| Run ID | 作者 | 问题 | 语言 | 测评结果 | Time | Memory | 代码长度 | 提交时间 |
|---|---|---|---|---|---|---|---|---|
| 127534 | 陈淮遇 | 输出第i行第j列 | C++ | Wrong Answer | 1 MS | 284 KB | 406 | 2025-07-28 14:50:16 |
#include<bits/stdc++.h> using namespace std; int main() { int a[100][100], b[100][100], m, n, sum = 0; cin >> n >> m; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) cin >> a[i][j]; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { cin >> b[i][j]; if (a[i][j] == b[i][j]) sum++; } printf("%.2f", sum * 1.0 / (m * n) * 100); return 0; }
------Input------
2 2
------Answer-----
21
------Your output-----
100.00