20
5 5
#####
#...#
#...#
#....
.....
3 3
###
###
###
5 5
#####
#####
#####
#####
#####
<27002 bytes omitted>
用户输出
5
25
25
25
22
37
5
27
7
57
4
9
0
1
1389
895
915
1
1079
1058
系统信息
Exited with return code 0
编号 | 题目 | 状态 | 分数 | 总时间 | 内存 | 代码 / 答案文件 | 提交者 | 提交时间 |
---|---|---|---|---|---|---|---|---|
#48317 | #1010. J. 团队配置,新奇的面试题 | Accepted | 100 | 15 ms | 396 K | C++ / 1.2 K | 少年班94刘千瑀 | 2021-05-13 13:32:52 |
#include <bits/stdc++.h>
using namespace std;
int n, m, N, M;
char a[255][255], b[255][255];
int main() {
ios::sync_with_stdio(0);
int t;
cin >> t;
while (t--) {
int ans = 0;
cin >> N >> M;
for (int i = 1; i <= N; i++) {
string s;
cin >> s;
for (int j = 1; j <= M; j++) {
a[i][j] = s[j - 1];
}
}
cin >> n >> m;
for (int i = 1; i <= n; i++) {
string s;
cin >> s;
for (int j = 1; j <= m; j++) {
b[i][j] = s[j - 1];
}
}
int c1 = n / 2;
int c2 = m / 2;
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++) {
bool bb = 0;
for (int k = 1; k <= n; k++) {
for (int l = 1; l <= m; l++) {
if (b[k][l] == '.')
continue;
if (k + i - c1 - 1 > 0 && k + i - c1 - 1 <= N && l + j - c2 - 1 > 0 &&
l + j - c2 - 1 <= M && a[k + i - c1 - 1][l + j - c2 - 1] == b[k][l]) {
bb = 1;
break;
}
}
}
if (!bb)
ans++;
}
}
cout << ans << "\n";
}
return 0;
}