CF1119C Ramesses and Corner Inversion 题解

前言

显然是一道shabi结论题

解法

求出 \(A\) , \(B\) 的异或矩阵,记录每一行,每一列异或和, 异或和为奇数显然不行；

Code

#include<bits/stdc++.h>

typedef long long ll;

const int maxn = 510;

int n, m, a[maxn][maxn], b[maxn][maxn], c[maxn][maxn], x[maxn], y[maxn];

int read()
{
    int x = 0, ch = getchar(), f = 1;
    while(!isdigit(ch)){if(ch == '-') f = -1; ch = getchar(); }
    while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return x * f;
}

int main()
{
//  freopen(".in","r",stdin); freopen(".out","w",stdout);
//    int n = (8^9);
//    int m = (9^10);
//    std::cout<<(8^9)<<" "<<n<<" "<<m<<" "<<(9^10);
    n = read(), m = read();
    for(int i = 1; i <= n; ++i)
    {
        for(int j = 1; j <= m; ++j)
        {
            a[i][j] = read();
        }
    }
    for(int i = 1; i <= n; ++i)
    {
        for(int j = 1; j <= m; ++j)
        {
            b[i][j] = read();
        } 
    }
    for(int i = 1; i <= n; ++i)
    {
        for(int j = 1; j <= m; ++j)
        {
            c[i][j] = a[i][j]^b[i][j];
            x[i] += c[i][j]; 
            y[j] += c[i][j];
        }
    }
    for(int i = 1; i <= n; ++i) if(x[i]&1){ puts("No"); return 0;}
    for(int i = 1; i <= m; ++i) if(y[i]&1){ puts("No"); return 0;}
    puts("Yes");
    return 0;
}