message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51 | instruction | 0 | 40,072 | 7 | 80,144 |
Tags: implementation, math, number theory
Correct Solution:
```
def GCD(a, b):
if a == 0:
return b
return GCD(b % a, a)
# Function to return LCM of two numbers
def LCM(a, b):
return (a * b) / GCD(a, b)
n,a,b,p,q=map(int,input().split())
lcm = LCM(a,b)
x=0
sum=0
if p>q :
x=int(n/a)
sum += (x*p)
x=int(n/b) - int(n/lcm)
sum += (x*q)
else :
x=int(n/b)
sum += (x*q)
x=int(n/a) - int(n/lcm)
sum += (x*p)
print(sum)
``` | output | 1 | 40,072 | 7 | 80,145 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51 | instruction | 0 | 40,073 | 7 | 80,146 |
Tags: implementation, math, number theory
Correct Solution:
```
n,a,b,p,q = map(int, input().strip().split())
def gcd(x, y):
if y==0:
return x
return gcd(y, x%y)
def lcm(x, y):
return (x*y)//gcd(x,y)
comm = lcm(a, b)
ans = (n//a - n//comm)*p
ans += (n//b - n//comm)*q
ans += (n//comm)*max(p, q)
print(ans)
``` | output | 1 | 40,073 | 7 | 80,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51 | instruction | 0 | 40,074 | 7 | 80,148 |
Tags: implementation, math, number theory
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
from fractions import gcd
def lcm(a, b):
return (a*b)//gcd(a,b)
(n, a, b, p, q) = (int(i) for i in input().split())
start = time.time()
if q > p:
(a, b, p, q) = (b, a, q, p)
ans = (n//a)*p + (n//b-n//lcm(a,b))*q
print(ans)
finish = time.time()
#print(finish - start)
``` | output | 1 | 40,074 | 7 | 80,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51
Submitted Solution:
```
def nod(a,b):
while a>0 and b>0:
a%=b
a,b=b,a
return a+b
def gcd(a,b):
return (a*b)//nod(a,b)
n,a,b,p,q=[int(i) for i in input().split()]
res1=(n//a)*p+(n//b-n//gcd(a,b))*q
res2=(n//b)*q+(n//a-n//gcd(a,b))*p
print(int(max(res1,res2)))
``` | instruction | 0 | 40,075 | 7 | 80,150 |
Yes | output | 1 | 40,075 | 7 | 80,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51
Submitted Solution:
```
from fractions import gcd
n,a,b,p,q=map(int,input().split())
x=n//(a*b//gcd(a,b))
print((n//a-x)*p+(n//b-x)*q+x*max(p, q))
``` | instruction | 0 | 40,076 | 7 | 80,152 |
Yes | output | 1 | 40,076 | 7 | 80,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51
Submitted Solution:
```
from math import gcd
n,a,b,p,q = map(int,input().split())
lcm = a * b // gcd(a,b);
ans = n // lcm * max(p,q)
ans += (n // a - n // lcm) * p
ans += (n // b - n // lcm) * q
print(ans)
``` | instruction | 0 | 40,077 | 7 | 80,154 |
Yes | output | 1 | 40,077 | 7 | 80,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51
Submitted Solution:
```
from sys import stdin, stdout
def gcd(a, b):
if not b:
return a
return gcd(b, a % b)
n, a, b, p, q = map(int, stdin.readline().split())
first, second, third = (n // a) * p, (n // b) * q, (n // (a * b // gcd(a, b))) * min(p, q)
stdout.write(str(first + second - third))
``` | instruction | 0 | 40,078 | 7 | 80,156 |
Yes | output | 1 | 40,078 | 7 | 80,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51
Submitted Solution:
```
primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089,2099,2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287,2293,2297,2309,2311,2333,2339,2341,2347,2351,2357,2371,2377,2381,2383,2389,2393,2399,2411,2417,2423,2437,2441,2447,2459,2467,2473,2477,2503,2521,2531,2539,2543,2549,2551,2557,2579,2591,2593,2609,2617,2621,2633,2647,2657,2659,2663,2671,2677,2683,2687,2689,2693,2699,2707,2711,2713,2719,2729,2731,2741,2749,2753,2767,2777,2789,2791,2797,2801,2803,2819,2833,2837,2843,2851,2857,2861,2879,2887,2897,2903,2909,2917,2927,2939,2953,2957,2963,2969,2971,2999,3001,3011,3019,3023,3037,3041,3049,3061,3067,3079,3083,3089,3109,3119,3121,3137,3163,3167,3169,3181,3187,3191,3203,3209,3217,3221,3229,3251,3253,3257,3259,3271,3299,3301,3307,3313,3319,3323,3329,3331,3343,3347,3359,3361,3371,3373,3389,3391,3407,3413,3433,3449,3457,3461,3463,3467,3469,3491,3499,3511,3517,3527,3529,3533,3539,3541,3547,3557,3559,3571,3581,3583,3593,3607,3613,3617,3623,3631,3637,3643,3659,3671,3673,3677,3691,3697,3701,3709,3719,3727,3733,3739,3761,3767,3769,3779,3793,3797,3803,3821,3823,3833,3847,3851,3853,3863,3877,3881,3889,3907,3911,3917,3919,3923,3929,3931,3943,3947,3967,3989,4001,4003,4007,4013,4019,4021,4027,4049,4051,4057,4073,4079,4091,4093,4099,4111,4127,4129,4133,4139,4153,4157,4159,4177,4201,4211,4217,4219,4229,4231,4241,4243,4253,4259,4261,4271,4273,4283,4289,4297,4327,4337,4339,4349,4357,4363,4373,4391,4397,4409,4421,4423,4441,4447,4451,4457,4463,4481,4483,4493,4507,4513,4517,4519,4523,4547,4549,4561,4567,4583,4591,4597,4603,4621,4637,4639,4643,4649,4651,4657,4663,4673,4679,4691,4703,4721,4723,4729,4733,4751,4759,4783,4787,4789,4793,4799,4801,4813,4817,4831,4861,4871,4877,4889,4903,4909,4919,4931,4933,4937,4943,4951,4957,4967,4969,4973,4987,4993,4999,5003,5009,5011,5021,5023,5039,5051,5059,5077,5081,5087,5099,5101,5107,5113,5119,5147,5153,5167,5171,5179,5189,5197,5209,5227,5231,5233,5237,5261,5273,5279,5281,5297,5303,5309,5323,5333,5347,5351,5381,5387,5393,5399,5407,5413,5417,5419,5431,5437,5441,5443,5449,5471,5477,5479,5483,5501,5503,5507,5519,5521,5527,5531,5557,5563,5569,5573,5581,5591,5623,5639,5641,5647,5651,5653,5657,5659,5669,5683,5689,5693,5701,5711,5717,5737,5741,5743,5749,5779,5783,5791,5801,5807,5813,5821,5827,5839,5843,5849,5851,5857,5861,5867,5869,5879,5881,5897,5903,5923,5927,5939,5953,5981,5987,6007,6011,6029,6037,6043,6047,6053,6067,6073,6079,6089,6091,6101,6113,6121,6131,6133,6143,6151,6163,6173,6197,6199,6203,6211,6217,6221,6229,6247,6257,6263,6269,6271,6277,6287,6299,6301,6311,6317,6323,6329,6337,6343,6353,6359,6361,6367,6373,6379,6389,6397,6421,6427,6449,6451,6469,6473,6481,6491,6521,6529,6547,6551,6553,6563,6569,6571,6577,6581,6599,6607,6619,6637,6653,6659,6661,6673,6679,6689,6691,6701,6703,6709,6719,6733,6737,6761,6763,6779,6781,6791,6793,6803,6823,6827,6829,6833,6841,6857,6863,6869,6871,6883,6899,6907,6911,6917,6947,6949,6959,6961,6967,6971,6977,6983,6991,6997,7001,7013,7019,7027,7039,7043,7057,7069,7079,7103,7109,7121,7127,7129,7151,7159,7177,7187,7193,7207,7211,7213,7219,7229,7237,7243,7247,7253,7283,7297,7307,7309,7321,7331,7333,7349,7351,7369,7393,7411,7417,7433,7451,7457,7459,7477,7481,7487,7489,7499,7507,7517,7523,7529,7537,7541,7547,7549,7559,7561,7573,7577,7583,7589,7591,7603,7607,7621,7639,7643,7649,7669,7673,7681,7687,7691,7699,7703,7717,7723,7727,7741,7753,7757,7759,7789,7793,7817,7823,7829,7841,7853,7867,7873,7877,7879,7883,7901,7907,7919,7927,7933,7937,7949,7951,7963,7993,8009,8011,8017,8039,8053,8059,8069,8081,8087,8089,8093,8101,8111,8117,8123,8147,8161,8167,8171,8179,8191,8209,8219,8221,8231,8233,8237,8243,8263,8269,8273,8287,8291,8293,8297,8311,8317,8329,8353,8363,8369,8377,8387,8389,8419,8423,8429,8431,8443,8447,8461,8467,8501,8513,8521,8527,8537,8539,8543,8563,8573,8581,8597,8599,8609,8623,8627,8629,8641,8647,8663,8669,8677,8681,8689,8693,8699,8707,8713,8719,8731,8737,8741,8747,8753,8761,8779,8783,8803,8807,8819,8821,8831,8837,8839,8849,8861,8863,8867,8887,8893,8923,8929,8933,8941,8951,8963,8969,8971,8999,9001,9007,9011,9013,9029,9041,9043,9049,9059,9067,9091,9103,9109,9127,9133,9137,9151,9157,9161,9173,9181,9187,9199,9203,9209,9221,9227,9239,9241,9257,9277,9281,9283,9293,9311,9319,9323,9337,9341,9343,9349,9371,9377,9391,9397,9403,9413,9419,9421,9431,9433,9437,9439,9461,9463,9467,9473,9479,9491,9497,9511,9521,9533,9539,9547,9551,9587,9601,9613,9619,9623,9629,9631,9643,9649,9661,9677,9679,9689,9697,9719,9721,9733,9739,9743,9749,9767,9769,9781,9787,9791,9803,9811,9817,9829,9833,9839,9851,9857,9859,9871,9883,9887,9901,9907,9923,9929,9931,9941,9949,9967,9973,10007,10009,10037,10039,10061,10067,10069,10079,10091,10093,10099,10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193,10211,10223,10243,10247,10253,10259,10267,10271,10273,10289,10301,10303,10313,10321,10331,10333,10337,10343,10357,10369,10391,10399,10427,10429,10433,10453,10457,10459,10463,10477,10487,10499,10501,10513,10529,10531,10559,10567,10589,10597,10601,10607,10613,10627,10631,10639,10651,10657,10663,10667,10687,10691,10709,10711,10723,10729,10733,10739,10753,10771,10781,10789,10799,10831,10837,10847,10853,10859,10861,10867,10883,10889,10891,10903,10909,10937,10939,10949,10957,10973,10979,10987,10993,11003,11027,11047,11057,11059,11069,11071,11083,11087,11093,11113,11117,11119,11131,11149,11159,11161,11171,11173,11177,11197,11213,11239,11243,11251,11257,11261,11273,11279,11287,11299,11311,11317,11321,11329,11351,11353,11369,11383,11393,11399,11411,11423,11437,11443,11447,11467,11471,11483,11489,11491,11497,11503,11519,11527,11549,11551,11579,11587,11593,11597,11617,11621,11633,11657,11677,11681,11689,11699,11701,11717,11719,11731,11743,11777,11779,11783,11789,11801,11807,11813,11821,11827,11831,11833,11839,11863,11867,11887,11897,11903,11909,11923,11927,11933,11939,11941,11953,11959,11969,11971,11981,11987,12007,12011,12037,12041,12043,12049,12071,12073,12097,12101,12107,12109,12113,12119,12143,12149,12157,12161,12163,12197,12203,12211,12227,12239,12241,12251,12253,12263,12269,12277,12281,12289,12301,12323,12329,12343,12347,12373,12377,12379,12391,12401,12409,12413,12421,12433,12437,12451,12457,12473,12479,12487,12491,12497,12503,12511,12517,12527,12539,12541,12547,12553,12569,12577,12583,12589,12601,12611,12613,12619,12637,12641,12647,12653,12659,12671,12689,12697,12703,12713,12721,12739,12743,12757,12763,12781,12791,12799,12809,12821,12823,12829,12841,12853,12889,12893,12899,12907,12911,12917,12919,12923,12941,12953,12959,12967,12973,12979,12983,13001,13003,13007,13009,13033,13037,13043,13049,13063,13093,13099,13103,13109,13121,13127,13147,13151,13159,13163,13171,13177,13183,13187,13217,13219,13229,13241,13249,13259,13267,13291,13297,13309,13313,13327,13331,13337,13339,13367,13381,13397,13399,13411,13417,13421,13441,13451,13457,13463,13469,13477,13487,13499,13513,13523,13537,13553,13567,13577,13591,13597,13613,13619,13627,13633,13649,13669,13679,13681,13687,13691,13693,13697,13709,13711,13721,13723,13729,13751,13757,13759,13763,13781,13789,13799,13807,13829,13831,13841,13859,13873,13877,13879,13883,13901,13903,13907,13913,13921,13931,13933,13963,13967,13997,13999,14009,14011,14029,14033,14051,14057,14071,14081,14083,14087,14107,14143,14149,14153,14159,14173,14177,14197,14207,14221,14243,14249,14251,14281,14293,14303,14321,14323,14327,14341,14347,14369,14387,14389,14401,14407,14411,14419,14423,14431,14437,14447,14449,14461,14479,14489,14503,14519,14533,14537,14543,14549,14551,14557,14561,14563,14591,14593,14621,14627,14629,14633,14639,14653,14657,14669,14683,14699,14713,14717,14723,14731,14737,14741,14747,14753,14759,14767,14771,14779,14783,14797,14813,14821,14827,14831,14843,14851,14867,14869,14879,14887,14891,14897,14923,14929,14939,14947,14951,14957,14969,14983,15013,15017,15031,15053,15061,15073,15077,15083,15091,15101,15107,15121,15131,15137,15139,15149,15161,15173,15187,15193,15199,15217,15227,15233,15241,15259,15263,15269,15271,15277,15287,15289,15299,15307,15313,15319,15329,15331,15349,15359,15361,15373,15377,15383,15391,15401,15413,15427,15439,15443,15451,15461,15467,15473,15493,15497,15511,15527,15541,15551,15559,15569,15581,15583,15601,15607,15619,15629,15641,15643,15647,15649,15661,15667,15671,15679,15683,15727,15731,15733,15737,15739,15749,15761,15767,15773,15787,15791,15797,15803,15809,15817,15823,15859,15877,15881,15887,15889,15901,15907,15913,15919,15923,15937,15959,15971,15973,15991,16001,16007,16033,16057,16061,16063,16067,16069,16073,16087,16091,16097,16103,16111,16127,16139,16141,16183,16187,16189,16193,16217,16223,16229,16231,16249,16253,16267,16273,16301,16319,16333,16339,16349,16361,16363,16369,16381,16411,16417,16421,16427,16433,16447,16451,16453,16477,16481,16487,16493,16519,16529,16547,16553,16561,16567,16573,16603,16607,16619,16631,16633,16649,16651,16657,16661,16673,16691,16693,16699,16703,16729,16741,16747,16759,16763,16787,16811,16823,16829,16831,16843,16871,16879,16883,16889,16901,16903,16921,16927,16931,16937,16943,16963,16979,16981,16987,16993,17011,17021,17027,17029,17033,17041,17047,17053,17077,17093,17099,17107,17117,17123,17137,17159,17167,17183,17189,17191,17203,17207,17209,17231,17239,17257,17291,17293,17299,17317,17321,17327,17333,17341,17351,17359,17377,17383,17387,17389,17393,17401,17417,17419,17431,17443,17449,17467,17471,17477,17483,17489,17491,17497,17509,17519,17539,17551,17569,17573,17579,17581,17597,17599,17609,17623,17627,17657,17659,17669,17681,17683,17707,17713,17729,17737,17747,17749,17761,17783,17789,17791,17807,17827,17837,17839,17851,17863,17881,17891,17903,17909,17911,17921,17923,17929,17939,17957,17959,17971,17977,17981,17987,17989,18013,18041,18043,18047,18049,18059,18061,18077,18089,18097,18119,18121,18127,18131,18133,18143,18149,18169,18181,18191,18199,18211,18217,18223,18229,18233,18251,18253,18257,18269,18287,18289,18301,18307,18311,18313,18329,18341,18353,18367,18371,18379,18397,18401,18413,18427,18433,18439,18443,18451,18457,18461,18481,18493,18503,18517,18521,18523,18539,18541,18553,18583,18587,18593,18617,18637,18661,18671,18679,18691,18701,18713,18719,18731,18743,18749,18757,18773,18787,18793,18797,18803,18839,18859,18869,18899,18911,18913,18917,18919,18947,18959,18973,18979,19001,19009,19013,19031,19037,19051,19069,19073,19079,19081,19087,19121,19139,19141,19157,19163,19181,19183,19207,19211,19213,19219,19231,19237,19249,19259,19267,19273,19289,19301,19309,19319,19333,19373,19379,19381,19387,19391,19403,19417,19421,19423,19427,19429,19433,19441,19447,19457,19463,19469,19471,19477,19483,19489,19501,19507,19531,19541,19543,19553,19559,19571,19577,19583,19597,19603,19609,19661,19681,19687,19697,19699,19709,19717,19727,19739,19751,19753,19759,19763,19777,19793,19801,19813,19819,19841,19843,19853,19861,19867,19889,19891,19913,19919,19927,19937,19949,19961,19963,19973,19979,19991,19993,19997,20011,20021,20023,20029,20047,20051,20063,20071,20089,20101,20107,20113,20117,20123,20129,20143,20147,20149,20161,20173,20177,20183,20201,20219,20231,20233,20249,20261,20269,20287,20297,20323,20327,20333,20341,20347,20353,20357,20359,20369,20389,20393,20399,20407,20411,20431,20441,20443,20477,20479,20483,20507,20509,20521,20533,20543,20549,20551,20563,20593,20599,20611,20627,20639,20641,20663,20681,20693,20707,20717,20719,20731,20743,20747,20749,20753,20759,20771,20773,20789,20807,20809,20849,20857,20873,20879,20887,20897,20899,20903,20921,20929,20939,20947,20959,20963,20981,20983,21001,21011,21013,21017,21019,21023,21031,21059,21061,21067,21089,21101,21107,21121,21139,21143,21149,21157,21163,21169,21179,21187,21191,21193,21211,21221,21227,21247,21269,21277,21283,21313,21317,21319,21323,21341,21347,21377,21379,21383,21391,21397,21401,21407,21419,21433,21467,21481,21487,21491,21493,21499,21503,21517,21521,21523,21529,21557,21559,21563,21569,21577,21587,21589,21599,21601,21611,21613,21617,21647,21649,21661,21673,21683,21701,21713,21727,21737,21739,21751,21757,21767,21773,21787,21799,21803,21817,21821,21839,21841,21851,21859,21863,21871,21881,21893,21911,21929,21937,21943,21961,21977,21991,21997,22003,22013,22027,22031,22037,22039,22051,22063,22067,22073,22079,22091,22093,22109,22111,22123,22129,22133,22147,22153,22157,22159,22171,22189,22193,22229,22247,22259,22271,22273,22277,22279,22283,22291,22303,22307,22343,22349,22367,22369,22381,22391,22397,22409,22433,22441,22447,22453,22469,22481,22483,22501,22511,22531,22541,22543,22549,22567,22571,22573,22613,22619,22621,22637,22639,22643,22651,22669,22679,22691,22697,22699,22709,22717,22721,22727,22739,22741,22751,22769,22777,22783,22787,22807,22811,22817,22853,22859,22861,22871,22877,22901,22907,22921,22937,22943,22961,22963,22973,22993,23003,23011,23017,23021,23027,23029,23039,23041,23053,23057,23059,23063,23071,23081,23087,23099,23117,23131,23143,23159,23167,23173,23189,23197,23201,23203,23209,23227,23251,23269,23279,23291,23293,23297,23311,23321,23327,23333,23339,23357,23369,23371,23399,23417,23431,23447,23459,23473,23497,23509,23531,23537,23539,23549,23557,23561,23563,23567,23581,23593,23599,23603,23609,23623,23627,23629,23633,23663,23669,23671,23677,23687,23689,23719,23741,23743,23747,23753,23761,23767,23773,23789,23801,23813,23819,23827,23831,23833,23857,23869,23873,23879,23887,23893,23899,23909,23911,23917,23929,23957,23971,23977,23981,23993,24001,24007,24019,24023,24029,24043,24049,24061,24071,24077,24083,24091,24097,24103,24107,24109,24113,24121,24133,24137,24151,24169,24179,24181,24197,24203,24223,24229,24239,24247,24251,24281,24317,24329,24337,24359,24371,24373,24379,24391,24407,24413,24419,24421,24439,24443,24469,24473,24481,24499,24509,24517,24527,24533,24547,24551,24571,24593,24611,24623,24631,24659,24671,24677,24683,24691,24697,24709,24733,24749,24763,24767,24781,24793,24799,24809,24821,24841,24847,24851,24859,24877,24889,24907,24917,24919,24923,24943,24953,24967,24971,24977,24979,24989,25013,25031,25033,25037,25057,25073,25087,25097,25111,25117,25121,25127,25147,25153,25163,25169,25171,25183,25189,25219,25229,25237,25243,25247,25253,25261,25301,25303,25307,25309,25321,25339,25343,25349,25357,25367,25373,25391,25409,25411,25423,25439,25447,25453,25457,25463,25469,25471,25523,25537,25541,25561,25577,25579,25583,25589,25601,25603,25609,25621,25633,25639,25643,25657,25667,25673,25679,25693,25703,25717,25733,25741,25747,25759,25763,25771,25793,25799,25801,25819,25841,25847,25849,25867,25873,25889,25903,25913,25919,25931,25933,25939,25943,25951,25969,25981,25997,25999,26003,26017,26021,26029,26041,26053,26083,26099,26107,26111,26113,26119,26141,26153,26161,26171,26177,26183,26189,26203,26209,26227,26237,26249,26251,26261,26263,26267,26293,26297,26309,26317,26321,26339,26347,26357,26371,26387,26393,26399,26407,26417,26423,26431,26437,26449,26459,26479,26489,26497,26501,26513,26539,26557,26561,26573,26591,26597,26627,26633,26641,26647,26669,26681,26683,26687,26693,26699,26701,26711,26713,26717,26723,26729,26731,26737,26759,26777,26783,26801,26813,26821,26833,26839,26849,26861,26863,26879,26881,26891,26893,26903,26921,26927,26947,26951,26953,26959,26981,26987,26993,27011,27017,27031,27043,27059,27061,27067,27073,27077,27091,27103,27107,27109,27127,27143,27179,27191,27197,27211,27239,27241,27253,27259,27271,27277,27281,27283,27299,27329,27337,27361,27367,27397,27407,27409,27427,27431,27437,27449,27457,27479,27481,27487,27509,27527,27529,27539,27541,27551,27581,27583,27611,27617,27631,27647,27653,27673,27689,27691,27697,27701,27733,27737,27739,27743,27749,27751,27763,27767,27773,27779,27791,27793,27799,27803,27809,27817,27823,27827,27847,27851,27883,27893,27901,27917,27919,27941,27943,27947,27953,27961,27967,27983,27997,28001,28019,28027,28031,28051,28057,28069,28081,28087,28097,28099,28109,28111,28123,28151,28163,28181,28183,28201,28211,28219,28229,28277,28279,28283,28289,28297,28307,28309,28319,28349,28351,28387,28393,28403,28409,28411,28429,28433,28439,28447,28463,28477,28493,28499,28513,28517,28537,28541,28547,28549,28559,28571,28573,28579,28591,28597,28603,28607,28619,28621,28627,28631,28643,28649,28657,28661,28663,28669,28687,28697,28703,28711,28723,28729,28751,28753,28759,28771,28789,28793,28807,28813,28817,28837,28843,28859,28867,28871,28879,28901,28909,28921,28927,28933,28949,28961,28979,29009,29017,29021,29023,29027,29033,29059,29063,29077,29101,29123,29129,29131,29137,29147,29153,29167,29173,29179,29191,29201,29207,29209,29221,29231,29243,29251,29269,29287,29297,29303,29311,29327,29333,29339,29347,29363,29383,29387,29389,29399,29401,29411,29423,29429,29437,29443,29453,29473,29483,29501,29527,29531,29537,29567,29569,29573,29581,29587,29599,29611,29629,29633,29641,29663,29669,29671,29683,29717,29723,29741,29753,29759,29761,29789,29803,29819,29833,29837,29851,29863,29867,29873,29879,29881,29917,29921,29927,29947,29959,29983,29989,30011,30013,30029,30047,30059,30071,30089,30091,30097,30103,30109,30113,30119,30133,30137,30139,30161,30169,30181,30187,30197,30203,30211,30223,30241,30253,30259,30269,30271,30293,30307,30313,30319,30323,30341,30347,30367,30389,30391,30403,30427,30431,30449,30467,30469,30491,30493,30497,30509,30517,30529,30539,30553,30557,30559,30577,30593,30631,30637,30643,30649,30661,30671,30677,30689,30697,30703,30707,30713,30727,30757,30763,30773,30781,30803,30809,30817,30829,30839,30841,30851,30853,30859,30869,30871,30881,30893,30911,30931,30937,30941,30949,30971,30977,30983,31013,31019,31033,31039,31051,31063,31069,31079,31081,31091,31121,31123,31139,31147,31151,31153,31159,31177,31181,31183,31189,31193,31219,31223,31231,31237,31247,31249,31253,31259,31267,31271,31277,31307,31319,31321,31327,31333,31337,31357,31379,31387,31391,31393,31397,31469,31477,31481,31489,31511,31513,31517,31531,31541,31543,31547,31567,31573,31583,31601,31607,31627]
def fact(x):
d= {}
for i in primes:
if x % i == 0:
d[i] = 0
while x % i == 0:
x //=i
d[i]+=1
if x == 1: break;
return d
n, a, b, p, q = map(int, input().split())
ad = fact(a)
bd = fact(b)
td = {}
t = 1
for key in ad:
if key in bd:
td[key] = max(ad[key], bd[key])
else:
td[key] = ad[key]
t*=key**td[key]
for key in bd:
if key in td: continue
td[key] = bd[key]
t*=key**td[key]
if p > q:
a1 = n//a
b1 = n//b - n//t
else:
b1 = n//b
a1 = n//a - n//t
print(a1*p+b1*q)
``` | instruction | 0 | 40,079 | 7 | 80,158 |
No | output | 1 | 40,079 | 7 | 80,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51
Submitted Solution:
```
import os, sys
from io import BytesIO, IOBase
mod=10**9+7
from math import gcd
input = lambda: sys.stdin.readline().rstrip("\r\n")
def pw(x,y):
ans=1
while y:
if y%2:
ans*=x
y-=1
ans%=mod
else:
x*=x
y//=2
x%=mod
return ans
n,a,b,p,q=map(int,input().split())
z=(a*b)//gcd(a,b)
t=n//z
ans=(n//a -t)*p +(n//b -t)*q
if a>b:
ans+=t*p
else:
ans+=t*q
print(ans)
``` | instruction | 0 | 40,080 | 7 | 80,160 |
No | output | 1 | 40,080 | 7 | 80,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51
Submitted Solution:
```
import bisect
from itertools import accumulate
import os
import sys
import math
from decimal import *
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def SieveOfEratosthenes(n):
prime=[]
primes = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (primes[p] == True):
prime.append(p)
for i in range(p * p, n+1, p):
primes[i] = False
p += 1
return prime
def primefactors(n):
fac=[]
while(n%2==0):
fac.append(2)
n=n//2
for i in range(3,int(math.sqrt(n))+2):
while(n%i==0):
fac.append(i)
n=n//i
if n>1:
fac.append(n)
return fac
def factors(n):
fac=set()
fac.add(1)
fac.add(n)
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
fac.add(i)
fac.add(n//i)
return list(fac)
#------------------------------------------------------code by AD18
n,a,b,p,q=map(int,input().split())
red=n//a
blue=n//b
both=(n//a)//b
if p>q:
blue=blue-both
print(blue*q+red*p)
elif q>p:
red=red-both
print(red*p+blue*q)
else:
red=red-both
print(red*p+blue*q)
``` | instruction | 0 | 40,081 | 7 | 80,162 |
No | output | 1 | 40,081 | 7 | 80,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Examples
Input
5 2 3 12 15
Output
39
Input
20 2 3 3 5
Output
51
Submitted Solution:
```
n,a,b,p,q=map(int,input().split())
da=n // a
db=n // b
dab=n // (a*b)
da -= dab
db -= dab
print(da * p + db * q + dab * max(p,q))
``` | instruction | 0 | 40,082 | 7 | 80,164 |
No | output | 1 | 40,082 | 7 | 80,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an image, that can be represented with a 2-d n by m grid of pixels. Each pixel of the image is either on or off, denoted by the characters "0" or "1", respectively. You would like to compress this image. You want to choose an integer k > 1 and split the image into k by k blocks. If n and m are not divisible by k, the image is padded with only zeros on the right and bottom so that they are divisible by k. Each pixel in each individual block must have the same value. The given image may not be compressible in its current state. Find the minimum number of pixels you need to toggle (after padding) in order for the image to be compressible for some k. More specifically, the steps are to first choose k, then the image is padded with zeros, then, we can toggle the pixels so it is compressible for this k. The image must be compressible in that state.
Input
The first line of input will contain two integers n, m (2 ≤ n, m ≤ 2 500), the dimensions of the image.
The next n lines of input will contain a binary string with exactly m characters, representing the image.
Output
Print a single integer, the minimum number of pixels needed to toggle to make the image compressible.
Example
Input
3 5
00100
10110
11001
Output
5
Note
We first choose k = 2.
The image is padded as follows:
001000
101100
110010
000000
We can toggle the image to look as follows:
001100
001100
000000
000000
We can see that this image is compressible for k = 2. | instruction | 0 | 40,110 | 7 | 80,220 |
Tags: brute force
Correct Solution:
```
#!/usr/bin/env python
# coding:utf-8
# Copyright (C) dirlt
from sys import stdin
def run(n, m, pixels):
ans = 1 << 30
acc = [[0] * (m + 1) for _ in range(n + 1)]
for i in range(n):
for j in range(m):
acc[i + 1][j + 1] = acc[i + 1][j] + int(pixels[i][j])
for j in range(m):
acc[i + 1][j + 1] += acc[i][j + 1]
# print(acc)
for k in range(2, max(n, m) + 1):
r, c = (n + k - 1) // k, (m + k - 1) // k
res = 0
for i in range(r):
for j in range(c):
x, y = i * k, j * k
x2, y2 = min(x + k - 1, n - 1), min(y + k - 1, m - 1)
zero = acc[x2 + 1][y2 + 1] - acc[x][y2 + 1] - acc[x2 + 1][y] + acc[x][y]
# print(x, y, k, zero, k * k - zero)
res += min(zero, k * k - zero)
# print(k, res)
ans = min(ans, res)
print(ans)
def main():
n, m = [int(x) for x in stdin.readline().split()]
pixels = []
for i in range(n):
pixels.append(stdin.readline().strip())
run(n, m, pixels)
if __name__ == '__main__':
import os
if os.path.exists('tmp.in'):
stdin = open('tmp.in')
main()
``` | output | 1 | 40,110 | 7 | 80,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an image, that can be represented with a 2-d n by m grid of pixels. Each pixel of the image is either on or off, denoted by the characters "0" or "1", respectively. You would like to compress this image. You want to choose an integer k > 1 and split the image into k by k blocks. If n and m are not divisible by k, the image is padded with only zeros on the right and bottom so that they are divisible by k. Each pixel in each individual block must have the same value. The given image may not be compressible in its current state. Find the minimum number of pixels you need to toggle (after padding) in order for the image to be compressible for some k. More specifically, the steps are to first choose k, then the image is padded with zeros, then, we can toggle the pixels so it is compressible for this k. The image must be compressible in that state.
Input
The first line of input will contain two integers n, m (2 ≤ n, m ≤ 2 500), the dimensions of the image.
The next n lines of input will contain a binary string with exactly m characters, representing the image.
Output
Print a single integer, the minimum number of pixels needed to toggle to make the image compressible.
Example
Input
3 5
00100
10110
11001
Output
5
Note
We first choose k = 2.
The image is padded as follows:
001000
101100
110010
000000
We can toggle the image to look as follows:
001100
001100
000000
000000
We can see that this image is compressible for k = 2. | instruction | 0 | 40,111 | 7 | 80,222 |
Tags: brute force
Correct Solution:
```
from sys import stdin as fin
# fin = open("ih2e2017a.in", "r")
# fout = open("ih2e2017a.in", "a")
# for i in range(2500):
# for j in range(2500):
# print('1' if (i % 2) == (j % 2) else '0', end='', file=fout)
# print(file=fout)
def check(x, y):
if x < 0 or y < 0:
return 0
else:
return d[x][y]
def check2(x, y):
if x < 0 or y < 0:
return 0
else:
return d[min(x, n - 1)][min(y, m - 1)]
def border(a, b):
d, m = divmod(a, b)
return b * (d + (m != 0))
# print('lol')
n, m = map(int, fin.readline().split())
# field = tuple(fin.readline().strip() for i in range(n))
field = fin.readlines()
# print("inp")
# print(len(field), len(field[0]))
d = [[None] * m for i in range(n)]
d[0][0] = int(field[0][0]) #field[0][0]
for i in range(n):
for j in range(m):
d[i][j] = check(i - 1, j) + check(i, j - 1) - check(i - 1, j - 1) + int(field[i][j])# field[i][j]
minv = float('inf')
for k in range(2, max(n, m)):
cv = 0
# print(k)
for i in range(k - 1, border(n, k), k):
for j in range(k - 1, border(m, k), k):
ccnt = check2(i, j) - check2(i - k, j) - check2(i, j - k) + check2(i - k, j - k)
cv += min(ccnt, k**2 - ccnt)
minv = min(minv, cv)
print(minv)
fin.close()
``` | output | 1 | 40,111 | 7 | 80,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an image, that can be represented with a 2-d n by m grid of pixels. Each pixel of the image is either on or off, denoted by the characters "0" or "1", respectively. You would like to compress this image. You want to choose an integer k > 1 and split the image into k by k blocks. If n and m are not divisible by k, the image is padded with only zeros on the right and bottom so that they are divisible by k. Each pixel in each individual block must have the same value. The given image may not be compressible in its current state. Find the minimum number of pixels you need to toggle (after padding) in order for the image to be compressible for some k. More specifically, the steps are to first choose k, then the image is padded with zeros, then, we can toggle the pixels so it is compressible for this k. The image must be compressible in that state.
Input
The first line of input will contain two integers n, m (2 ≤ n, m ≤ 2 500), the dimensions of the image.
The next n lines of input will contain a binary string with exactly m characters, representing the image.
Output
Print a single integer, the minimum number of pixels needed to toggle to make the image compressible.
Example
Input
3 5
00100
10110
11001
Output
5
Note
We first choose k = 2.
The image is padded as follows:
001000
101100
110010
000000
We can toggle the image to look as follows:
001100
001100
000000
000000
We can see that this image is compressible for k = 2.
Submitted Solution:
```
import itertools
n, m = [int(x) for x in input().split()]
table = []
for _ in range(n):
table.append([int(x) for x in list(input())])
amount = {}
for i in range(2, max(n, m)+1):
amount[i] = 0
for outter_j, outter_k in list(itertools.product(range(n//i), range(m//i))):
aofi = 0
aofz = 0
for inner_j, inner_k in list(itertools.product(range(i), range(i))):
if table[i*outter_j + inner_j][i*outter_k + inner_k] == 0:
aofz += 1
else:
aofi += 1
# print('to {} in ({}, {}) sector + {}'.format(i, i*outter_j, i*outter_k, min(aofi, aofz)))
amount[i] += min(aofi, aofz)
if n%i != 0:
for outter_k in range(m//i):
aofi = 0
aofz = 0
for inner_j, inner_k in list(itertools.product(range(n - i*(n//i)), range(i))):
if table[i*(n//i) + inner_j][i*outter_k + inner_k] == 0:
aofz += 1
else:
aofi += 1
aofz += i**2 - (n - i*(n//i))*i
# print('to {} in ({}, {}) sector + {}'.format(i, i*(n//i), i*outter_k, min(aofi, aofz)))
amount[i] += min(aofi, aofz)
if m%i != 0:
for outter_j in range(n//i):
aofi = 0
aofz = 0
for inner_j, inner_k in list(itertools.product(range(i), range(m - i*(m//i)))):
if table[i*outter_j + inner_j][i*(m//i) + inner_k] == 0:
aofz += 1
else:
aofi += 1
aofz += i**2 - i*(m - i*(m//i))
# print('to {} in ({}, {}) sector + {}'.format(i, i*outter_j, i*(m//i), min(aofi, aofz)))
amount[i] += min(aofi, aofz)
if n%i != 0 and m%i != 0:
aoif = 0
aofz = 0
for inner_j, inner_k in list(itertools.product(range(n - i*(n//i)), range(m - i*(m//i)))):
if table[i*(n//i) + inner_j][i*(m//i) + inner_k] == 0:
aofz += 1
else:
aofi += 1
aofz += i**2 - (n - i*(n//i))*(m - i*(m//i))
# print('to {} in outer sector + {}'.format(i, min(aofi, aofz)))
amount[i] += min(aofi, aofz)
print(min(amount.values()))
``` | instruction | 0 | 40,116 | 7 | 80,232 |
No | output | 1 | 40,116 | 7 | 80,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image> | instruction | 0 | 40,511 | 7 | 81,022 |
Tags: dp, greedy, math
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
dp = [0 for _ in range(2)]
pre = -1
for i in range(n):
if i%2 == 0:
dp[0] += a[i]//2
dp[1] += a[i] - a[i]//2
else:
dp[0] += a[i] - a[i]//2
dp[1] += a[i]//2
print(min(dp[0], dp[1]))
``` | output | 1 | 40,511 | 7 | 81,023 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image> | instruction | 0 | 40,512 | 7 | 81,024 |
Tags: dp, greedy, math
Correct Solution:
```
n = int(input())
A = list(map(int, input().split()))
x, y = 0, 0
for a in A:
x += a // 2
y += (a + 1) // 2
x, y = y, x
print(min(x, y))
``` | output | 1 | 40,512 | 7 | 81,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image> | instruction | 0 | 40,513 | 7 | 81,026 |
Tags: dp, greedy, math
Correct Solution:
```
n = int(input())
l = list(map(int, input().split()))
r = 0
b = 0
for i in range(len(l)):
if l[i]%2 == 1:
if i%2 == 1:
r += int(l[i]/2) + 1
b += int(l[i]/2)
else:
r += int(l[i]/2)
b += int(l[i]/2) + 1
else:
r += int(l[i]/2)
b += int(l[i]/2)
print(min(r, b))
``` | output | 1 | 40,513 | 7 | 81,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image> | instruction | 0 | 40,514 | 7 | 81,028 |
Tags: dp, greedy, math
Correct Solution:
```
N = int(input())
A = list(map(int,input().split()))
num_white = 0
num_black = 0
for i in range(N):
if i % 2 == 0:
num_white += (A[i]+1)//2
num_black += (A[i]-(A[i]+1)//2)
else:
num_black += (A[i]+1)//2
num_white += (A[i]-(A[i]+1)//2)
print(min(num_white,num_black))
``` | output | 1 | 40,514 | 7 | 81,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image> | instruction | 0 | 40,515 | 7 | 81,030 |
Tags: dp, greedy, math
Correct Solution:
```
# f=open('asdf.txt','r')
# n=int(f.readline())
# array=list(map(int,f.readline().strip().split()))
# from time import *
# import sys
# s = time()
n = int(input())
array = list(map(int, input().split()))
c1, c2 = 0, 0
for i in range(n):
a = array[i]
if a % 2 == 0:
c1 += a // 2
c2 += a // 2
else:
c1 += a // 2
c2 += a // 2
if i % 2 == 0:
c1 += 1
else:
c2 += 1
print (min(c1, c2))
# print (time() - s, file=sys.stderr)
``` | output | 1 | 40,515 | 7 | 81,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image> | instruction | 0 | 40,516 | 7 | 81,032 |
Tags: dp, greedy, math
Correct Solution:
```
# n, m = map(int, input().split())
# a = list(map(int, input().split()))
# b = list(map(int, input().split()))
# d_a = {}
# for i in a:
# if (i%m) not in d_a.keys():
# d_a[i % m] = 1
# else:
# d_a[i % m] += 1
# d_b = {}
# for i in b:
# if (i%m) not in d_b.keys():
# d_b[i % m] = 1
# else:
# d_b[i % m] += 1
# res_a = list(d_a.items())
# res_b = list(d_b.items())
# res_a = sorted(res_a, key=lambda tup: tup[1], reverse=True)
# res_b = sorted(res_b, key=lambda tup: tup[1], reverse=True)
# print(abs(res_a[0][0] - res_b[0][0]))
#
# n, k = map(int, input().split())
# x = input()
# s = x[:k] * n
# s = s[:n]
# if int(s) >= int(x):
# print(n)
# print(s)
# else:
# ans = str(int(x[:k]) + 1) * (n // k + 1)
# print(n)
# print(ans[:n])
n = int(input())
a = list(map(int, input().split()))
k = 1
white = 0
black = 0
for i in a:
black += (i // 2) + k * (i % 2)
white += (i // 2) + (1 - k) * (i % 2)
k = 1-k
print(min(black, white))
# n,k,t = map(float, input().split())
# x = [0 for i in range(int(n))]
# y = [0 for i in range(int(n))]
# c = [0 for i in range(int(n))]
# dist = []
# for i in range(int(n)):
# a, b, cl = map(float, input().split())
# x[i] = a
# y[i] = b
# c[i] = int(cl)
# for i in range(int(t)):
# dist = []
# clst = [0 for j in range(11)]
# mx = 0
# ans = 0
# a, b = map(float, input().split())
# for j in range(int(n)):
# dist.append([((x[j] - a) ** 2 + (y[j] - b) ** 2) ** (1/2), c[j]])
# dist = sorted(dist, key=lambda tup: tup[0])
# for j in range(int(k)):
# clst[dist[j][1]] += 1
# for j in range(1, len(clst)):
# if clst[j] > mx:
# mx = clst[j]
# ans = j
# print(ans)
#
#
#
# print(8000**5)
# s = input()
# k = 0
# for i in range(len(s) // 2):
# if s[i] != s[-i-1]:
# k = k + 1
# if k == 0:
# print(0)
# else:
# print(k-1)
#
# def check_one(s):
# for j in range((len(s) + 1) // 2, len(s)):
# if s[0:j] == s[len(s) - j:len(s)+1]:
# return 1
# return 0
#
#
# st = input()
# k = 0
# for i in range(2, len(st) + 1):
# k += check_one(st[0:i])
# print(k)
``` | output | 1 | 40,516 | 7 | 81,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image> | instruction | 0 | 40,517 | 7 | 81,034 |
Tags: dp, greedy, math
Correct Solution:
```
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
n = int(input())
a = [int(item) for item in input().split()] + [0]
black = 0; white = 0
for i, item in enumerate(a):
if item % 2 == 0:
black += item // 2
white += item // 2
else:
if i % 2 == 0:
black += (item + 1) // 2
white += item // 2
else:
white += (item + 1) // 2
black += item // 2
print(min(black, white))
``` | output | 1 | 40,517 | 7 | 81,035 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image> | instruction | 0 | 40,518 | 7 | 81,036 |
Tags: dp, greedy, math
Correct Solution:
```
###############################
# https://codeforces.com/contest/1269/problem/D
# 2021/01/13
# WenhuZhang
################################
from sys import stdin
import collections
import copy
n= int(stdin.readline())
a = list(map(int, stdin.readline().split()))
ans =0
left=[]
for i in range(len(a)):
# print(ans, left)
ans += a[i]//2
if a[i]%2 :
if not len(left):
left.append(i)
elif (i-left[-1])%2:
# print(i,a)
left.pop()
ans+=1
else:
left.append(i)
print(ans)
``` | output | 1 | 40,518 | 7 | 81,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).
<image>
Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.
Input
The first line of the input contains a single number n (1 ≤ n ≤ 26) — the number of colors of beads. The second line contains after n positive integers ai — the quantity of beads of i-th color. It is guaranteed that the sum of ai is at least 2 and does not exceed 100 000.
Output
In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.
Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.
Examples
Input
3
4 2 1
Output
1
abacaba
Input
1
4
Output
4
aaaa
Input
2
1 1
Output
0
ab
Note
In the first sample a necklace can have at most one beautiful cut. The example of such a necklace is shown on the picture.
In the second sample there is only one way to compose a necklace. | instruction | 0 | 40,812 | 7 | 81,624 |
Tags: constructive algorithms, math
Correct Solution:
```
#!/usr/bin/python3
import sys, functools, fractions
def make_simple(ball_per_period, letters):
output = []
for letter, count in zip(letters, ball_per_period):
output.append(letter * count)
return ''.join(output)
def no_solution(balls, letters):
print(0)
print(make_simple(balls, letters))
sys.exit(0)
def make_palindrom(ball_per_period, letters):
palindrom = [None] * sum(ball_per_period)
c = 0
for index, b in enumerate(ball_per_period):
if b % 2 == 1:
if len(palindrom) % 2 == 0:
return None
center = len(palindrom) // 2
if palindrom[center] is not None: # already filled
return None
palindrom[center] = letters[index]
for _ in range(b // 2):
palindrom[c] = letters[index]
palindrom[len(palindrom) - c - 1] = letters[index]
c += 1
return ''.join(palindrom)
def main():
n = int(input())
balls = [int(i) for i in input().split()]
letters = [chr(i + ord('a')) for i in range(26)]
gcd = functools.reduce(fractions.gcd, balls)
period = sum(balls) // gcd
ball_per_period = [i // gcd for i in balls]
palindrom = [0] * period
odd_count = 0
odd_index = None
for index, b in enumerate(ball_per_period):
if b % 2 == 1:
odd_count += 1
odd_index = index
if gcd % 2 == 0:
print(gcd)
simple = make_simple(ball_per_period, letters)
simple_rev = simple[::-1]
for i in range(0, gcd, 2):
print(simple, end='')
print(simple_rev, end='')
print()
sys.exit(0)
if odd_count <= 1:
print(gcd)
palindrom = make_palindrom(ball_per_period, letters)
print(''.join(palindrom) * gcd)
else:
no_solution(balls, letters)
main()
``` | output | 1 | 40,812 | 7 | 81,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).
<image>
Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.
Input
The first line of the input contains a single number n (1 ≤ n ≤ 26) — the number of colors of beads. The second line contains after n positive integers ai — the quantity of beads of i-th color. It is guaranteed that the sum of ai is at least 2 and does not exceed 100 000.
Output
In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.
Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.
Examples
Input
3
4 2 1
Output
1
abacaba
Input
1
4
Output
4
aaaa
Input
2
1 1
Output
0
ab
Note
In the first sample a necklace can have at most one beautiful cut. The example of such a necklace is shown on the picture.
In the second sample there is only one way to compose a necklace. | instruction | 0 | 40,813 | 7 | 81,626 |
Tags: constructive algorithms, math
Correct Solution:
```
from fractions import gcd
from functools import reduce
LETTERS = 'abcdefghijklmnopqrstuvwxyz'
def necklace_odd(a):
oi = next(i for i, ai in enumerate(a) if ai%2)
o = a[oi]
g = reduce(gcd, a)
s = [LETTERS[i] * (a[i]//(2*g)) for i in range(len(a)) if i != oi]
return g, (''.join(s) + (LETTERS[oi]*(o//g)) + ''.join(reversed(s))) * g
def necklace_even(a):
g = reduce(gcd, a)//2
s = [LETTERS[i]*(a[i]//(2*g)) for i in range(len(a))]
return 2*g, (''.join(s) + ''.join(reversed(s))) * g
def necklace(a):
if len(a) == 1:
return a[0], LETTERS[0]*a[0]
nodd = sum(ai%2 for ai in a)
if nodd > 1:
return 0, ''.join(LETTERS[i]*a[i] for i in range(len(a)))
return (necklace_odd if nodd else necklace_even)(a)
if __name__ == '__main__':
n = int(input())
a = list(map(int, input().split()))
assert len(a) == n
k, e = necklace(a)
print(k)
print(e)
# Made By Mostafa_Khaled
``` | output | 1 | 40,813 | 7 | 81,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).
<image>
Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.
Input
The first line of the input contains a single number n (1 ≤ n ≤ 26) — the number of colors of beads. The second line contains after n positive integers ai — the quantity of beads of i-th color. It is guaranteed that the sum of ai is at least 2 and does not exceed 100 000.
Output
In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.
Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.
Examples
Input
3
4 2 1
Output
1
abacaba
Input
1
4
Output
4
aaaa
Input
2
1 1
Output
0
ab
Note
In the first sample a necklace can have at most one beautiful cut. The example of such a necklace is shown on the picture.
In the second sample there is only one way to compose a necklace. | instruction | 0 | 40,814 | 7 | 81,628 |
Tags: constructive algorithms, math
Correct Solution:
```
from fractions import gcd
from functools import reduce
LETTERS = 'abcdefghijklmnopqrstuvwxyz'
def necklace_odd(a):
oi = next(i for i, ai in enumerate(a) if ai%2)
o = a[oi]
g = reduce(gcd, a)
s = [LETTERS[i] * (a[i]//(2*g)) for i in range(len(a)) if i != oi]
return g, (''.join(s) + (LETTERS[oi]*(o//g)) + ''.join(reversed(s))) * g
def necklace_even(a):
g = reduce(gcd, a)//2
s = [LETTERS[i]*(a[i]//(2*g)) for i in range(len(a))]
return 2*g, (''.join(s) + ''.join(reversed(s))) * g
def necklace(a):
if len(a) == 1:
return a[0], LETTERS[0]*a[0]
nodd = sum(ai%2 for ai in a)
if nodd > 1:
return 0, ''.join(LETTERS[i]*a[i] for i in range(len(a)))
return (necklace_odd if nodd else necklace_even)(a)
if __name__ == '__main__':
n = int(input())
a = list(map(int, input().split()))
assert len(a) == n
k, e = necklace(a)
print(k)
print(e)
``` | output | 1 | 40,814 | 7 | 81,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).
<image>
Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.
Input
The first line of the input contains a single number n (1 ≤ n ≤ 26) — the number of colors of beads. The second line contains after n positive integers ai — the quantity of beads of i-th color. It is guaranteed that the sum of ai is at least 2 and does not exceed 100 000.
Output
In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.
Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.
Examples
Input
3
4 2 1
Output
1
abacaba
Input
1
4
Output
4
aaaa
Input
2
1 1
Output
0
ab
Note
In the first sample a necklace can have at most one beautiful cut. The example of such a necklace is shown on the picture.
In the second sample there is only one way to compose a necklace. | instruction | 0 | 40,815 | 7 | 81,630 |
Tags: constructive algorithms, math
Correct Solution:
```
import math
#import fractions
from functools import reduce
n = int(input())
odd = -1
beads = [int(x) for x in input().split()]
for i in range(n):
if beads[i]%2:
if odd >= 0:
print(0)
print(''.join(chr(ord('a') + i)*beads[i] for i in range(n)))
break
else:
odd = i
else:
gcd = reduce(lambda x,y: math.gcd(x,y), beads)
print(gcd)
if odd >= 0:
s = ''.join(chr(ord('a') + i)*(beads[i]//(2*gcd)) for i in range(n) if i != odd)
p = s + chr(ord('a') + odd)*(beads[odd]//gcd) + s[::-1]
print(p*gcd)
else:
s = ''.join(chr(ord('a') + i)*(beads[i]//gcd) for i in range(n))
p = s + s[::-1]
print(p*(gcd//2))
``` | output | 1 | 40,815 | 7 | 81,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).
<image>
Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.
Input
The first line of the input contains a single number n (1 ≤ n ≤ 26) — the number of colors of beads. The second line contains after n positive integers ai — the quantity of beads of i-th color. It is guaranteed that the sum of ai is at least 2 and does not exceed 100 000.
Output
In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.
Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.
Examples
Input
3
4 2 1
Output
1
abacaba
Input
1
4
Output
4
aaaa
Input
2
1 1
Output
0
ab
Note
In the first sample a necklace can have at most one beautiful cut. The example of such a necklace is shown on the picture.
In the second sample there is only one way to compose a necklace.
Submitted Solution:
```
import sys, functools, fractions
n = int(input())
balls = [int(i) for i in input().split()]
letters = [chr(i + ord('a')) for i in range(26)]
def no_solution():
print(0)
output = []
for letter, count in zip(letters, balls):
output.append(letter * count)
print(''.join(output))
sys.exit(0)
gcd = functools.reduce(fractions.gcd, balls)
period = sum(balls) // gcd
ball_per_period = [i // gcd for i in balls]
palindrom = [0] * period
odd_count = 0
odd_index = None
for index, b in enumerate(ball_per_period):
if b % 2 == 1:
odd_count += 1
odd_index = index
if odd_count > 1:
no_solution()
if odd_count > 0 and len(palindrom) % 2 != 1:
no_solution()
if not(odd_index is None):
palindrom[len(palindrom) // 2] = letters[odd_index]
ball_per_period[index] -= 1
c = 0
for index, b in enumerate(ball_per_period):
for i in range(b // 2):
palindrom[c] = letters[index]
palindrom[len(palindrom) - c - 1] = letters[index]
c += 1
print(gcd)
palindrom = ''.join(palindrom)
for i in range(gcd):
print(palindrom, end='')
print()
``` | instruction | 0 | 40,816 | 7 | 81,632 |
No | output | 1 | 40,816 | 7 | 81,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).
<image>
Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.
Input
The first line of the input contains a single number n (1 ≤ n ≤ 26) — the number of colors of beads. The second line contains after n positive integers ai — the quantity of beads of i-th color. It is guaranteed that the sum of ai is at least 2 and does not exceed 100 000.
Output
In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.
Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.
Examples
Input
3
4 2 1
Output
1
abacaba
Input
1
4
Output
4
aaaa
Input
2
1 1
Output
0
ab
Note
In the first sample a necklace can have at most one beautiful cut. The example of such a necklace is shown on the picture.
In the second sample there is only one way to compose a necklace.
Submitted Solution:
```
import math
from functools import reduce
n = int(input())
odd = -1
beads = [int(x) for x in input().split()]
for i in range(n):
if beads[i]%2:
if odd >= 0:
print(0)
print(''.join(chr(ord('a') + i)*beads[i] for i in range(n)))
break
else:
odd = i
else:
if odd >= 0:
print(1)
s = ''.join(chr(ord('a') + i)*(beads[i]//2) for i in range(n) if i != odd)
print(s + chr(ord('a') + odd) + s[::-1])
else:
gcd = reduce(lambda x,y: math.gcd(x,y), beads)
print(gcd)
s = ''.join(chr(ord('a') + i)*(beads[i]//gcd) for i in range(n))
print((s+s[::-1])*(gcd//2))
``` | instruction | 0 | 40,817 | 7 | 81,634 |
No | output | 1 | 40,817 | 7 | 81,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).
<image>
Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.
Input
The first line of the input contains a single number n (1 ≤ n ≤ 26) — the number of colors of beads. The second line contains after n positive integers ai — the quantity of beads of i-th color. It is guaranteed that the sum of ai is at least 2 and does not exceed 100 000.
Output
In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.
Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.
Examples
Input
3
4 2 1
Output
1
abacaba
Input
1
4
Output
4
aaaa
Input
2
1 1
Output
0
ab
Note
In the first sample a necklace can have at most one beautiful cut. The example of such a necklace is shown on the picture.
In the second sample there is only one way to compose a necklace.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from sys import stdin
from collections import defaultdict
from operator import itemgetter
def main():
n = int(stdin.readline())
xs = list(map(int, stdin.readline().split()))
run(xs)
def run(xs):
odd = 0
m = -1
for x in xs:
if x % 2 == 1:
odd += 1
else:
m = min(m, x) if m > -1 else x
if odd > 1:
print(0)
return ppAny(xs)
if odd == 1:
print(1)
return ppOdd(xs)
print(m)
ppEven(xs)
def ppAny(xs):
s = ""
y = ord('a') - 1
for x in xs:
y += 1
s += chr(y) * x
print(s)
def ppOdd(xs):
l, r, m = "", "", ""
y = ord('a') - 1
for x in xs:
y += 1
c = chr(y)
if x % 2 == 0:
l = c*(x//2) + l
r = r + c*(x//2)
else:
m = c * x
print(l + m + r)
def ppEven(xs):
n = sum(xs)
l = len(xs)
s = ""
while n > 0:
y = ord('a') - 1
for i in range(l):
y += 1
if xs[i] > 0:
xs[i] -= 2
n -= 2
s += chr(y) * 2
print(s)
if __name__=="__main__":
main()
``` | instruction | 0 | 40,818 | 7 | 81,636 |
No | output | 1 | 40,818 | 7 | 81,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).
<image>
Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.
Input
The first line of the input contains a single number n (1 ≤ n ≤ 26) — the number of colors of beads. The second line contains after n positive integers ai — the quantity of beads of i-th color. It is guaranteed that the sum of ai is at least 2 and does not exceed 100 000.
Output
In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.
Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.
Examples
Input
3
4 2 1
Output
1
abacaba
Input
1
4
Output
4
aaaa
Input
2
1 1
Output
0
ab
Note
In the first sample a necklace can have at most one beautiful cut. The example of such a necklace is shown on the picture.
In the second sample there is only one way to compose a necklace.
Submitted Solution:
```
from fractions import gcd
from functools import reduce
LETTERS = 'abcdefghijklmnopqrstuvwxyz'
def necklace_odd(a):
oi = next(i for i, ai in enumerate(a) if ai%2)
o = a[oi]
if all(ai % o == 0 for ai in a):
s = [LETTERS[i]*(a[i]//(2*o)) for i in range(len(a)) if i != oi]
return o, (''.join(s) + LETTERS[oi] + ''.join(reversed(s))) * o
else:
s = [LETTERS[i]*(a[i]//2) for i in range(len(a)) if i != oi]
return 1, ''.join(s) + LETTERS[oi]*a[oi] + ''.join(reversed(s))
def necklace_even(a):
g = reduce(gcd, a)//2
s = [LETTERS[i]*(a[i]//(2*g)) for i in range(len(a))]
return 2*g, (''.join(s) + ''.join(reversed(s))) * g
def necklace(a):
if len(a) == 1:
return a[0], LETTERS[0]*a[0]
nodd = sum(ai%2 for ai in a)
if nodd > 1:
return 0, ''.join(LETTERS[i]*a[i] for i in range(len(a)))
return (necklace_odd if nodd else necklace_even)(a)
if __name__ == '__main__':
n = int(input())
a = list(map(int, input().split()))
assert len(a) == n
k, e = necklace(a)
print(k)
print(e)
``` | instruction | 0 | 40,819 | 7 | 81,638 |
No | output | 1 | 40,819 | 7 | 81,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1 | instruction | 0 | 40,880 | 7 | 81,760 |
Tags: brute force, constructive algorithms
Correct Solution:
```
m=[input() for i in range(8)]
r=sum([1 for i in range(8) if m[i]=='B'*8 ])
c=sum([1 for i in range(8) if sum([1 for k in range(8) if m[k][i]=='B'])==8])
if max(r,c)==8:print(8)
else:print(r+c)
``` | output | 1 | 40,880 | 7 | 81,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1 | instruction | 0 | 40,881 | 7 | 81,762 |
Tags: brute force, constructive algorithms
Correct Solution:
```
board = [[0]*8]*8
for i in range(8):
aux = input()
for j in range(8):
board[i] = aux
count = 0
for i in board:
if i.count('B') == 8:
count += 1
if count == 8:
print(8)
exit(0)
for i in range(8):
aux_count = 0
for j in range(8):
if board[j][i] == 'B':
aux_count += 1
if aux_count == 8:
count += 1
print(count)
# 1512827419058
``` | output | 1 | 40,881 | 7 | 81,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1 | instruction | 0 | 40,882 | 7 | 81,764 |
Tags: brute force, constructive algorithms
Correct Solution:
```
brd = [input() for i in range(8)]
ans = 0
for i in range(8):
if all([brd[i][j] == "B" for j in range(8)]):
ans += 1
if all([brd[j][i] == "B" for j in range(8)]):
ans += 1
if ans == 16:
ans = 8
print(ans)
``` | output | 1 | 40,882 | 7 | 81,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1 | instruction | 0 | 40,883 | 7 | 81,766 |
Tags: brute force, constructive algorithms
Correct Solution:
```
m1 = []
m2 = ["" for i in range(8)]
for i in range(8):
row = input()
m1.append(row)
for j in range(8):
m2[j] += row[j]
cnt1 = m1.count("B" * 8)
cnt2 = m2.count("B" * 8)
if cnt1 == 8:
print(8)
else:
print(cnt1 + cnt2)
``` | output | 1 | 40,883 | 7 | 81,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1 | instruction | 0 | 40,884 | 7 | 81,768 |
Tags: brute force, constructive algorithms
Correct Solution:
```
def rr(): return input().rstrip()
def rri(): return int(rr())
def rrl(): return list(map(int, rr().split()))
def rrt(): return tuple(map(int, rr().split()))
from collections import defaultdict
def mus(d=lambda: 0): return defaultdict(lambda: defaultdict(d))
def dd0(d=lambda: 0): return defaultdict(d)
def ms(x, y, d=0): return [[0]*y for i in range(x)]
def ar(x, d=0): return [d]*x
def ppm(m, n=0, x=0, y=0): print("\n".join(("\t".join((str(m[j][i]) for j in range(y or n))) for i in range(x or n))))
def ppa(a, n): print("\t".join(map(str, a[0:n])))
def ppl(a=0, x=[1]): x[0]=a or (x[0]+1); print("-"+"- -"*14+"-# "+str(x[0]-2)+" #-"+"- -"*14+"-")
def fltn(a): (i for sub in a for i in sub)
def yn(x): return "YES" if x else "NO"
from itertools import accumulate
from itertools import groupby
def grp(x): return ((i, sum(1 for _ in g)) for i, g in groupby(x))
import math
def rnar(): return (*rrl(), rrl())
def rn(): return (*rrl(),)
def dpp(*args): print(*args)
def read():
l = [rr() for _ in range(8)]
return (l,)
def solve(l):
vert = 99
for row in l:
vert = min(vert, row.count("B"))
horiz = 99
for x in range(8):
cur_b = 0
for y in range(8):
if l[y][x] == "B":
cur_b += 1
horiz = min(horiz, cur_b)
if (vert == 8): horiz = 0
ans = vert + horiz
return ans
if __name__ == "__main__":
#test_count = rri()
test_count = 1
for _ in range(test_count):
input_data = read()
result = solve(*input_data)
print(result)
``` | output | 1 | 40,884 | 7 | 81,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1 | instruction | 0 | 40,885 | 7 | 81,770 |
Tags: brute force, constructive algorithms
Correct Solution:
```
s = []
for i in range(8):
s.append(input())
num = 0
for i in s:
if i == 'BBBBBBBB':
num = num + 1
else:
w = i
if num == 8:
print(8)
else:
for i in w:
if i == 'B':
num = num + 1
print(num)
``` | output | 1 | 40,885 | 7 | 81,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1 | instruction | 0 | 40,886 | 7 | 81,772 |
Tags: brute force, constructive algorithms
Correct Solution:
```
# 7A Kalevitch ans Chess
# original problem: http://codeforces.com/problemset/problem/7/A
def k_chess():
board = ''.join([input().strip() for i in range(8)])
count = 0
# columns
for i in range(8):
for j in range(0,57,8):
if board[i+j] == 'W': break
else: count +=1
if count == 8: print(count); return
# rows
for i in range(0,57,8):
for j in range(8):
if board[i+j] == 'W': break
else: count +=1
print(count); return
k_chess()
``` | output | 1 | 40,886 | 7 | 81,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1 | instruction | 0 | 40,887 | 7 | 81,774 |
Tags: brute force, constructive algorithms
Correct Solution:
```
import cmath
d={}
c=0;ans=0;
for i in range (1,9) :
x=list("0"+input());
if "W" in x:
c=1
else :
ans=ans+1
for j in range (1,9) :
z=complex(i,j)
d[z]=x[j]
if (c==0):
exit (print("8"))
for i in range (1,9):
w=0
for j in range (1,9):
if d[complex(j,i)]=="B" :
w=w+1
if (8==w):
ans=ans+1
print(ans)
``` | output | 1 | 40,887 | 7 | 81,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1
Submitted Solution:
```
b=0
a=0
for i in range(8):
k = input()
n = k.count("B")
if n<8:
a = n
else:
b+=1
print(a+b)
``` | instruction | 0 | 40,888 | 7 | 81,776 |
Yes | output | 1 | 40,888 | 7 | 81,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1
Submitted Solution:
```
l = []
m = []
a = ""
for j in range(0,8):
a = input()
m.append(a)
a = ""
pr = 0
tot = 0
for i in range(0,8):
if m[i].count('B') == 8:
tot += 1
for i in range(0,8):
for j in range(0,8):
if m[j][i] == 'B':
pr += 1
if pr == 8:
tot += 1
pr = 0
if tot == 16:
print(8)
else:
print(tot)
# 1512819581428
``` | instruction | 0 | 40,889 | 7 | 81,778 |
Yes | output | 1 | 40,889 | 7 | 81,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1
Submitted Solution:
```
a = []
for i in range(0, 8):
a.append(input())
t1 = 0
t2 = 0
for i in range(0, 8):
if a[i].count('B') == 8:
t1 += 1
else:
t2 = a[i].count('B')
print(t1+t2)
``` | instruction | 0 | 40,890 | 7 | 81,780 |
Yes | output | 1 | 40,890 | 7 | 81,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1
Submitted Solution:
```
a,s=[],0
for i in range(8):
b=list(input())
if b.count('W')==0:s=s+1
a.append(b)
if s==8:print(s)
else:
k=a[0].count('B')
for i in range(len(a)):
if a[i].count('B')<k:k=a[i].count('B')
print(s+k)
``` | instruction | 0 | 40,891 | 7 | 81,782 |
Yes | output | 1 | 40,891 | 7 | 81,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1
Submitted Solution:
```
#!/usr/bin/env python3
grid = [input() for _ in range(8)]
print(min(15, grid[0].count('B') + [x[0] for x in grid].count('B')))
``` | instruction | 0 | 40,892 | 7 | 81,784 |
No | output | 1 | 40,892 | 7 | 81,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1
Submitted Solution:
```
board,columns,rows = [],[],[]
for i in range(8):
line = []
for char in input():
line.append(True) if char == "B" else line.append(False);
board.append(line)
if False not in line: rows.append(i)
#columns
for i in range(8):
white = False
for j in range(8):
if not board[j][i]: white = True
if not white:
columns.append(i)
#the bruteforce
brute = [[False for i in range(8)] for j in range(8)]
strikes = 0
for line in rows:
strikes +=1
brute[line] = [True for i in range(8)]
if brute == board:
print(strikes)
break;
if brute != board:
for column in columns:
strikes+=1
for row in brute:
row[column] = True
if brute == board:
print(strikes)
break;
``` | instruction | 0 | 40,893 | 7 | 81,786 |
No | output | 1 | 40,893 | 7 | 81,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1
Submitted Solution:
```
def main():
hs = []
t, f = 0, 0
for i in range(8):
hs.append( input().split() )
for j in range(8):
check = 0
trak = 0
for jj in range(8):
if hs[j][0][jj] == "W": check+=1
elif hs[j][0][jj] == "B": trak+=1
if hs[jj][0][j] == "W": check+=1
elif hs[jj][0][j] == "B": trak+=1
if trak >=7:t+=1
if check >=7:f+=1
print( min(f, t) )
if __name__ == "__main__": main()
``` | instruction | 0 | 40,894 | 7 | 81,788 |
No | output | 1 | 40,894 | 7 | 81,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
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
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.
Examples
Input
WWWBWWBW
BBBBBBBB
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
WWWBWWBW
Output
3
Input
WWWWWWWW
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
WWWWWWWW
Output
1
Submitted Solution:
```
data = [input() for _ in range(8)]
ans = sum([1 for i in range(8) if all([1 if j == "B" else 0 for j in data[i]])])
ans += sum([1 for i in range(8) if all(1 if k == "B" else 0 for k in [data[j][i] for j in range(8)])])
print(ans)
``` | instruction | 0 | 40,895 | 7 | 81,790 |
No | output | 1 | 40,895 | 7 | 81,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Luba bought a monitor. Monitor is a rectangular matrix of size n × m. But then she started to notice that some pixels cease to work properly. Luba thinks that the monitor will become broken the first moment when it contains a square k × k consisting entirely of broken pixels. She knows that q pixels are already broken, and for each of them she knows the moment when it stopped working. Help Luba to determine when the monitor became broken (or tell that it's still not broken even after all q pixels stopped working).
Input
The first line contains four integer numbers n, m, k, q (1 ≤ n, m ≤ 500, 1 ≤ k ≤ min(n, m), 0 ≤ q ≤ n·m) — the length and width of the monitor, the size of a rectangle such that the monitor is broken if there is a broken rectangle with this size, and the number of broken pixels.
Each of next q lines contain three integer numbers xi, yi, ti (1 ≤ xi ≤ n, 1 ≤ yi ≤ m, 0 ≤ t ≤ 109) — coordinates of i-th broken pixel (its row and column in matrix) and the moment it stopped working. Each pixel is listed at most once.
We consider that pixel is already broken at moment ti.
Output
Print one number — the minimum moment the monitor became broken, or "-1" if it's still not broken after these q pixels stopped working.
Examples
Input
2 3 2 5
2 1 8
2 2 8
1 2 1
1 3 4
2 3 2
Output
8
Input
3 3 2 5
1 2 2
2 2 1
2 3 5
3 2 10
2 1 100
Output
-1 | instruction | 0 | 40,903 | 7 | 81,806 |
Tags: binary search, data structures
Correct Solution:
```
from sys import stdout
from sys import stdin
def get():
return stdin.readline().strip()
def getf():
return [int(i) for i in get().split()]
def put(a, end = "\n"):
stdout.write(str(a) + end)
def putf(a, sep = " ", end = "\n"):
stdout.write(sep.join(map(str, a)) + end)
def isbroken(n, m, mt, k, qu, q):
a = [[1]*m for i in range(n)]
for i in range(q):
x, y, t = qu[i]
if(t <= mt):
a[x - 1][y - 1] = 0
dp =[[0]*(m + 1) for i in range(n + 1)]
dp[1][1] = a[0][0]
for j in range(2, m + 1):
dp[1][j] = dp[1][j - 1] + a[0][j - 1]
for i in range(2, n + 1):
dp[i][1] = dp[i - 1][1] + a[i - 1][0]
for i in range(2, n + 1):
for j in range(2, m + 1):
dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + a[i - 1][j - 1]
#[print(*i) for i in dp]
for i in range(k, n + 1):
for j in range(k, m + 1):
sm = dp[i][j] - dp[i - k][j] - dp[i][j - k] + dp[i - k][j - k]
if(sm == 0):
return 1
return 0
def main():
n, m, k, q = getf()
qu = [getf() for i in range(q)]
l, r = -1, 10**9 + 1
while(l + 1 < r):
mt = (l + r)//2
if(isbroken(n, m, mt, k, qu, q)):
r = mt
else:
l = mt
if(r != 10**9 + 1):
put(r)
else:
put(-1)
main()
``` | output | 1 | 40,903 | 7 | 81,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Luba bought a monitor. Monitor is a rectangular matrix of size n × m. But then she started to notice that some pixels cease to work properly. Luba thinks that the monitor will become broken the first moment when it contains a square k × k consisting entirely of broken pixels. She knows that q pixels are already broken, and for each of them she knows the moment when it stopped working. Help Luba to determine when the monitor became broken (or tell that it's still not broken even after all q pixels stopped working).
Input
The first line contains four integer numbers n, m, k, q (1 ≤ n, m ≤ 500, 1 ≤ k ≤ min(n, m), 0 ≤ q ≤ n·m) — the length and width of the monitor, the size of a rectangle such that the monitor is broken if there is a broken rectangle with this size, and the number of broken pixels.
Each of next q lines contain three integer numbers xi, yi, ti (1 ≤ xi ≤ n, 1 ≤ yi ≤ m, 0 ≤ t ≤ 109) — coordinates of i-th broken pixel (its row and column in matrix) and the moment it stopped working. Each pixel is listed at most once.
We consider that pixel is already broken at moment ti.
Output
Print one number — the minimum moment the monitor became broken, or "-1" if it's still not broken after these q pixels stopped working.
Examples
Input
2 3 2 5
2 1 8
2 2 8
1 2 1
1 3 4
2 3 2
Output
8
Input
3 3 2 5
1 2 2
2 2 1
2 3 5
3 2 10
2 1 100
Output
-1 | instruction | 0 | 40,904 | 7 | 81,808 |
Tags: binary search, data structures
Correct Solution:
```
from sys import stdout
from sys import stdin
def get():
return stdin.readline().strip()
def getf():
return [int(i) for i in get().split()]
def put(a, end = "\n"):
stdout.write(str(a) + end)
def putf(a, sep = " ", end = "\n"):
stdout.write(sep.join([str(i) for i in a]) + end)
def check(lmt, qr, k, n, m, q):
d = [[0] * m for i in range(n)]
for i in range(q):
ti, x, y = qr[i]
if(ti <= lmt):
d[x][y] = 1
dp = [[0] * (m + 1) for i in range(n + 1)]
dp[1][1] = d[0][0]
for i in range(2, n + 1):
dp[i][1] = dp[i - 1][1] + d[i - 1][0]
for j in range(2, m + 1):
dp[1][j] = dp[1][j - 1] + d[0][j - 1]
for i in range(2, n + 1):
for j in range(2, m + 1):
dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + d[i - 1][j - 1]
for i in range(k, n + 1):
for j in range(k, m + 1):
sm = dp[i][j] - dp[i - k][j] - dp[i][j - k] + dp[i - k][j - k]
if(sm == k * k):
return True
def main():
n, m, k, q = getf()
qr = []
t0 = -1
for i in range(q):
xi, yi, ti = getf()
t0 = max(t0, ti)
qr.append([ti, xi - 1, yi - 1])
l, r = -1, 10 ** 9 + 7
while(l + 1 < r):
lmt = (l + r) // 2
f = check(lmt, qr, k, n, m, q)
if(f == True):
r = lmt
else:
l = lmt
if(q == 0):
print(-1)
return 0
if(r == 10 ** 9 + 7):
put(-1)
else:
put(r)
main()
``` | output | 1 | 40,904 | 7 | 81,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Luba bought a monitor. Monitor is a rectangular matrix of size n × m. But then she started to notice that some pixels cease to work properly. Luba thinks that the monitor will become broken the first moment when it contains a square k × k consisting entirely of broken pixels. She knows that q pixels are already broken, and for each of them she knows the moment when it stopped working. Help Luba to determine when the monitor became broken (or tell that it's still not broken even after all q pixels stopped working).
Input
The first line contains four integer numbers n, m, k, q (1 ≤ n, m ≤ 500, 1 ≤ k ≤ min(n, m), 0 ≤ q ≤ n·m) — the length and width of the monitor, the size of a rectangle such that the monitor is broken if there is a broken rectangle with this size, and the number of broken pixels.
Each of next q lines contain three integer numbers xi, yi, ti (1 ≤ xi ≤ n, 1 ≤ yi ≤ m, 0 ≤ t ≤ 109) — coordinates of i-th broken pixel (its row and column in matrix) and the moment it stopped working. Each pixel is listed at most once.
We consider that pixel is already broken at moment ti.
Output
Print one number — the minimum moment the monitor became broken, or "-1" if it's still not broken after these q pixels stopped working.
Examples
Input
2 3 2 5
2 1 8
2 2 8
1 2 1
1 3 4
2 3 2
Output
8
Input
3 3 2 5
1 2 2
2 2 1
2 3 5
3 2 10
2 1 100
Output
-1 | instruction | 0 | 40,905 | 7 | 81,810 |
Tags: binary search, data structures
Correct Solution:
```
import sys
from operator import itemgetter
n, m, k, q = map(int, input().split())
query = sorted((list(map(int, line.split()))
for line in sys.stdin), key=itemgetter(2))
times = [q[2] for q in query]
def solve(ti):
imos = [[0]*(m+1) for _ in range(n+1)]
for i in range(ti):
imos[query[i][0]][query[i][1]] += 1
for i in range(n+1):
for j in range(m):
imos[i][j+1] += imos[i][j]
for j in range(m+1):
for i in range(n):
imos[i+1][j] += imos[i][j]
for i in range(k, n+1):
for j in range(k, m+1):
if imos[i][j] - imos[i-k][j] - imos[i][j-k] + imos[i-k][j-k] == k*k:
return True
return False
inf = len(times)+1
ok, ng = inf, 0
while abs(ok - ng) > 1:
mid = (ok + ng) >> 1
if mid >= k*k and solve(mid):
ok = mid
else:
ng = mid
print(times[ok-1] if ok != inf else -1)
``` | output | 1 | 40,905 | 7 | 81,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Luba bought a monitor. Monitor is a rectangular matrix of size n × m. But then she started to notice that some pixels cease to work properly. Luba thinks that the monitor will become broken the first moment when it contains a square k × k consisting entirely of broken pixels. She knows that q pixels are already broken, and for each of them she knows the moment when it stopped working. Help Luba to determine when the monitor became broken (or tell that it's still not broken even after all q pixels stopped working).
Input
The first line contains four integer numbers n, m, k, q (1 ≤ n, m ≤ 500, 1 ≤ k ≤ min(n, m), 0 ≤ q ≤ n·m) — the length and width of the monitor, the size of a rectangle such that the monitor is broken if there is a broken rectangle with this size, and the number of broken pixels.
Each of next q lines contain three integer numbers xi, yi, ti (1 ≤ xi ≤ n, 1 ≤ yi ≤ m, 0 ≤ t ≤ 109) — coordinates of i-th broken pixel (its row and column in matrix) and the moment it stopped working. Each pixel is listed at most once.
We consider that pixel is already broken at moment ti.
Output
Print one number — the minimum moment the monitor became broken, or "-1" if it's still not broken after these q pixels stopped working.
Examples
Input
2 3 2 5
2 1 8
2 2 8
1 2 1
1 3 4
2 3 2
Output
8
Input
3 3 2 5
1 2 2
2 2 1
2 3 5
3 2 10
2 1 100
Output
-1
Submitted Solution:
```
n,m,k,q=list(map(int,input().split()))
qry=[list(map(int,input().split())) for _ in range(q)]
qry=sorted(qry,key= lambda x:x[2],reverse=True)
#print(qry)
low=0
high=q-1
def func(t):
arr=[[0]*(m) for _ in range(n)]
for i in range(t,q):
arr[qry[i][0]-1][qry[i][1]-1]=1
precal=[[0]*(m+1) for _ in range(n+1)]
for i in range(1,n+1):
for j in range(1,m+1):
precal[i][j]=precal[i-1][j]+precal[i][j-1]+arr[i-1][j-1]-precal[i-1][j-1]
for i in range(1,n-k+2):
for j in range(1,m-k+2):
s=precal[i+k-1][j+k-1]+precal[i-1][j-1]-precal[i+k-1][j-1]-precal[i-1][j+k-1]
#print(s)
if s==k*k:
return True
return False
ans=-1
while(low<=high):
mid=(low+high)//2
check=func(mid)
#print(chec)
if check:
if mid==q-1:
ans=mid
break
elif func(mid-1):
low=high+1
ans=mid
else:
ans=mid
break
else:
high=mid-1
#print(ans)
if ans!=-1:
print(qry[ans][2])
else:
print(ans)
``` | instruction | 0 | 40,907 | 7 | 81,814 |
No | output | 1 | 40,907 | 7 | 81,815 |
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