A 矩阵乘法
思路：
1° 牛客机器太快了，暴力能过。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 5000 5 6 int n, p, m; 7 int a[N][100], b[100][N], ans[N][N]; 8 9 void Run() 10 { 11 while (scanf("%d%d%d", &n, &p, &m) != EOF) 12 { 13 for (int i = 1; i <= n; ++i) 14 for (int j = 1; j <= p; ++j) 15 scanf("%X", &a[i][j]); 16 for (int j = 1; j <= m; ++j) 17 for (int i = 1; i <= p; ++i) 18 scanf("%1d", &b[i][j]); 19 for (int i = 1; i <= n; ++i) 20 for (int j = 1; j <= m; ++j) 21 for (int k = 1; k <= p; ++k) 22 ans[i][j] += a[i][k] * b[k][j]; 23 int res = 0; 24 for (int i = 1; i <= n; ++i) 25 for (int j = 1; j <= m; ++j) 26 res ^= ans[i][j]; 27 printf("%d\n", res); 28 } 29 } 30 31 int main() 32 { 33 #ifdef LOCAL 34 freopen("Test.in", "r", stdin); 35 #endif 36 37 Run(); 38 return 0; 39 }