CodeForces 1983B Corner Twist

合集 - Codeforces(21)

1.CodeForces 1992E Novice's Mistake2024-07-162.CodeForces 1935A Entertainment in MAC2024-06-163.CodeForces 1992C Gorilla and Permutation2024-07-164.CodeForces 1992D Test of Love2024-07-165.CodeForces 1992B Angry Monk2024-07-166.CodeForces 1992A Only Pluses2024-07-16

7.CodeForces 1983B Corner Twist2024-07-16

8.CodeForces 1983A Array Divisibility2024-07-169.CodeForces 1983C Have Your Cake and Eat It Too2024-07-1610.CodeForces 1883B Chemistry2024-07-2611.CodeForces 1883A Morning2024-07-2612.CodeForces 1883C Raspberries2024-07-2813.CodeForces 1883D In Love2024-07-2814.CodeForces 1883E Look Back2024-07-2815.CodeForces 1883F You Are So Beautiful2024-07-2816.CodeForces 1883G1 Dances (Easy version)2024-07-2817.CodeForces 908B New Year and Buggy Bot2024-08-0118.CodeForces 908C New Year and Curling2024-08-0119.CodeForces 1132B Discounts2024-08-0120.CodeForces 1619D New Year's Problem2024-08-0121.Codeforces Round 964 (Div. 4)2024-08-07

收起

题目链接：CodeForces 1983B【Corner Twist】

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思路

       可以发现操作一次，被操作位置的对应每一横行和每一纵行的加减数都是3，所以可以根据网格a和b的横纵状态确定是否通过操作使得网格a到达网格b。

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代码

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 5e2 + 10;
 
int a[N][N], b[N][N];
int main() {
  int t;
  cin >> t;
  while (t--) {
    memset(a, 0, sizeof a);
    memset(b, 0, sizeof b);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      string s;
      cin >> s;
      for (int j = 0; j < m; j++) {
        a[i][j + 1] = s[j] - '0';
        a[i][j + 1] = a[i][j + 1] + a[i - 1][j + 1] + a[i][j] - a[i - 1][j];
      }
    }
    
    for (int i = 1; i <= n; i++) {
      a[i][m + 1] = a[i][m] - a[i - 1][m];
    }
    for (int i = 1; i <= m; i++) {
      a[n + 1][i] = a[n][i] - a[n][i - 1];
    }
 
    for (int i = 1; i <= n; i++) {
      string s;
      cin >> s;
      for (int j = 0; j < m; j++) {
        b[i][j + 1] = s[j] - '0';
        b[i][j + 1] = b[i][j + 1] + b[i - 1][j + 1] + b[i][j] - b[i - 1][j];
      }
    }
 
    for (int i = 1; i <= n; i++) {
      b[i][m + 1] = b[i][m] - b[i - 1][m];
    }
    for (int i = 1; i <= m; i++) {
      b[n + 1][i] = b[n][i] - b[n][i - 1];
    }
    
    int flag = 0;
    for (int i = 1; i <= n; i++) {
      b[i][m + 1] = b[i][m] - b[i - 1][m];
      if ((b[i][m + 1] - a[i][m + 1]) % 3 != 0) {
        cout << "NO" << endl;
        flag = 1;
        break;
      }
    }
 
    if (flag) {
      continue;
    }
 
    for (int i = 1; i <= m; i++) {
      b[n + 1][i] = b[n][i] - b[n][i - 1];
      if ((b[n + 1][i] - a[n + 1][i]) % 3 != 0) {
        cout << "NO" << endl;
        flag = 1;
        break;
      }
    }
 
    if (flag) {
      continue;
    }
 
    cout << "YES" << endl;
  }
  
  return 0;
}