给定n，m（200），构造一个n*m的矩阵a，使得每个4*4的子矩阵，左上角2*2的子矩阵的异或和等于右下角的，左下角的异或和等于右上角的

题目

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e18 + 5, maxm = 4e4 + 5, mod = 998244353, N = 1e6;
int a[505][505], b[maxn];
// bool vis[maxn];
int n, m;
string s;
int f[maxn];bool check(){for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){if(i + 4 - 1 > n || j + 4 - 1 > m) continue;int t = a[i][j] ^ a[i][j + 1] ^ a[i + 1][j] ^ a[i + 1][j + 1];int t2 = a[i + 2][j + 2] ^ a[i + 2][j + 1 + 2] ^ a[i + 1 + 2][j + 2] ^ a[i + 1 + 2][j + 1 + 2];if(t != t2){cout << i << ' ' << j << '\n';cout << t << ' ' << t2 << '\n';return 0;}// cout << i << ' ' << j << ' ' << t << '\n';t = a[i][j + 2] ^ a[i][j + 1 + 2] ^ a[i + 1][j + 2] ^ a[i + 1][j + 1 + 2];t2 = a[i + 2][j] ^ a[i + 2][j + 1] ^ a[i + 1 + 2][j] ^ a[i + 1 + 2][j + 1];if((t != t2)){cout << i << ' ' << j  << '\n';return 0;}}}return 1;
}
void ff(int x){for(int j = 15; j >= 0; j--){cout << (x >> j & 1);}cout << '\n';
}
void solve(){int res = 0;int k;int x;int q;cin >> n >> m;int add = 0;int d2 = 1LL << 30;for(int i = 1; i <= n; i += 2){add = 0;for(int j = 1; j <= m; j += 2){a[i][j] = 0 + add + (i - 1) / 2 * d2;a[i][j + 1] = 1 + add + (i - 1) / 2 * d2;a[i + 1][j] = 2 + add + (i - 1) / 2 * d2;a[i + 1][j + 1] = 3 + add + (i - 1) / 2 * d2;add += 4;}}/*把 2 * 2 的子矩阵为一组，按4*k,   4*k+14*k+2, 4*k+3 的方式构造，每一组内异或和为0，但是为了矩阵的每个数字都不同，每一行都要加上d2，第i行加(i - 1) / 2个d2*/cout << n * m << '\n';for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){cout <<  a[i][j] << ' ';}cout << '\n';}}signed main(){ios::sync_with_stdio(0);cin.tie(0);// fac[0] = 1;// for(int i = 1; i <= N; i++){//     fac[i] = fac[i - 1] * i % mod;// }// inv[N] = qpow(fac[N], mod - 2);// for(int i = N - 1; i >= 0; i--){//     inv[i] = inv[i + 1] * (i + 1) % mod;// }int T = 1;cin >> T;while (T--){solve();}return 0;
}