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 |
---|---|---|---|---|---|---|---|---|
#31875 | #1010. J. 团队配置,新奇的面试题 | Accepted | 100 | 19 ms | 304 K | C++ 17 / 1.3 K | 2n8asm | 2020-07-05 17:31:12 |
#ifdef D
#include "hpp.hpp"
#else
#include <bits/stdc++.h>
#endif
using namespace std;
class Main {
vector<vector<short>> A;
vector<vector<short>> a;
short M, N, T = 0, m, n;
short AT(short i, short j) const { return 0 <= i && i < N && 0 <= j && j < M ? A[i][j] : false; }
void in(vector<vector<short>> &a, short &m, short &n) {
char c;
cin >> n >> m;
a.resize(n, vector<short>(m, 0));
for (auto &r : a) {
for (short &b : r) {
cin >> c;
b = c == '#';
}
}
}
void ok(short x, short y) {
short I, J, i, j;
for (I = x - (n >> 1), i = 0; i < n; ++I, ++i) {
for (J = y - (m >> 1), j = 0; j < m; ++J, ++j) {
if (AT(I, J) && a[i][j]) {
return;
}
}
}
++T;
}
public:
Main() {
in(A, M, N);
in(a, m, n);
}
void main() {
short i, j;
for (i = 0; i < N; ++i) {
for (j = 0; j < M; ++j) {
ok(i, j);
}
}
cout << T << endl;
}
};
int main() {
short t;
for (cin >> t; t; --t) {
Main main;
main.main();
}
}