 0 0 0 1 0 0 0 1",
"output": "5"
},
{
"input": "10\n1 0 0 1 0 0 1 1 0 1",
"output": "8"
},
{
"input": "27\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "39\n1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1",
"output": "39"
},
{
"input": "48\n1 0 1 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 1 0 0 1 0 1 0 0 1 0 0 1 0 0 1 0 1 0 1 0 0 1 0 0 1",
"output": "39"
},
{
"input": "71\n0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0",
"output": "59"
},
{
"input": "99\n1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1",
"output": "99"
},
{
"input": "99\n1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1",
"output": "99"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "99"
},
{
"input": "193\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "1"
},
{
"input": "213\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "5"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "5\n0 0 0 0 1",
"output": "1"
},
{
"input": "12\n0 1 1 0 1 1 0 1 1 0 0 0",
"output": "8"
},
{
"input": "3\n1 0 0",
"output": "1"
},
{
"input": "5\n0 0 1 0 0",
"output": "1"
},
{
"input": "3\n0 1 0",
"output": "1"
},
{
"input": "6\n1 1 0 0 0 0",
"output": "2"
},
{
"input": "4\n0 1 0 0",
"output": "1"
},
{
"input": "5\n1 1 1 0 0",
"output": "3"
},
{
"input": "4\n1 0 0 0",
"output": "1"
},
{
"input": "6\n1 1 1 1 0 0",
"output": "4"
}
] | 1,586,873,365
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 13
| 109
| 0
|
n = int(input())
lst = list(map(int, input().split()))
i = 0
count = 0
while i < n-1:
if lst[i] == 1:
count += 1
if lst[i+1] != 1 and i+1 != n-1:
count += 1
i += 1
if lst[i] == 1:
count += 1
print(count)
|
Title: Inbox (100500)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Over time, Alexey's mail box got littered with too many letters. Some of them are read, while others are unread.
Alexey's mail program can either show a list of all letters or show the content of a single letter. As soon as the program shows the content of an unread letter, it becomes read letter (if the program shows the content of a read letter nothing happens). In one click he can do any of the following operations:
- Move from the list of letters to the content of any single letter.- Return to the list of letters from single letter viewing mode.- In single letter viewing mode, move to the next or to the previous letter in the list. You cannot move from the first letter to the previous one or from the last letter to the next one.
The program cannot delete the letters from the list or rearrange them.
Alexey wants to read all the unread letters and go watch football. Now he is viewing the list of all letters and for each letter he can see if it is read or unread. What minimum number of operations does Alexey need to perform to read all unread letters?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of letters in the mailbox.
The second line contains *n* space-separated integers (zeros and ones) — the state of the letter list. The *i*-th number equals either 1, if the *i*-th number is unread, or 0, if the *i*-th letter is read.
Output Specification:
Print a single number — the minimum number of operations needed to make all the letters read.
Demo Input:
['5\n0 1 0 1 0\n', '5\n1 1 0 0 1\n', '2\n0 0\n']
Demo Output:
['3\n', '4\n', '0\n']
Note:
In the first sample Alexey needs three operations to cope with the task: open the second letter, move to the third one, move to the fourth one.
In the second sample the action plan: open the first letter, move to the second letter, return to the list, open the fifth letter.
In the third sample all letters are already read.
|
```python
n = int(input())
lst = list(map(int, input().split()))
i = 0
count = 0
while i < n-1:
if lst[i] == 1:
count += 1
if lst[i+1] != 1 and i+1 != n-1:
count += 1
i += 1
if lst[i] == 1:
count += 1
print(count)
```
| 0
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,609,700,084
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 218
| 0
|
A = input()
B = input()
poczatek = 0
koniec = len(A) - 1
warunek = True
if len(A) == len(B):
for x in range(len(A)):
if A[poczatek] != B[koniec]:
warunek = False
poczatek += 1
koniec -= 1
if warunek == True:
print('YES')
else:
print('NO')
else:
print('NO')
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
A = input()
B = input()
poczatek = 0
koniec = len(A) - 1
warunek = True
if len(A) == len(B):
for x in range(len(A)):
if A[poczatek] != B[koniec]:
warunek = False
poczatek += 1
koniec -= 1
if warunek == True:
print('YES')
else:
print('NO')
else:
print('NO')
```
| 3.9455
|
912
|
B
|
New Year's Eve
|
PROGRAMMING
| 1,300
|
[
"bitmasks",
"constructive algorithms",
"number theory"
] | null | null |
Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.
|
The sole string contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1018).
|
Output one number — the largest possible xor-sum.
|
[
"4 3\n",
"6 6\n"
] |
[
"7\n",
"7\n"
] |
In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.
| 1,000
|
[
{
"input": "4 3",
"output": "7"
},
{
"input": "6 6",
"output": "7"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "1022 10",
"output": "1023"
},
{
"input": "415853337373441 52",
"output": "562949953421311"
},
{
"input": "75 12",
"output": "127"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1152921504606846975"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000000000000 2",
"output": "1152921504606846975"
},
{
"input": "49194939 22",
"output": "67108863"
},
{
"input": "228104606 17",
"output": "268435455"
},
{
"input": "817034381 7",
"output": "1073741823"
},
{
"input": "700976748 4",
"output": "1073741823"
},
{
"input": "879886415 9",
"output": "1073741823"
},
{
"input": "18007336 10353515",
"output": "33554431"
},
{
"input": "196917003 154783328",
"output": "268435455"
},
{
"input": "785846777 496205300",
"output": "1073741823"
},
{
"input": "964756444 503568330",
"output": "1073741823"
},
{
"input": "848698811 317703059",
"output": "1073741823"
},
{
"input": "676400020444788 1",
"output": "676400020444788"
},
{
"input": "502643198528213 1",
"output": "502643198528213"
},
{
"input": "815936580997298686 684083143940282566",
"output": "1152921504606846975"
},
{
"input": "816762824175382110 752185261508428780",
"output": "1152921504606846975"
},
{
"input": "327942415253132295 222598158321260499",
"output": "576460752303423487"
},
{
"input": "328768654136248423 284493129147496637",
"output": "576460752303423487"
},
{
"input": "329594893019364551 25055600080496801",
"output": "576460752303423487"
},
{
"input": "921874985256864012 297786684518764536",
"output": "1152921504606846975"
},
{
"input": "922701224139980141 573634416190460758",
"output": "1152921504606846975"
},
{
"input": "433880815217730325 45629641110945892",
"output": "576460752303423487"
},
{
"input": "434707058395813749 215729375494216481",
"output": "576460752303423487"
},
{
"input": "435533301573897173 34078453236225189",
"output": "576460752303423487"
},
{
"input": "436359544751980597 199220719961060641",
"output": "576460752303423487"
},
{
"input": "437185783635096725 370972992240105630",
"output": "576460752303423487"
},
{
"input": "438012026813180149 111323110116193830",
"output": "576460752303423487"
},
{
"input": "438838269991263573 295468957052046146",
"output": "576460752303423487"
},
{
"input": "439664513169346997 46560240538186155",
"output": "576460752303423487"
},
{
"input": "440490752052463125 216165966013438147",
"output": "576460752303423487"
},
{
"input": "441316995230546549 401964286420555423",
"output": "576460752303423487"
},
{
"input": "952496582013329437 673506882352402278",
"output": "1152921504606846975"
},
{
"input": "1000000000000000000 1",
"output": "1000000000000000000"
},
{
"input": "2147483647 1",
"output": "2147483647"
},
{
"input": "2147483647 2",
"output": "2147483647"
},
{
"input": "2147483647 31",
"output": "2147483647"
},
{
"input": "8 2",
"output": "15"
},
{
"input": "3 3",
"output": "3"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "10 2",
"output": "15"
},
{
"input": "288230376151711743 2",
"output": "288230376151711743"
},
{
"input": "5 2",
"output": "7"
},
{
"input": "576460752303423487 2",
"output": "576460752303423487"
},
{
"input": "36028797018963967 123",
"output": "36028797018963967"
},
{
"input": "1125899906842623 2",
"output": "1125899906842623"
},
{
"input": "576460752303423489 5",
"output": "1152921504606846975"
},
{
"input": "288230376151711743 3",
"output": "288230376151711743"
},
{
"input": "36028797018963967 345",
"output": "36028797018963967"
},
{
"input": "18014398509481984 30",
"output": "36028797018963967"
},
{
"input": "8 8",
"output": "15"
},
{
"input": "8 1",
"output": "8"
}
] | 1,552,464,174
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 124
| 0
|
import sys
n,k = [int(s) for s in input().split()]
if k == 1:
print(n)
else:
copy = n
array = []
while n != 0:
array.append(n % 2)
n = n // 2
array = array[::-1]
num = 0
sum_ = 0
#print(array)
for i in range(len(array)):
if not array[i]:
num += (2 ** (len(array) - i - 1))
#print(2 ** i)
sum_ += (2 ** i)
if num < copy:
print(sum_)
else:
print(copy)
|
Title: New Year's Eve
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.
Input Specification:
The sole string contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1018).
Output Specification:
Output one number — the largest possible xor-sum.
Demo Input:
['4 3\n', '6 6\n']
Demo Output:
['7\n', '7\n']
Note:
In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.
|
```python
import sys
n,k = [int(s) for s in input().split()]
if k == 1:
print(n)
else:
copy = n
array = []
while n != 0:
array.append(n % 2)
n = n // 2
array = array[::-1]
num = 0
sum_ = 0
#print(array)
for i in range(len(array)):
if not array[i]:
num += (2 ** (len(array) - i - 1))
#print(2 ** i)
sum_ += (2 ** i)
if num < copy:
print(sum_)
else:
print(copy)
```
| 3
|
|
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": "4\n< 10 Y\n> -6 Y\n< 10 Y\n< -10 N",
"output": "-5"
},
{
"input": "1\n< 1 N",
"output": "1361956"
},
{
"input": "1\n<= 1 Y",
"output": "-1998638045"
},
{
"input": "1\n> 1 N",
"output": "-1998638045"
},
{
"input": "1\n>= 1 Y",
"output": "1361956"
},
{
"input": "4\n< 1 N\n< 1 N\n< 1 N\n<= 1 Y",
"output": "1"
},
{
"input": "4\n< 1 N\n>= 1 Y\n< 1 N\n< 1 N",
"output": "1361956"
},
{
"input": "4\n> 1 N\n<= 1 Y\n<= 1 Y\n> 1 N",
"output": "-1998638045"
},
{
"input": "4\n>= 1 Y\n> 1 N\n>= 1 Y\n>= 1 Y",
"output": "1"
},
{
"input": "4\n<= 9 Y\n< 3 Y\n< 2 Y\n< 2 Y",
"output": "-1998638045"
},
{
"input": "4\n< 0 N\n< -7 N\n>= 8 N\n>= -5 Y",
"output": "3"
},
{
"input": "4\n<= -6 N\n<= -8 N\n<= 3 Y\n<= 7 Y",
"output": "-2"
},
{
"input": "4\n>= 7 N\n<= -1 N\n>= 5 N\n<= -10 N",
"output": "0"
},
{
"input": "4\n> 5 N\n>= -5 Y\n> -9 Y\n> -9 Y",
"output": "-4"
},
{
"input": "10\n<= -60 N\n>= -59 Y\n> 22 Y\n> 95 N\n<= 91 Y\n> 77 Y\n>= -59 Y\n> -25 Y\n> -22 Y\n>= 52 Y",
"output": "85"
},
{
"input": "10\n>= -18 Y\n>= -35 Y\n> -94 Y\n< -23 N\n< -69 N\n< -68 N\n< 82 Y\n> 92 N\n< 29 Y\n>= -25 Y",
"output": "18"
},
{
"input": "10\n>= 18 Y\n<= -32 N\n>= 85 N\n<= 98 Y\n<= -43 N\n<= -79 N\n>= 97 N\n< -38 N\n< -55 N\n<= -93 N",
"output": "64"
},
{
"input": "10\n<= 2 Y\n< -33 Y\n> 6 N\n> -6 N\n< -28 Y\n> -62 Y\n< 57 Y\n<= 24 Y\n> 23 N\n> -25 N",
"output": "-54"
},
{
"input": "10\n<= -31 N\n>= 66 N\n<= 0 Y\n> -95 Y\n< 27 Y\n< -42 N\n> 3 N\n< 6 Y\n>= -42 Y\n> -70 Y",
"output": "-29"
},
{
"input": "10\n>= 54 N\n<= -52 N\n>= 64 N\n> 65 N\n< 37 Y\n> -84 Y\n>= -94 Y\n>= -95 Y\n> -72 Y\n<= 18 N",
"output": "22"
},
{
"input": "10\n> -24 N\n<= -5 Y\n<= -33 Y\n> 45 N\n> -59 Y\n> -21 N\n<= -48 N\n> 40 N\n< 12 Y\n>= 14 N",
"output": "-47"
},
{
"input": "10\n>= 91 Y\n>= -68 Y\n< 92 N\n>= -15 Y\n> 51 Y\n<= 14 N\n> 17 Y\n< 94 Y\n>= 49 Y\n> -36 Y",
"output": "93"
},
{
"input": "1\n< -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n< 1 Y",
"output": "-1998638045"
},
{
"input": "1\n>= -999999999 Y",
"output": "-998638044"
},
{
"input": "1\n> 100000 Y",
"output": "1461956"
},
{
"input": "1\n<= 999999999 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 1000000000 N",
"output": "1001361956"
},
{
"input": "4\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n>= 1000000000 Y",
"output": "1001361955"
},
{
"input": "1\n<= 999999999 N",
"output": "1001361955"
},
{
"input": "1\n<= 100 Y",
"output": "-1998638045"
},
{
"input": "1\n> 1000000000 Y",
"output": "1001361956"
},
{
"input": "1\n<= 1 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n<= -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n<= -999999999 Y",
"output": "-1998638045"
},
{
"input": "1\n> 100 Y",
"output": "1362056"
},
{
"input": "2\n< -1000000000 Y\n< 3 Y",
"output": "-1998638045"
},
{
"input": "1\n<= -1000000 Y",
"output": "-1998638045"
},
{
"input": "8\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 15 Y",
"output": "-1998638045"
},
{
"input": "1\n< 1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n> 10000000 Y",
"output": "11361956"
},
{
"input": "1\n< 0 Y",
"output": "-1998638045"
},
{
"input": "1\n< 100 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 5 Y",
"output": "-1998638045"
},
{
"input": "3\n> 5 Y\n> 0 Y\n< 4 Y",
"output": "Impossible"
},
{
"input": "1\n>= -1000000000 N",
"output": "-1998638045"
},
{
"input": "3\n>= 1 Y\n<= 1 Y\n> 10 Y",
"output": "Impossible"
},
{
"input": "2\n> 1 Y\n< 2 Y",
"output": "Impossible"
},
{
"input": "3\n>= 5 Y\n<= 5 Y\n< 5 Y",
"output": "Impossible"
},
{
"input": "2\n>= 5 N\n> 5 Y",
"output": "Impossible"
},
{
"input": "3\n>= 4 Y\n> 4 Y\n<= 4 Y",
"output": "Impossible"
},
{
"input": "2\n>= 4 Y\n> 4 Y",
"output": "1361960"
}
] | 1,398,433,333
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 77
| 0
|
n = int(input())
mn = -100000000000000000
mx = 1000000000000000000
for i in range(n):
znak, m, ans = map(str, input().split())
if znak == '>':
if ans == 'Y':
mn = max(mn, int(m) + 1)
else:
mx = min(mx, int(m))
if znak == '<':
if ans == 'Y':
mx = min(mx, int(m) - 1)
else:
mn = max(mn, int(m))
if znak == '>=':
if ans == 'Y':
mn = max(mn, int(m))
else:
mx = min(mx, int(m) - 1)
if znak == '<=':
if ans == 'Y':
mx = min(mx, int(m))
else:
mn = max(mn, int(m) + 1)
if mx < mn:
print("Impossible")
else:
print((mx + mn) // 2)
|
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
n = int(input())
mn = -100000000000000000
mx = 1000000000000000000
for i in range(n):
znak, m, ans = map(str, input().split())
if znak == '>':
if ans == 'Y':
mn = max(mn, int(m) + 1)
else:
mx = min(mx, int(m))
if znak == '<':
if ans == 'Y':
mx = min(mx, int(m) - 1)
else:
mn = max(mn, int(m))
if znak == '>=':
if ans == 'Y':
mn = max(mn, int(m))
else:
mx = min(mx, int(m) - 1)
if znak == '<=':
if ans == 'Y':
mx = min(mx, int(m))
else:
mn = max(mn, int(m) + 1)
if mx < mn:
print("Impossible")
else:
print((mx + mn) // 2)
```
| 0
|
|
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 94 8 16 73 85",
"output": "YES"
},
{
"input": "4\n25 35 7 31\n70 37 43 35",
"output": "YES"
},
{
"input": "10\n15 26 15 14 14 39 40 4 25 39\n27 72 16 44 69 48 53 17 63 42",
"output": "NO"
},
{
"input": "5\n22 5 19 16 32\n26 10 43 38 37",
"output": "NO"
},
{
"input": "5\n32 4 22 40 26\n39 20 36 98 44",
"output": "YES"
},
{
"input": "6\n18 25 3 10 13 37\n38 73 19 35 24 37",
"output": "YES"
},
{
"input": "2\n2 2\n2 2",
"output": "YES"
},
{
"input": "2\n2 5\n2 5",
"output": "YES"
},
{
"input": "2\n1000 1008\n10000 2352",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5\n1 2 3 4 11",
"output": "YES"
},
{
"input": "4\n1 0 0 0\n2 0 0 0",
"output": "YES"
},
{
"input": "2\n0 0\n1 2",
"output": "YES"
},
{
"input": "3\n9 13 4\n10 14 5",
"output": "NO"
},
{
"input": "2\n0 0\n1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 1\n1 1 2 3 4",
"output": "NO"
},
{
"input": "2\n0 0\n0 0",
"output": "YES"
},
{
"input": "3\n5 1 1\n5 5 5",
"output": "YES"
}
] | 1,530,269,413
| 313
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 93
| 0
|
n=int(input())
l=list(map(int,input().strip().split()))
l1=list(map(int,input().strip().split()))
s=sum(l)
l.sort()
r=l1[-1]+l1[-2]
if s<=r:
print ("YES")
else:
print ("NO")
|
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())
l=list(map(int,input().strip().split()))
l1=list(map(int,input().strip().split()))
s=sum(l)
l.sort()
r=l1[-1]+l1[-2]
if s<=r:
print ("YES")
else:
print ("NO")
```
| 0
|
|
825
|
C
|
Multi-judge Solving
|
PROGRAMMING
| 1,600
|
[
"greedy",
"implementation"
] | null | null |
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
|
The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
|
[
"3 3\n2 1 9\n",
"4 20\n10 3 6 3\n"
] |
[
"1\n",
"0\n"
] |
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
| 0
|
[
{
"input": "3 3\n2 1 9",
"output": "1"
},
{
"input": "4 20\n10 3 6 3",
"output": "0"
},
{
"input": "1 1000000000\n1",
"output": "0"
},
{
"input": "1 1\n3",
"output": "1"
},
{
"input": "50 100\n74 55 33 5 83 24 75 59 30 36 13 4 62 28 96 17 6 35 45 53 33 11 37 93 34 79 61 72 13 31 44 75 7 3 63 46 18 16 44 89 62 25 32 12 38 55 75 56 61 82",
"output": "0"
},
{
"input": "100 10\n246 286 693 607 87 612 909 312 621 37 801 558 504 914 416 762 187 974 976 123 635 488 416 659 988 998 93 662 92 749 889 78 214 786 735 625 921 372 713 617 975 119 402 411 878 138 548 905 802 762 940 336 529 373 745 835 805 880 816 94 166 114 475 699 974 462 61 337 555 805 968 815 392 746 591 558 740 380 668 29 881 151 387 986 174 923 541 520 998 947 535 651 103 584 664 854 180 852 726 93",
"output": "1"
},
{
"input": "2 1\n1 1000000000",
"output": "29"
},
{
"input": "29 2\n1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 16383 32767 65535 131071 262143 524287 1048575 2097151 4194303 8388607 16777215 33554431 67108863 134217727 268435455 536870911",
"output": "27"
},
{
"input": "1 1\n1000000000",
"output": "29"
},
{
"input": "7 6\n4 20 16 14 3 17 4",
"output": "1"
},
{
"input": "2 1\n3 6",
"output": "1"
},
{
"input": "1 1\n20",
"output": "4"
},
{
"input": "5 2\n86 81 53 25 18",
"output": "4"
},
{
"input": "4 1\n88 55 14 39",
"output": "4"
},
{
"input": "3 1\n2 3 6",
"output": "0"
},
{
"input": "3 2\n4 9 18",
"output": "1"
},
{
"input": "5 3\n6 6 6 13 27",
"output": "2"
},
{
"input": "5 1\n23 8 83 26 18",
"output": "4"
},
{
"input": "3 1\n4 5 6",
"output": "1"
},
{
"input": "3 1\n1 3 6",
"output": "1"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "3 2\n4 5 6",
"output": "0"
},
{
"input": "5 1\n100 200 400 1000 2000",
"output": "7"
},
{
"input": "2 1\n1 4",
"output": "1"
},
{
"input": "4 1\n2 4 8 32",
"output": "1"
},
{
"input": "2 10\n21 42",
"output": "1"
},
{
"input": "3 3\n1 7 13",
"output": "1"
},
{
"input": "3 1\n1 4 6",
"output": "1"
},
{
"input": "2 2\n2 8",
"output": "1"
},
{
"input": "1 1\n4",
"output": "1"
},
{
"input": "2 2\n8 16",
"output": "1"
},
{
"input": "3 1\n4 8 16",
"output": "1"
},
{
"input": "3 1\n3 6 9",
"output": "1"
},
{
"input": "2 1\n4 8",
"output": "1"
},
{
"input": "2 2\n7 14",
"output": "1"
},
{
"input": "1 4\n9",
"output": "1"
},
{
"input": "5 3\n1024 4096 16384 65536 536870913",
"output": "24"
},
{
"input": "2 5\n10 11",
"output": "0"
},
{
"input": "2 2\n3 6",
"output": "0"
},
{
"input": "2 2\n8 11",
"output": "1"
},
{
"input": "3 19905705\n263637263 417905394 108361057",
"output": "3"
},
{
"input": "4 25\n100 11 1 13",
"output": "1"
},
{
"input": "10 295206008\n67980321 440051990 883040288 135744260 96431758 242465794 576630162 972797687 356406646 547451696",
"output": "0"
},
{
"input": "4 2\n45 44 35 38",
"output": "4"
},
{
"input": "1 2\n9",
"output": "2"
},
{
"input": "3 6\n13 26 52",
"output": "1"
},
{
"input": "9 30111088\n824713578 11195876 458715185 731769293 680826358 189542586 550198537 860586039 101083021",
"output": "2"
},
{
"input": "3 72014068\n430005292 807436976 828082746",
"output": "2"
},
{
"input": "3 165219745\n737649884 652879952 506420386",
"output": "1"
},
{
"input": "2 60669400\n95037700 337255240",
"output": "1"
},
{
"input": "4 28\n34 1 86 90",
"output": "1"
},
{
"input": "2 1\n5 10",
"output": "2"
},
{
"input": "2 1\n4 1000000000",
"output": "28"
},
{
"input": "2 1\n2 3",
"output": "0"
},
{
"input": "2 1\n3 5",
"output": "1"
},
{
"input": "3 3\n1 5 20",
"output": "1"
},
{
"input": "9 1\n1 2 4 9 15 32 64 128 1024",
"output": "4"
}
] | 1,500,726,965
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 4,608,000
|
def list_input():
return list(map(int,input().split()))
def map_input():
return map(int,input().split())
def map_string():
return input().split()
def dist(a,b,k):
a = max(a,k)
a *= 2
cnt = 0
while a < b:
a *= 2
cnt += 1
return cnt
n,k = map_input()
a = list_input()
ans = 0
a.sort()
for i in range(1,n):
ans += dist(a[i-1],a[i],k)
print(ans)
|
Title: Multi-judge Solving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input Specification:
The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Demo Input:
['3 3\n2 1 9\n', '4 20\n10 3 6 3\n']
Demo Output:
['1\n', '0\n']
Note:
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
|
```python
def list_input():
return list(map(int,input().split()))
def map_input():
return map(int,input().split())
def map_string():
return input().split()
def dist(a,b,k):
a = max(a,k)
a *= 2
cnt = 0
while a < b:
a *= 2
cnt += 1
return cnt
n,k = map_input()
a = list_input()
ans = 0
a.sort()
for i in range(1,n):
ans += dist(a[i-1],a[i],k)
print(ans)
```
| 0
|
|
965
|
A
|
Paper Airplanes
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
|
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
|
Print a single integer — the minimum number of packs they should buy.
|
[
"5 3 2 3\n",
"5 3 100 1\n"
] |
[
"4\n",
"5\n"
] |
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets.
| 500
|
[
{
"input": "5 3 2 3",
"output": "4"
},
{
"input": "5 3 100 1",
"output": "5"
},
{
"input": "10000 10000 1 1",
"output": "100000000"
},
{
"input": "1 1 10000 10000",
"output": "1"
},
{
"input": "300 300 21 23",
"output": "196"
},
{
"input": "300 2 37 51",
"output": "6"
},
{
"input": "2 400 23 57",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10000 10000 3 2",
"output": "16670000"
},
{
"input": "1 2 1 2",
"output": "1"
},
{
"input": "1 1 2 2",
"output": "1"
},
{
"input": "1 1 10 10",
"output": "1"
},
{
"input": "5324 5439 32 13",
"output": "69622"
},
{
"input": "9000 1 2432 1",
"output": "9000"
},
{
"input": "230 1234 9124 23",
"output": "10"
},
{
"input": "11 1 1 1",
"output": "11"
},
{
"input": "6246 8489 1227 9",
"output": "4858"
},
{
"input": "9 20 5 7",
"output": "6"
}
] | 1,584,166,422
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 18
| 124
| 0
|
import math as m
k,n,s,p=map(int,input().split())
tot=n*k
tota=m.ceil(tot/s)
tote=(m.ceil(tota/k))*k
print(m.ceil(tote/p))
|
Title: Paper Airplanes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
Input Specification:
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
Output Specification:
Print a single integer — the minimum number of packs they should buy.
Demo Input:
['5 3 2 3\n', '5 3 100 1\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets.
|
```python
import math as m
k,n,s,p=map(int,input().split())
tot=n*k
tota=m.ceil(tot/s)
tote=(m.ceil(tota/k))*k
print(m.ceil(tote/p))
```
| 3
|
|
110
|
A
|
Nearly Lucky Number
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Nearly Lucky Number
|
2
|
256
|
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number.
|
The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
|
Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes).
|
[
"40047\n",
"7747774\n",
"1000000000000000000\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO".
| 500
|
[
{
"input": "40047",
"output": "NO"
},
{
"input": "7747774",
"output": "YES"
},
{
"input": "1000000000000000000",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "4",
"output": "NO"
},
{
"input": "474404774",
"output": "NO"
},
{
"input": "4744000695826",
"output": "YES"
},
{
"input": "10000000004744744",
"output": "YES"
},
{
"input": "446486416781684178",
"output": "YES"
},
{
"input": "999999999",
"output": "NO"
},
{
"input": "7777",
"output": "YES"
},
{
"input": "87414417444",
"output": "NO"
},
{
"input": "111222333444555667",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "4700",
"output": "NO"
},
{
"input": "3794555488744477",
"output": "NO"
},
{
"input": "444444444444444444",
"output": "NO"
},
{
"input": "474447447774444774",
"output": "NO"
},
{
"input": "777777777777777",
"output": "NO"
},
{
"input": "34777745021000000",
"output": "NO"
},
{
"input": "963",
"output": "NO"
},
{
"input": "855474448854788540",
"output": "NO"
},
{
"input": "999999999999994744",
"output": "YES"
},
{
"input": "400000000474",
"output": "YES"
},
{
"input": "123456789123456789",
"output": "YES"
},
{
"input": "740577777584945874",
"output": "NO"
},
{
"input": "7777777",
"output": "YES"
},
{
"input": "4444000111222333",
"output": "YES"
},
{
"input": "9847745885202111",
"output": "YES"
},
{
"input": "123456000000",
"output": "NO"
},
{
"input": "4744447444444",
"output": "NO"
},
{
"input": "7477",
"output": "YES"
},
{
"input": "4747477",
"output": "YES"
},
{
"input": "777777777444444444",
"output": "NO"
}
] | 1,694,838,219
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 154
| 2,764,800
|
n=int(input(""))
x= (n)
luckyno=0
for digit in x:
if digit=='4' or digit=='7':
luckyno += 1
if luckyno==4 or luckyno==7:
print("YES")
else:
print("NO")
|
Title: Nearly Lucky Number
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number.
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output Specification:
Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes).
Demo Input:
['40047\n', '7747774\n', '1000000000000000000\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO".
|
```python
n=int(input(""))
x= (n)
luckyno=0
for digit in x:
if digit=='4' or digit=='7':
luckyno += 1
if luckyno==4 or luckyno==7:
print("YES")
else:
print("NO")
```
| -1
|
265
|
A
|
Colorful Stones (Simplified Edition)
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
|
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
|
Print the final 1-based position of Liss in a single line.
|
[
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] |
[
"2\n",
"3\n",
"15\n"
] |
none
| 500
|
[
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,690,935,124
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 19
| 154
| 0
|
a=input()
b=input()
e=0
for i in range(len(b)):
if b[i]==a[e]:
e+=1
print(e+1)
|
Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none
|
```python
a=input()
b=input()
e=0
for i in range(len(b)):
if b[i]==a[e]:
e+=1
print(e+1)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
|
In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
|
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.
|
[
"5\n1 3 2 1 2\n",
"6\n2 1 2 2 4 1\n"
] |
[
"3\n",
"2\n"
] |
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
| 0
|
[
{
"input": "5\n1 3 2 1 2",
"output": "3"
},
{
"input": "6\n2 1 2 2 4 1",
"output": "2"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n200000",
"output": "200000"
},
{
"input": "2\n2018 2017",
"output": "2018"
},
{
"input": "5\n100 1000 1000 1000 1000",
"output": "100"
},
{
"input": "8\n200000 1 200000 1 200000 1 200000 2",
"output": "1"
},
{
"input": "5\n20000 1 2 3 4",
"output": "20000"
},
{
"input": "2\n200000 1",
"output": "200000"
},
{
"input": "3\n2 100 2",
"output": "100"
},
{
"input": "2\n10 1",
"output": "10"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "3\n5 5 5",
"output": "5"
},
{
"input": "2\n5 1",
"output": "5"
}
] | 1,698,508,119
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
n = int(input())
cofas = list(map(int, input().split()))
print(set(cofas[::-1]).pop())
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
Input Specification:
In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
Output Specification:
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.
Demo Input:
['5\n1 3 2 1 2\n', '6\n2 1 2 2 4 1\n']
Demo Output:
['3\n', '2\n']
Note:
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
|
```python
n = int(input())
cofas = list(map(int, input().split()))
print(set(cofas[::-1]).pop())
```
| 0
|
|
785
|
A
|
Anton and Polyhedrons
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
|
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
|
[
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] |
[
"42\n",
"28\n"
] |
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
| 500
|
[
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron",
"output": "256"
},
{
"input": "1\nTetrahedron",
"output": "4"
},
{
"input": "1\nCube",
"output": "6"
},
{
"input": "1\nOctahedron",
"output": "8"
},
{
"input": "1\nDodecahedron",
"output": "12"
},
{
"input": "1\nIcosahedron",
"output": "20"
},
{
"input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron",
"output": "340"
}
] | 1,677,942,862
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 1,076
| 9,113,600
|
n = int(input())
a=0
for i in range(n):
inp = input()
if(inp=="Tetrahedron"):a+=4
if(inp=="Cube"):a+=6
if(inp=="Octahedron"):a+=8
if(inp=="Dodecahedron"):a+=12
if(inp=="Icosahedron"):a+=20
print(a)
|
Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
|
```python
n = int(input())
a=0
for i in range(n):
inp = input()
if(inp=="Tetrahedron"):a+=4
if(inp=="Cube"):a+=6
if(inp=="Octahedron"):a+=8
if(inp=="Dodecahedron"):a+=12
if(inp=="Icosahedron"):a+=20
print(a)
```
| 3
|
|
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\n916 897\n-651 -808\n-748 301\n-734 414\n-651 -973\n-734 897\n916 -550\n-758 414\n916 180\n-758 -808\n-758 -973\n125 -550\n125 -973\n125 301\n916 414\n-748 -808\n-651 301\n-734 301\n-307 897\n-651 -550\n-651 414\n125 -808\n-748 -550\n916 -808",
"output": "7"
},
{
"input": "1\n487 550",
"output": "0"
},
{
"input": "10\n990 -396\n990 736\n990 646\n990 -102\n990 -570\n990 155\n990 528\n990 489\n990 268\n990 676",
"output": "0"
},
{
"input": "30\n507 836\n525 836\n-779 196\n507 -814\n525 -814\n525 42\n525 196\n525 -136\n-779 311\n507 -360\n525 300\n507 578\n507 311\n-779 836\n507 300\n525 -360\n525 311\n-779 -360\n-779 578\n-779 300\n507 42\n525 578\n-779 379\n507 196\n525 379\n507 379\n-779 -814\n-779 42\n-779 -136\n507 -136",
"output": "8"
},
{
"input": "25\n890 -756\n890 -188\n-37 -756\n-37 853\n523 998\n-261 853\n-351 853\n-351 -188\n523 -756\n-261 -188\n-37 998\n523 -212\n-351 998\n-37 -188\n-351 -756\n-37 -212\n890 998\n890 -212\n523 853\n-351 -212\n-261 -212\n-261 998\n-261 -756\n890 853\n523 -188",
"output": "9"
},
{
"input": "21\n-813 -11\n486 254\n685 254\n-708 254\n-55 -11\n-671 -191\n486 -11\n-671 -11\n685 -11\n685 -191\n486 -191\n-55 254\n-708 -11\n-813 254\n-708 -191\n41 -11\n-671 254\n-813 -191\n41 254\n-55 -191\n41 -191",
"output": "5"
},
{
"input": "4\n1 0\n2 0\n1 1\n1 -1",
"output": "0"
}
] | 1,670,522,902
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 92
| 0
|
n = int(input())
sx = []
sy = []
count = 0
for i in range(n):
s = list(map(int, input().split()))
x, y = s[0], s[1]
sx.append(x)
sy.append(y)
mAxx = max(sx)
mInx = min(sx)
mAxy = max(sy)
mIny = min(sy)
for i in range(len(sx)):
if sx.count(sx[i]) >= 3 and sx[i] != mAxx and sx[i] != mInx:
if sy.count(sy[i]) >= 3 and sy[i] != mAxy and sy[i] != mIny:
count += 1
print(count)
|
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())
sx = []
sy = []
count = 0
for i in range(n):
s = list(map(int, input().split()))
x, y = s[0], s[1]
sx.append(x)
sy.append(y)
mAxx = max(sx)
mInx = min(sx)
mAxy = max(sy)
mIny = min(sy)
for i in range(len(sx)):
if sx.count(sx[i]) >= 3 and sx[i] != mAxx and sx[i] != mInx:
if sy.count(sy[i]) >= 3 and sy[i] != mAxy and sy[i] != mIny:
count += 1
print(count)
```
| 0
|
|
16
|
B
|
Burglar and Matches
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] |
B. Burglar and Matches
|
0
|
64
|
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
|
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
|
Output the only number — answer to the problem.
|
[
"7 3\n5 10\n2 5\n3 6\n",
"3 3\n1 3\n2 2\n3 1\n"
] |
[
"62\n",
"7\n"
] |
none
| 0
|
[
{
"input": "7 3\n5 10\n2 5\n3 6",
"output": "62"
},
{
"input": "3 3\n1 3\n2 2\n3 1",
"output": "7"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "1 2\n1 9\n1 6",
"output": "9"
},
{
"input": "1 10\n1 1\n1 9\n1 3\n1 9\n1 7\n1 10\n1 4\n1 7\n1 3\n1 1",
"output": "10"
},
{
"input": "2 1\n2 1",
"output": "2"
},
{
"input": "2 2\n2 4\n1 4",
"output": "8"
},
{
"input": "2 3\n1 7\n1 2\n1 5",
"output": "12"
},
{
"input": "4 1\n2 2",
"output": "4"
},
{
"input": "4 2\n1 10\n4 4",
"output": "22"
},
{
"input": "4 3\n1 4\n6 4\n1 7",
"output": "19"
},
{
"input": "5 1\n10 5",
"output": "25"
},
{
"input": "5 2\n3 9\n2 2",
"output": "31"
},
{
"input": "5 5\n2 9\n3 1\n2 1\n1 8\n2 8",
"output": "42"
},
{
"input": "5 10\n1 3\n1 2\n1 9\n1 10\n1 1\n1 5\n1 10\n1 2\n1 3\n1 7",
"output": "41"
},
{
"input": "10 1\n9 4",
"output": "36"
},
{
"input": "10 2\n14 3\n1 3",
"output": "30"
},
{
"input": "10 7\n4 8\n1 10\n1 10\n1 2\n3 3\n1 3\n1 10",
"output": "71"
},
{
"input": "10 10\n1 8\n2 10\n1 9\n1 1\n1 9\n1 6\n1 4\n2 5\n1 2\n1 4",
"output": "70"
},
{
"input": "10 4\n1 5\n5 2\n1 9\n3 3",
"output": "33"
},
{
"input": "100 5\n78 6\n29 10\n3 6\n7 3\n2 4",
"output": "716"
},
{
"input": "1000 7\n102 10\n23 6\n79 4\n48 1\n34 10\n839 8\n38 4",
"output": "8218"
},
{
"input": "10000 10\n336 2\n2782 5\n430 10\n1893 7\n3989 10\n2593 8\n165 6\n1029 2\n2097 4\n178 10",
"output": "84715"
},
{
"input": "100000 3\n2975 2\n35046 4\n61979 9",
"output": "703945"
},
{
"input": "1000000 4\n314183 9\n304213 4\n16864 5\n641358 9",
"output": "8794569"
},
{
"input": "10000000 10\n360313 10\n416076 1\n435445 9\n940322 7\n1647581 7\n4356968 10\n3589256 2\n2967933 5\n2747504 7\n1151633 3",
"output": "85022733"
},
{
"input": "100000000 7\n32844337 7\n11210848 7\n47655987 1\n33900472 4\n9174763 2\n32228738 10\n29947408 5",
"output": "749254060"
},
{
"input": "200000000 10\n27953106 7\n43325979 4\n4709522 1\n10975786 4\n67786538 8\n48901838 7\n15606185 6\n2747583 1\n100000000 1\n633331 3",
"output": "1332923354"
},
{
"input": "200000000 9\n17463897 9\n79520463 1\n162407 4\n41017993 8\n71054118 4\n9447587 2\n5298038 9\n3674560 7\n20539314 5",
"output": "996523209"
},
{
"input": "200000000 8\n6312706 6\n2920548 2\n16843192 3\n1501141 2\n13394704 6\n10047725 10\n4547663 6\n54268518 6",
"output": "630991750"
},
{
"input": "200000000 7\n25621043 2\n21865270 1\n28833034 1\n22185073 5\n100000000 2\n13891017 9\n61298710 8",
"output": "931584598"
},
{
"input": "200000000 6\n7465600 6\n8453505 10\n4572014 8\n8899499 3\n86805622 10\n64439238 6",
"output": "1447294907"
},
{
"input": "200000000 5\n44608415 6\n100000000 9\n51483223 9\n44136047 1\n52718517 1",
"output": "1634907859"
},
{
"input": "200000000 4\n37758556 10\n100000000 6\n48268521 3\n20148178 10",
"output": "1305347138"
},
{
"input": "200000000 3\n65170000 7\n20790088 1\n74616133 4",
"output": "775444620"
},
{
"input": "200000000 2\n11823018 6\n100000000 9",
"output": "970938108"
},
{
"input": "200000000 1\n100000000 6",
"output": "600000000"
},
{
"input": "200000000 10\n12097724 9\n41745972 5\n26982098 9\n14916995 7\n21549986 7\n3786630 9\n8050858 7\n27994924 4\n18345001 5\n8435339 5",
"output": "1152034197"
},
{
"input": "200000000 10\n55649 8\n10980981 9\n3192542 8\n94994808 4\n3626106 1\n100000000 6\n5260110 9\n4121453 2\n15125061 4\n669569 6",
"output": "1095537357"
},
{
"input": "10 20\n1 7\n1 7\n1 8\n1 3\n1 10\n1 7\n1 7\n1 9\n1 3\n1 1\n1 2\n1 1\n1 3\n1 10\n1 9\n1 8\n1 8\n1 6\n1 7\n1 5",
"output": "83"
},
{
"input": "10000000 20\n4594 7\n520836 8\n294766 6\n298672 4\n142253 6\n450626 1\n1920034 9\n58282 4\n1043204 1\n683045 1\n1491746 5\n58420 4\n451217 2\n129423 4\n246113 5\n190612 8\n912923 6\n473153 6\n783733 6\n282411 10",
"output": "54980855"
},
{
"input": "200000000 20\n15450824 5\n839717 10\n260084 8\n1140850 8\n28744 6\n675318 3\n25161 2\n5487 3\n6537698 9\n100000000 5\n7646970 9\n16489 6\n24627 3\n1009409 5\n22455 1\n25488456 4\n484528 9\n32663641 3\n750968 4\n5152 6",
"output": "939368573"
},
{
"input": "200000000 20\n16896 2\n113 3\n277 2\n299 7\n69383562 2\n3929 8\n499366 4\n771846 5\n9 4\n1278173 7\n90 2\n54 7\n72199858 10\n17214 5\n3 10\n1981618 3\n3728 2\n141 8\n2013578 9\n51829246 5",
"output": "1158946383"
},
{
"input": "200000000 20\n983125 2\n7453215 9\n9193588 2\n11558049 7\n28666199 1\n34362244 1\n5241493 5\n15451270 4\n19945845 8\n6208681 3\n38300385 7\n6441209 8\n21046742 7\n577198 10\n3826434 8\n9764276 8\n6264675 7\n8567063 3\n3610303 4\n2908232 3",
"output": "1131379312"
},
{
"input": "10 15\n1 6\n2 6\n3 4\n1 3\n1 2\n1 5\n1 6\n1 2\n2 9\n1 10\n1 3\n1 7\n1 8\n1 2\n2 9",
"output": "79"
},
{
"input": "10000000 15\n111 5\n914124 3\n3 9\n177790 1\n2352 3\n32138 9\n104477 1\n1223 4\n18 6\n6655580 4\n57643 10\n94309 2\n37 1\n227002 10\n1733193 7",
"output": "45116295"
},
{
"input": "200000000 15\n7069868 1\n5567826 8\n2310059 10\n13539782 7\n38420939 4\n29911411 8\n52256316 1\n12265839 9\n2074265 1\n24896428 9\n72470695 5\n3236301 1\n3890243 2\n65168965 8\n65724 6",
"output": "1489289257"
},
{
"input": "200000000 15\n12044094 7\n2475138 10\n944451 7\n4854766 2\n3809145 10\n7727571 2\n43908937 6\n2745883 1\n427511 2\n100000000 5\n190914 6\n554889 3\n288798 4\n1848572 5\n893874 3",
"output": "961871671"
},
{
"input": "200000000 15\n6334191 7\n1927941 4\n5175933 10\n468389 1\n433043 10\n6863198 5\n7480646 4\n14774279 10\n2921129 8\n18325627 7\n6973152 9\n8277324 9\n21522856 2\n2058070 1\n2444742 4",
"output": "664376069"
}
] | 1,661,262,310
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 24
| 500
| 0
|
#Read input
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
#read inputs
n,m = inlt()
n_matchboxes = []
size_matchboxes = []
for i in range(m):
a,b = inlt()
n_matchboxes.append(a)
size_matchboxes.append(b)
# Find criminals
def find_max(n,m,n_matchboxes,size_matchboxes):
n_max = 0
sort_indexes = [i[0] for i in sorted(enumerate(size_matchboxes), key=lambda x:x[1],reverse=True)]
n_s = []
s_s = []
for i in sort_indexes:
n_s.append(n_matchboxes[i])
s_s.append(size_matchboxes[i])
n_max = 0
n_boxes = 0
k = 0
while n_boxes<n and k<m:
n_max += s_s[k]
n_boxes +=1
n_s[k] = n_s[k] -1
if n_s[k] == 0:
k+=1
return n_max
n_max = find_max(n,m,n_matchboxes,size_matchboxes)
print(n_max)
|
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
#Read input
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
#read inputs
n,m = inlt()
n_matchboxes = []
size_matchboxes = []
for i in range(m):
a,b = inlt()
n_matchboxes.append(a)
size_matchboxes.append(b)
# Find criminals
def find_max(n,m,n_matchboxes,size_matchboxes):
n_max = 0
sort_indexes = [i[0] for i in sorted(enumerate(size_matchboxes), key=lambda x:x[1],reverse=True)]
n_s = []
s_s = []
for i in sort_indexes:
n_s.append(n_matchboxes[i])
s_s.append(size_matchboxes[i])
n_max = 0
n_boxes = 0
k = 0
while n_boxes<n and k<m:
n_max += s_s[k]
n_boxes +=1
n_s[k] = n_s[k] -1
if n_s[k] == 0:
k+=1
return n_max
n_max = find_max(n,m,n_matchboxes,size_matchboxes)
print(n_max)
```
| 0
|
514
|
C
|
Watto and Mechanism
|
PROGRAMMING
| 2,000
|
[
"binary search",
"data structures",
"hashing",
"string suffix structures",
"strings"
] | null | null |
Watto, the owner of a spare parts store, has recently got an order for the mechanism that can process strings in a certain way. Initially the memory of the mechanism is filled with *n* strings. Then the mechanism should be able to process queries of the following type: "Given string *s*, determine if the memory of the mechanism contains string *t* that consists of the same number of characters as *s* and differs from *s* in exactly one position".
Watto has already compiled the mechanism, all that's left is to write a program for it and check it on the data consisting of *n* initial lines and *m* queries. He decided to entrust this job to you.
|
The first line contains two non-negative numbers *n* and *m* (0<=≤<=*n*<=≤<=3·105, 0<=≤<=*m*<=≤<=3·105) — the number of the initial strings and the number of queries, respectively.
Next follow *n* non-empty strings that are uploaded to the memory of the mechanism.
Next follow *m* non-empty strings that are the queries to the mechanism.
The total length of lines in the input doesn't exceed 6·105. Each line consists only of letters 'a', 'b', 'c'.
|
For each query print on a single line "YES" (without the quotes), if the memory of the mechanism contains the required string, otherwise print "NO" (without the quotes).
|
[
"2 3\naaaaa\nacacaca\naabaa\nccacacc\ncaaac\n"
] |
[
"YES\nNO\nNO\n"
] |
none
| 2,000
|
[
{
"input": "2 3\naaaaa\nacacaca\naabaa\nccacacc\ncaaac",
"output": "YES\nNO\nNO"
},
{
"input": "1 5\nacbacbacb\ncbacbacb\nacbacbac\naacbacbacb\nacbacbacbb\nacbaabacb",
"output": "NO\nNO\nNO\nNO\nYES"
},
{
"input": "5 4\nab\ncacab\ncbabc\nacc\ncacab\nabc\naa\nacbca\ncb",
"output": "YES\nYES\nNO\nYES"
},
{
"input": "9 9\ncaccbcacabccba\naacbcbcaabacbcbcba\nbabccaaacccacbb\ncaaabcaacbababbabbb\nabbaccacabacaaaa\nbccbccababcaacb\ncaacbcaacbababbabbb\nbcacababbbcaaca\nccbbcbababbccaab\nbbcbccababcaacb\naacccbabbacbabacaca\nbbcbcccbabcaacb\nacbacacbcacc\ncaaabcaaabacabbabbb\nabbbabaaaba\naacccbcaabacbcbcba\nabbaccacabbcaaaa\naaccbbcabbacbcbcba",
"output": "YES\nNO\nNO\nNO\nNO\nNO\nYES\nYES\nNO"
},
{
"input": "1 1\nbbbbbbbaaaabbbbbaabbbba\naaabbbabbbbbbbaabbabbbb",
"output": "NO"
}
] | 1,690,107,125
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 5
| 62
| 2,457,600
|
st = set()
n, m = list(map(int, input().split()))
for i in range(n):
word = input()
for letter in range(len(word)):
newWord = list(word)
for pp in ['a', 'b', 'c']:
newWord[letter] = pp
st.add(''.join(newWord))
for i in range(m):
if input() in st:
print("YES")
else:
print("NO")
|
Title: Watto and Mechanism
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Watto, the owner of a spare parts store, has recently got an order for the mechanism that can process strings in a certain way. Initially the memory of the mechanism is filled with *n* strings. Then the mechanism should be able to process queries of the following type: "Given string *s*, determine if the memory of the mechanism contains string *t* that consists of the same number of characters as *s* and differs from *s* in exactly one position".
Watto has already compiled the mechanism, all that's left is to write a program for it and check it on the data consisting of *n* initial lines and *m* queries. He decided to entrust this job to you.
Input Specification:
The first line contains two non-negative numbers *n* and *m* (0<=≤<=*n*<=≤<=3·105, 0<=≤<=*m*<=≤<=3·105) — the number of the initial strings and the number of queries, respectively.
Next follow *n* non-empty strings that are uploaded to the memory of the mechanism.
Next follow *m* non-empty strings that are the queries to the mechanism.
The total length of lines in the input doesn't exceed 6·105. Each line consists only of letters 'a', 'b', 'c'.
Output Specification:
For each query print on a single line "YES" (without the quotes), if the memory of the mechanism contains the required string, otherwise print "NO" (without the quotes).
Demo Input:
['2 3\naaaaa\nacacaca\naabaa\nccacacc\ncaaac\n']
Demo Output:
['YES\nNO\nNO\n']
Note:
none
|
```python
st = set()
n, m = list(map(int, input().split()))
for i in range(n):
word = input()
for letter in range(len(word)):
newWord = list(word)
for pp in ['a', 'b', 'c']:
newWord[letter] = pp
st.add(''.join(newWord))
for i in range(m):
if input() in st:
print("YES")
else:
print("NO")
```
| 0
|
|
12
|
B
|
Correct Solution?
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] |
B. Correct Solution?
|
2
|
256
|
One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said:
—Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes.
—No problem! — said Bob and immediately gave her an answer.
Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict.
|
The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes.
|
Print OK if Bob's answer is correct and WRONG_ANSWER otherwise.
|
[
"3310\n1033\n",
"4\n5\n"
] |
[
"OK\n",
"WRONG_ANSWER\n"
] |
none
| 0
|
[
{
"input": "3310\n1033",
"output": "OK"
},
{
"input": "4\n5",
"output": "WRONG_ANSWER"
},
{
"input": "40\n04",
"output": "WRONG_ANSWER"
},
{
"input": "12\n12",
"output": "OK"
},
{
"input": "432\n234",
"output": "OK"
},
{
"input": "17109\n01179",
"output": "WRONG_ANSWER"
},
{
"input": "888\n888",
"output": "OK"
},
{
"input": "912\n9123",
"output": "WRONG_ANSWER"
},
{
"input": "0\n00",
"output": "WRONG_ANSWER"
},
{
"input": "11110\n1111",
"output": "WRONG_ANSWER"
},
{
"input": "7391\n1397",
"output": "WRONG_ANSWER"
},
{
"input": "201\n102",
"output": "OK"
},
{
"input": "111111111\n111111111",
"output": "OK"
},
{
"input": "32352320\n22203335",
"output": "WRONG_ANSWER"
},
{
"input": "1000000000\n1",
"output": "WRONG_ANSWER"
},
{
"input": "994321\n123499",
"output": "OK"
},
{
"input": "10101\n10101",
"output": "WRONG_ANSWER"
},
{
"input": "666\n0666",
"output": "WRONG_ANSWER"
},
{
"input": "2\n02",
"output": "WRONG_ANSWER"
},
{
"input": "313\n133",
"output": "OK"
},
{
"input": "987235645\n234556789",
"output": "OK"
},
{
"input": "90812\n010289",
"output": "WRONG_ANSWER"
},
{
"input": "123\n321",
"output": "WRONG_ANSWER"
},
{
"input": "707\n770",
"output": "WRONG_ANSWER"
},
{
"input": "77774444\n47474747",
"output": "WRONG_ANSWER"
},
{
"input": "1270\n1027",
"output": "OK"
},
{
"input": "320\n23",
"output": "WRONG_ANSWER"
},
{
"input": "123456789\n123456789",
"output": "OK"
},
{
"input": "918273645\n546372819",
"output": "WRONG_ANSWER"
},
{
"input": "300000003\n30000003",
"output": "WRONG_ANSWER"
},
{
"input": "0\n0",
"output": "OK"
},
{
"input": "0\n7",
"output": "WRONG_ANSWER"
}
] | 1,630,765,538
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 154
| 6,963,200
|
def transform(n):
m = ''+n
m = sorted(m)
for d in range(len(m)):
if ord(m[d]) > ord('0'):
wait = m[d]
m[d] = m[0]
m[0] = wait
break
return m
def verdict(m, t):
return m == t
def main():
# input
n = input()
m = input()
# proccessing
t = transform(n)
t = ''.join(t)
# output
if verdict(m, t):
print('OK')
else:
print('WRONG_ANSWER')
if __name__=='__main__':
main()
|
Title: Correct Solution?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said:
—Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes.
—No problem! — said Bob and immediately gave her an answer.
Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict.
Input Specification:
The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes.
Output Specification:
Print OK if Bob's answer is correct and WRONG_ANSWER otherwise.
Demo Input:
['3310\n1033\n', '4\n5\n']
Demo Output:
['OK\n', 'WRONG_ANSWER\n']
Note:
none
|
```python
def transform(n):
m = ''+n
m = sorted(m)
for d in range(len(m)):
if ord(m[d]) > ord('0'):
wait = m[d]
m[d] = m[0]
m[0] = wait
break
return m
def verdict(m, t):
return m == t
def main():
# input
n = input()
m = input()
# proccessing
t = transform(n)
t = ''.join(t)
# output
if verdict(m, t):
print('OK')
else:
print('WRONG_ANSWER')
if __name__=='__main__':
main()
```
| 3.94853
|
817
|
B
|
Makes And The Product
|
PROGRAMMING
| 1,500
|
[
"combinatorics",
"implementation",
"math",
"sortings"
] | null | null |
After returning from the army Makes received a gift — an array *a* consisting of *n* positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (*i*,<= *j*,<= *k*) (*i*<=<<=*j*<=<<=*k*), such that *a**i*·*a**j*·*a**k* is minimum possible, are there in the array? Help him with it!
|
The first line of input contains a positive integer number *n* (3<=≤<=*n*<=≤<=105) — the number of elements in array *a*. The second line contains *n* positive integer numbers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of a given array.
|
Print one number — the quantity of triples (*i*,<= *j*,<= *k*) such that *i*,<= *j* and *k* are pairwise distinct and *a**i*·*a**j*·*a**k* is minimum possible.
|
[
"4\n1 1 1 1\n",
"5\n1 3 2 3 4\n",
"6\n1 3 3 1 3 2\n"
] |
[
"4\n",
"2\n",
"1\n"
] |
In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.
In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.
In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices.
| 0
|
[
{
"input": "4\n1 1 1 1",
"output": "4"
},
{
"input": "5\n1 3 2 3 4",
"output": "2"
},
{
"input": "6\n1 3 3 1 3 2",
"output": "1"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "4\n1 1 2 2",
"output": "2"
},
{
"input": "3\n1 3 1",
"output": "1"
},
{
"input": "11\n1 2 2 2 2 2 2 2 2 2 2",
"output": "45"
},
{
"input": "5\n1 2 2 2 2",
"output": "6"
},
{
"input": "6\n1 2 2 3 3 4",
"output": "1"
},
{
"input": "8\n1 1 2 2 2 3 3 3",
"output": "3"
},
{
"input": "6\n1 2 2 2 2 3",
"output": "6"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "6\n1 2 2 2 3 3",
"output": "3"
},
{
"input": "6\n1 2 2 2 2 2",
"output": "10"
},
{
"input": "4\n1 2 2 2",
"output": "3"
},
{
"input": "5\n1 2 3 2 3",
"output": "1"
},
{
"input": "6\n2 2 3 3 3 3",
"output": "4"
},
{
"input": "6\n1 2 2 2 5 6",
"output": "3"
},
{
"input": "10\n1 2 2 2 2 2 2 2 2 2",
"output": "36"
},
{
"input": "3\n2 1 2",
"output": "1"
},
{
"input": "5\n1 2 3 3 3",
"output": "3"
},
{
"input": "6\n1 2 2 2 4 5",
"output": "3"
},
{
"input": "4\n1 2 2 3",
"output": "1"
},
{
"input": "10\n2 2 2 2 2 1 2 2 2 2",
"output": "36"
},
{
"input": "7\n2 2 2 3 3 3 1",
"output": "3"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "5\n1 1 2 2 2",
"output": "3"
},
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "9\n2 2 3 3 3 3 3 3 3",
"output": "7"
},
{
"input": "5\n1 1 2 2 3",
"output": "2"
},
{
"input": "4\n1 1 3 3",
"output": "2"
},
{
"input": "4\n33554432 33554432 67108864 33554432",
"output": "1"
},
{
"input": "6\n2 2 2 1 2 2",
"output": "10"
},
{
"input": "10\n1 2 1 2 3 2 3 2 2 2",
"output": "6"
},
{
"input": "10\n9 6 4 7 1 8 9 5 9 4",
"output": "1"
},
{
"input": "4\n5 7 2 7",
"output": "2"
},
{
"input": "3\n7 6 7",
"output": "1"
},
{
"input": "6\n3 2 8 2 5 3",
"output": "2"
},
{
"input": "3\n5 9 5",
"output": "1"
},
{
"input": "5\n6 3 7 6 3",
"output": "2"
},
{
"input": "9\n10 10 4 10 7 9 6 7 3",
"output": "1"
},
{
"input": "5\n9 10 10 3 8",
"output": "1"
},
{
"input": "5\n2 9 5 10 5",
"output": "1"
},
{
"input": "9\n7 1 9 6 6 8 3 1 3",
"output": "2"
},
{
"input": "5\n3 4 4 4 5",
"output": "3"
},
{
"input": "3\n3 1 3",
"output": "1"
},
{
"input": "8\n3 2 2 5 2 2 1 2",
"output": "10"
}
] | 1,517,128,216
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 18
| 1,606
| 12,902,400
|
n = int(input())
a = input().split()
for i in range(n):
a[i] = int(a[i])
from math import factorial
'''def merge(a,b):
c=[]
a1=0
b1=0
while a1!=len(a) and b1!=len(b):
if a[a1]<b[b1]:
c.append(a[a1])
a1+=1
else:
c.append(b[b1])
b1+=1
c.extend(a[a1:])
c.extend(b[b1:])
return c
def merge_sort(a):
b=len(a)
if b==1:
return a
m=merge_sort(a[:b//2])
n=merge_sort(a[b//2:])
return merge(m,n)'''
def delete(l, n):
for i in range(len(l)):
if l[i] != n:
return l[i:]
def C(n, k):
return factorial(n) // (factorial(k) * factorial(n - k))
a.sort()
mi = a[0]
c = a.count(mi)
if c >= 3:
print(C(c, 3))
elif c == 2:
a = delete(a, mi)
mi = a[0]
c = a.count(mi)
print(c)
elif c == 1:
a = delete(a, mi)
mi = a[0]
c = a.count(mi)
if c == 1:
a = delete(a, mi)
mi = a[0]
c = a.count(mi)
print(c)
elif c == 2:
print(1)
elif c >= 3:
a = delete(a, mi)
mi = a[0]
c = a.count(mi)
print(c * (c - 1) // 2)
|
Title: Makes And The Product
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After returning from the army Makes received a gift — an array *a* consisting of *n* positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (*i*,<= *j*,<= *k*) (*i*<=<<=*j*<=<<=*k*), such that *a**i*·*a**j*·*a**k* is minimum possible, are there in the array? Help him with it!
Input Specification:
The first line of input contains a positive integer number *n* (3<=≤<=*n*<=≤<=105) — the number of elements in array *a*. The second line contains *n* positive integer numbers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of a given array.
Output Specification:
Print one number — the quantity of triples (*i*,<= *j*,<= *k*) such that *i*,<= *j* and *k* are pairwise distinct and *a**i*·*a**j*·*a**k* is minimum possible.
Demo Input:
['4\n1 1 1 1\n', '5\n1 3 2 3 4\n', '6\n1 3 3 1 3 2\n']
Demo Output:
['4\n', '2\n', '1\n']
Note:
In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.
In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.
In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices.
|
```python
n = int(input())
a = input().split()
for i in range(n):
a[i] = int(a[i])
from math import factorial
'''def merge(a,b):
c=[]
a1=0
b1=0
while a1!=len(a) and b1!=len(b):
if a[a1]<b[b1]:
c.append(a[a1])
a1+=1
else:
c.append(b[b1])
b1+=1
c.extend(a[a1:])
c.extend(b[b1:])
return c
def merge_sort(a):
b=len(a)
if b==1:
return a
m=merge_sort(a[:b//2])
n=merge_sort(a[b//2:])
return merge(m,n)'''
def delete(l, n):
for i in range(len(l)):
if l[i] != n:
return l[i:]
def C(n, k):
return factorial(n) // (factorial(k) * factorial(n - k))
a.sort()
mi = a[0]
c = a.count(mi)
if c >= 3:
print(C(c, 3))
elif c == 2:
a = delete(a, mi)
mi = a[0]
c = a.count(mi)
print(c)
elif c == 1:
a = delete(a, mi)
mi = a[0]
c = a.count(mi)
if c == 1:
a = delete(a, mi)
mi = a[0]
c = a.count(mi)
print(c)
elif c == 2:
print(1)
elif c >= 3:
a = delete(a, mi)
mi = a[0]
c = a.count(mi)
print(c * (c - 1) // 2)
```
| -1
|
|
82
|
A
|
Double Cola
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] |
A. Double Cola
|
1
|
256
|
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
|
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
|
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
|
[
"1\n",
"6\n",
"1802\n"
] |
[
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,637,341,589
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 12
| 1,000
| 1,843,200
|
n = int(input())
queue =["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
i = 1
while i < n :
queue.append(queue[0])
queue.append(queue[0])
del queue[0]
i += 1
print(queue[0])
|
Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none
|
```python
n = int(input())
queue =["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
i = 1
while i < n :
queue.append(queue[0])
queue.append(queue[0])
del queue[0]
i += 1
print(queue[0])
```
| 0
|
82
|
A
|
Double Cola
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] |
A. Double Cola
|
1
|
256
|
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
|
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
|
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
|
[
"1\n",
"6\n",
"1802\n"
] |
[
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,685,607,552
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 46
| 0
|
def find_name(n):
rounds = 1
while rounds * 5 < n:
n -= rounds * 5
rounds *= 2
position = (n - 1) // rounds
names = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
return names[position]
n = int(input())
result = find_name(n)
print(result)
|
Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none
|
```python
def find_name(n):
rounds = 1
while rounds * 5 < n:
n -= rounds * 5
rounds *= 2
position = (n - 1) // rounds
names = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
return names[position]
n = int(input())
result = find_name(n)
print(result)
```
| 3.977
|
34
|
B
|
Sale
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] |
B. Sale
|
2
|
256
|
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
|
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
|
[
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] |
[
"8\n",
"7\n"
] |
none
| 1,000
|
[
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,611,593,132
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 154
| 0
|
n,m=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
#print(l)
sum=0
i=0
while i<n and l[i]<0 and i<=(m-1):
sum+=l[i]
i+=1
print(abs(sum))
|
Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none
|
```python
n,m=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
#print(l)
sum=0
i=0
while i<n and l[i]<0 and i<=(m-1):
sum+=l[i]
i+=1
print(abs(sum))
```
| 3.9615
|
817
|
D
|
Imbalanced Array
|
PROGRAMMING
| 1,900
|
[
"data structures",
"divide and conquer",
"dsu",
"sortings"
] | null | null |
You are given an array *a* consisting of *n* elements. The imbalance value of some subsegment of this array is the difference between the maximum and minimum element from this segment. The imbalance value of the array is the sum of imbalance values of all subsegments of this array.
For example, the imbalance value of array [1,<=4,<=1] is 9, because there are 6 different subsegments of this array:
- [1] (from index 1 to index 1), imbalance value is 0; - [1,<=4] (from index 1 to index 2), imbalance value is 3; - [1,<=4,<=1] (from index 1 to index 3), imbalance value is 3; - [4] (from index 2 to index 2), imbalance value is 0; - [4,<=1] (from index 2 to index 3), imbalance value is 3; - [1] (from index 3 to index 3), imbalance value is 0;
You have to determine the imbalance value of the array *a*.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — size of the array *a*.
The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=106) — elements of the array.
|
Print one integer — the imbalance value of *a*.
|
[
"3\n1 4 1\n"
] |
[
"9\n"
] |
none
| 0
|
[
{
"input": "3\n1 4 1",
"output": "9"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "10\n1 4 4 3 5 2 4 2 4 5",
"output": "123"
},
{
"input": "10\n9 6 8 5 5 2 8 9 2 2",
"output": "245"
},
{
"input": "30\n4 5 2 2 5 2 3 4 3 3 2 1 3 4 4 5 3 3 1 5 2 3 5 4 5 4 4 3 5 2",
"output": "1480"
},
{
"input": "30\n2 2 9 1 10 8 3 3 1 4 6 10 2 2 1 4 1 1 1 1 1 2 4 7 6 7 5 10 8 9",
"output": "3147"
},
{
"input": "30\n6 19 12 6 25 24 12 2 24 14 10 10 24 19 11 29 10 22 7 1 9 1 2 27 7 24 20 25 20 28",
"output": "10203"
},
{
"input": "100\n9 9 72 55 14 8 55 58 35 67 3 18 73 92 41 49 15 60 18 66 9 26 97 47 43 88 71 97 19 34 48 96 79 53 8 24 69 49 12 23 77 12 21 88 66 9 29 13 61 69 54 77 41 13 4 68 37 74 7 6 29 76 55 72 89 4 78 27 29 82 18 83 12 4 32 69 89 85 66 13 92 54 38 5 26 56 17 55 29 4 17 39 29 94 3 67 85 98 21 14",
"output": "426927"
}
] | 1,600,492,818
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 42
| 1,107
| 210,022,400
|
import sys
from operator import lt, le, gt, ge
n = int(sys.stdin.buffer.readline().decode('utf-8'))
a = list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
min_rng = [[0, 0] for _ in range(n)]
max_rng = [[0, 0] for _ in range(n)]
for j, (st, itr, op1, op2) in enumerate(
((-1, range(n), lt, gt), (n, range(n-1, -1, -1), le, ge))):
max_stack = [(st, 10**9)]
min_stack = [(st, -1)]
for i in itr:
while op1(max_stack[-1][1], a[i]):
max_stack.pop()
while op2(min_stack[-1][1], a[i]):
min_stack.pop()
max_rng[i][j] = max_stack[-1][0]
min_rng[i][j] = min_stack[-1][0]
max_stack.append((i, a[i]))
min_stack.append((i, a[i]))
ans = sum(a[i] * ((max_rng[i][1] - i) * (i - max_rng[i][0]) -
(min_rng[i][1] - i) * (i - min_rng[i][0])) for i in range(n))
print(ans)
|
Title: Imbalanced Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a* consisting of *n* elements. The imbalance value of some subsegment of this array is the difference between the maximum and minimum element from this segment. The imbalance value of the array is the sum of imbalance values of all subsegments of this array.
For example, the imbalance value of array [1,<=4,<=1] is 9, because there are 6 different subsegments of this array:
- [1] (from index 1 to index 1), imbalance value is 0; - [1,<=4] (from index 1 to index 2), imbalance value is 3; - [1,<=4,<=1] (from index 1 to index 3), imbalance value is 3; - [4] (from index 2 to index 2), imbalance value is 0; - [4,<=1] (from index 2 to index 3), imbalance value is 3; - [1] (from index 3 to index 3), imbalance value is 0;
You have to determine the imbalance value of the array *a*.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — size of the array *a*.
The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=106) — elements of the array.
Output Specification:
Print one integer — the imbalance value of *a*.
Demo Input:
['3\n1 4 1\n']
Demo Output:
['9\n']
Note:
none
|
```python
import sys
from operator import lt, le, gt, ge
n = int(sys.stdin.buffer.readline().decode('utf-8'))
a = list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
min_rng = [[0, 0] for _ in range(n)]
max_rng = [[0, 0] for _ in range(n)]
for j, (st, itr, op1, op2) in enumerate(
((-1, range(n), lt, gt), (n, range(n-1, -1, -1), le, ge))):
max_stack = [(st, 10**9)]
min_stack = [(st, -1)]
for i in itr:
while op1(max_stack[-1][1], a[i]):
max_stack.pop()
while op2(min_stack[-1][1], a[i]):
min_stack.pop()
max_rng[i][j] = max_stack[-1][0]
min_rng[i][j] = min_stack[-1][0]
max_stack.append((i, a[i]))
min_stack.append((i, a[i]))
ans = sum(a[i] * ((max_rng[i][1] - i) * (i - max_rng[i][0]) -
(min_rng[i][1] - i) * (i - min_rng[i][0])) for i in range(n))
print(ans)
```
| 3
|
|
887
|
B
|
Cubes for Masha
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation"
] | null | null |
Absent-minded Masha got set of *n* cubes for her birthday.
At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural *x* such she can make using her new cubes all integers from 1 to *x*.
To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.
The number can't contain leading zeros. It's not required to use all cubes to build a number.
Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.
|
In first line integer *n* is given (1<=≤<=*n*<=≤<=3) — the number of cubes, Masha got for her birthday.
Each of next *n* lines contains 6 integers *a**i**j* (0<=≤<=*a**i**j*<=≤<=9) — number on *j*-th face of *i*-th cube.
|
Print single integer — maximum number *x* such Masha can make any integers from 1 to *x* using her cubes or 0 if Masha can't make even 1.
|
[
"3\n0 1 2 3 4 5\n6 7 8 9 0 1\n2 3 4 5 6 7\n",
"3\n0 1 3 5 6 8\n1 2 4 5 7 8\n2 3 4 6 7 9\n"
] |
[
"87",
"98"
] |
In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8.
| 1,000
|
[
{
"input": "3\n0 1 2 3 4 5\n6 7 8 9 0 1\n2 3 4 5 6 7",
"output": "87"
},
{
"input": "3\n0 1 3 5 6 8\n1 2 4 5 7 8\n2 3 4 6 7 9",
"output": "98"
},
{
"input": "3\n0 1 2 3 4 5\n0 1 2 3 4 5\n0 1 2 3 4 5",
"output": "5"
},
{
"input": "3\n1 2 3 7 8 9\n9 8 7 1 2 3\n7 9 2 3 1 8",
"output": "3"
},
{
"input": "1\n5 2 2 5 6 7",
"output": "0"
},
{
"input": "1\n7 6 5 8 9 0",
"output": "0"
},
{
"input": "1\n2 5 9 6 7 9",
"output": "0"
},
{
"input": "1\n6 3 1 9 4 9",
"output": "1"
},
{
"input": "1\n1 9 8 3 7 8",
"output": "1"
},
{
"input": "2\n1 7 2 0 4 3\n5 2 3 6 1 0",
"output": "7"
},
{
"input": "2\n6 0 1 7 2 9\n1 3 4 6 7 0",
"output": "4"
},
{
"input": "2\n8 6 4 1 2 0\n7 8 5 3 2 1",
"output": "8"
},
{
"input": "2\n0 8 6 2 1 3\n5 2 7 1 0 9",
"output": "3"
},
{
"input": "2\n0 9 5 7 6 2\n8 6 2 7 1 4",
"output": "2"
},
{
"input": "3\n5 0 7 6 2 1\n2 7 4 6 1 9\n0 2 6 1 7 5",
"output": "2"
},
{
"input": "3\n0 6 2 9 5 4\n3 8 0 1 6 9\n6 9 0 1 5 2",
"output": "6"
},
{
"input": "3\n5 6 2 9 3 5\n5 4 1 5 9 8\n4 4 2 0 3 5",
"output": "6"
},
{
"input": "3\n0 1 9 1 0 8\n9 9 3 5 6 2\n9 3 9 9 7 3",
"output": "3"
},
{
"input": "3\n2 5 7 4 2 7\n1 5 5 9 0 3\n8 2 0 1 5 1",
"output": "5"
},
{
"input": "1\n4 6 9 8 2 7",
"output": "0"
},
{
"input": "1\n5 3 8 0 2 6",
"output": "0"
},
{
"input": "1\n7 9 5 0 4 6",
"output": "0"
},
{
"input": "1\n4 0 9 6 3 1",
"output": "1"
},
{
"input": "1\n7 9 2 5 0 4",
"output": "0"
},
{
"input": "1\n0 7 6 3 2 4",
"output": "0"
},
{
"input": "1\n9 8 1 6 5 7",
"output": "1"
},
{
"input": "1\n7 3 6 9 8 1",
"output": "1"
},
{
"input": "1\n3 9 1 7 4 5",
"output": "1"
},
{
"input": "1\n8 6 0 9 4 2",
"output": "0"
},
{
"input": "1\n8 2 7 4 1 0",
"output": "2"
},
{
"input": "1\n8 3 5 4 2 9",
"output": "0"
},
{
"input": "1\n0 8 7 1 3 2",
"output": "3"
},
{
"input": "1\n6 2 8 5 1 3",
"output": "3"
},
{
"input": "1\n6 0 7 5 4 8",
"output": "0"
},
{
"input": "1\n6 2 8 4 5 1",
"output": "2"
},
{
"input": "1\n4 3 8 9 2 3",
"output": "0"
},
{
"input": "1\n8 1 9 2 9 7",
"output": "2"
},
{
"input": "1\n3 7 7 6 4 2",
"output": "0"
},
{
"input": "1\n1 4 5 7 0 5",
"output": "1"
},
{
"input": "2\n6 6 4 7 9 0\n2 1 2 8 6 4",
"output": "2"
},
{
"input": "2\n5 3 2 9 8 2\n0 7 4 8 1 8",
"output": "5"
},
{
"input": "2\n5 7 4 2 1 9\n2 2 7 1 1 8",
"output": "2"
},
{
"input": "2\n9 3 3 6 7 2\n6 2 9 1 5 9",
"output": "3"
},
{
"input": "2\n2 0 5 7 0 8\n4 5 1 5 4 9",
"output": "2"
},
{
"input": "2\n2 6 8 1 3 1\n2 1 3 8 6 7",
"output": "3"
},
{
"input": "2\n4 3 8 6 0 1\n4 7 1 8 9 0",
"output": "1"
},
{
"input": "2\n0 2 9 1 8 5\n0 7 4 3 2 5",
"output": "5"
},
{
"input": "2\n1 7 6 9 2 5\n1 6 7 0 9 2",
"output": "2"
},
{
"input": "2\n0 2 9 8 1 7\n6 7 4 3 2 5",
"output": "9"
},
{
"input": "2\n3 6 8 9 5 0\n6 7 0 8 2 3",
"output": "0"
},
{
"input": "2\n5 1 2 3 0 8\n3 6 7 4 9 2",
"output": "9"
},
{
"input": "2\n7 8 6 1 4 5\n8 6 4 3 2 5",
"output": "8"
},
{
"input": "2\n2 3 5 1 9 6\n1 6 8 7 3 9",
"output": "3"
},
{
"input": "2\n1 7 8 6 0 9\n3 2 1 7 4 9",
"output": "4"
},
{
"input": "2\n2 4 0 3 7 6\n3 2 8 7 1 5",
"output": "8"
},
{
"input": "2\n6 5 2 7 1 3\n3 7 8 1 0 9",
"output": "3"
},
{
"input": "2\n5 8 4 7 1 2\n0 8 6 2 4 9",
"output": "2"
},
{
"input": "2\n8 0 6 5 1 4\n7 1 0 8 3 4",
"output": "1"
},
{
"input": "2\n2 3 9 1 6 7\n2 5 4 3 0 6",
"output": "7"
},
{
"input": "3\n9 4 3 0 2 6\n7 0 5 3 3 9\n1 0 7 4 6 7",
"output": "7"
},
{
"input": "3\n3 8 5 1 5 5\n1 5 7 2 6 9\n4 3 4 8 8 9",
"output": "9"
},
{
"input": "3\n7 7 2 5 3 2\n3 0 0 6 4 4\n1 2 1 1 9 1",
"output": "7"
},
{
"input": "3\n8 1 6 8 6 8\n7 0 2 5 8 4\n5 2 0 3 1 9",
"output": "32"
},
{
"input": "3\n2 7 4 0 7 1\n5 5 4 9 1 4\n2 1 7 5 1 7",
"output": "2"
},
{
"input": "3\n4 4 5 0 6 6\n7 1 6 9 5 4\n5 0 4 0 3 9",
"output": "1"
},
{
"input": "3\n9 4 3 3 9 3\n1 0 3 4 5 3\n2 9 6 2 4 1",
"output": "6"
},
{
"input": "3\n3 8 3 5 5 5\n3 0 1 6 6 3\n0 4 3 7 2 4",
"output": "8"
},
{
"input": "3\n4 1 0 8 0 2\n1 5 3 5 0 7\n7 7 2 7 2 2",
"output": "5"
},
{
"input": "3\n8 1 8 2 7 1\n9 1 9 9 4 7\n0 0 9 0 4 0",
"output": "2"
},
{
"input": "3\n4 6 0 3 9 2\n8 6 9 0 7 2\n6 9 3 2 5 7",
"output": "0"
},
{
"input": "3\n5 1 2 9 6 4\n9 0 6 4 2 8\n4 6 2 8 3 7",
"output": "10"
},
{
"input": "3\n9 3 1 8 4 6\n6 9 1 2 0 7\n8 9 1 5 0 3",
"output": "21"
},
{
"input": "3\n7 1 3 0 2 4\n2 4 3 0 9 5\n1 9 8 0 6 5",
"output": "65"
},
{
"input": "3\n9 4 6 2 7 0\n3 7 1 9 6 4\n6 1 0 8 7 2",
"output": "4"
},
{
"input": "3\n2 7 3 6 4 5\n0 2 1 9 4 8\n8 6 9 5 4 0",
"output": "10"
},
{
"input": "3\n2 6 3 7 1 0\n9 1 2 4 7 6\n1 4 8 7 6 2",
"output": "4"
},
{
"input": "3\n5 4 8 1 6 7\n0 9 3 5 8 6\n2 4 7 8 1 3",
"output": "21"
},
{
"input": "3\n7 2 1 3 6 9\n0 3 8 4 7 6\n1 4 5 8 7 0",
"output": "21"
},
{
"input": "3\n8 6 0 5 4 9\n1 8 5 3 9 7\n7 4 5 1 6 8",
"output": "1"
},
{
"input": "1\n0 1 2 3 4 5",
"output": "5"
},
{
"input": "3\n0 1 1 2 2 3\n4 5 6 7 8 9\n3 4 5 6 7 8",
"output": "9"
},
{
"input": "2\n0 1 2 3 4 5\n6 7 8 9 1 2",
"output": "29"
},
{
"input": "3\n0 1 2 3 4 5\n6 7 8 9 1 2\n3 4 5 6 7 8",
"output": "98"
},
{
"input": "3\n0 1 1 2 2 3\n4 5 6 7 8 9\n3 4 5 6 7 1",
"output": "19"
},
{
"input": "2\n0 1 2 3 4 5\n6 7 8 9 6 6",
"output": "9"
},
{
"input": "2\n0 1 2 3 4 5\n4 5 6 7 8 9",
"output": "9"
},
{
"input": "2\n1 8 9 1 1 0\n2 3 4 5 6 7",
"output": "9"
},
{
"input": "2\n0 1 2 3 4 5\n9 8 7 6 5 4",
"output": "9"
},
{
"input": "3\n2 3 4 5 6 7\n3 4 5 6 7 8\n9 1 2 3 4 5",
"output": "9"
},
{
"input": "3\n1 1 2 3 4 5\n6 7 8 9 0 2\n3 4 5 6 7 8",
"output": "10"
},
{
"input": "3\n1 1 1 0 2 3\n4 5 6 7 8 9\n0 0 0 0 0 0",
"output": "10"
},
{
"input": "3\n0 1 2 2 4 5\n6 7 8 9 0 1\n3 3 4 5 6 7",
"output": "21"
},
{
"input": "3\n2 0 1 3 4 5\n6 7 8 9 1 1\n3 4 5 6 6 7",
"output": "19"
},
{
"input": "3\n1 1 1 1 1 1\n0 2 3 4 5 6\n7 8 9 2 3 4",
"output": "10"
},
{
"input": "2\n0 1 2 3 4 5\n6 6 6 7 8 9",
"output": "9"
},
{
"input": "3\n3 4 5 6 8 9\n1 1 1 1 1 1\n1 2 4 5 7 0",
"output": "19"
}
] | 1,552,395,732
| 732
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 48
| 124
| 0
|
n = int(input())
a = [list(map(int, input().split())) for _ in range(n)]
fa = [i for b in a for i in b]
if 1 not in fa:
print(0)
exit()
ka = {i: fa.count(i) for i in range(10)}
for i in range(1, 1000):
rq = list(map(int, str(i)))
for j in rq:
if rq.count(j) > ka[j]:
print(i - 1)
exit()
print(999)
|
Title: Cubes for Masha
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Absent-minded Masha got set of *n* cubes for her birthday.
At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural *x* such she can make using her new cubes all integers from 1 to *x*.
To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.
The number can't contain leading zeros. It's not required to use all cubes to build a number.
Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.
Input Specification:
In first line integer *n* is given (1<=≤<=*n*<=≤<=3) — the number of cubes, Masha got for her birthday.
Each of next *n* lines contains 6 integers *a**i**j* (0<=≤<=*a**i**j*<=≤<=9) — number on *j*-th face of *i*-th cube.
Output Specification:
Print single integer — maximum number *x* such Masha can make any integers from 1 to *x* using her cubes or 0 if Masha can't make even 1.
Demo Input:
['3\n0 1 2 3 4 5\n6 7 8 9 0 1\n2 3 4 5 6 7\n', '3\n0 1 3 5 6 8\n1 2 4 5 7 8\n2 3 4 6 7 9\n']
Demo Output:
['87', '98']
Note:
In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8.
|
```python
n = int(input())
a = [list(map(int, input().split())) for _ in range(n)]
fa = [i for b in a for i in b]
if 1 not in fa:
print(0)
exit()
ka = {i: fa.count(i) for i in range(10)}
for i in range(1, 1000):
rq = list(map(int, str(i)))
for j in rq:
if rq.count(j) > ka[j]:
print(i - 1)
exit()
print(999)
```
| 0
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,689,882,447
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
def f(wrd):
h ='hello'
for a in h:
ln = len(wrd)
for i in range(ln):
if wrd[i] == a:
wrd = wrd[i:]
print(wrd,i)
break
else:
return 'NO'
return 'YES'
wrd = input()
print(f(wrd))
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
def f(wrd):
h ='hello'
for a in h:
ln = len(wrd)
for i in range(ln):
if wrd[i] == a:
wrd = wrd[i:]
print(wrd,i)
break
else:
return 'NO'
return 'YES'
wrd = input()
print(f(wrd))
```
| 0
|
Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.