P2670 [NOIP2015 普及组] 扫雷游戏

题目传送门

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
char a[N][N];
int dx[] = {0, 0, -1, 1, -1, 1, -1, 1}; //上下左右，左上，右上，左下，右下
int dy[] = {-1, 1, 0, 0, -1, -1, 1, 1};
int n, m;

int main() {
    cin >> n >> m;
    //计入地雷地图
    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++) {
            if (a[i][j] == '*') printf("%c", '*');
            else {
                int cnt = 0;
                for (int k = 0; k < 8; k++)
                    //当前点周边的*的个数是多少个？
                    if (a[i + dx[k]][j + dy[k]] == '*') cnt++;
                printf("%d", cnt);
            }
        }
        printf("\n");
    }
    return 0;
}