题解 | 扫雷

#include <iostream>
using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    int arr[2000][2000] = { 0 };
    int arr1[2000][2000] = { 0 };
    for (int i = 1; i <= n; i++) 
    {
        for (int j = 1; j <= m; j++)
        {
            char y;
            cin >> y;
            if (y == '*') arr[i][j] = 1;
            else arr[i][j] = 0;
        }
    }
    for (int i = 1; i <= n; i++) 
    {
        for (int j = 1; j <= m; j++)
            arr1[i][j] = arr[i - 1][j - 1] + arr[i - 1][j] + arr[i - 1][j + 1] + arr[i][j - 1] + arr[i][j + 1] + arr[i + 1][j - 1] + arr[i + 1][j] + arr[i + 1][j + 1];
    }
    for (int i = 1; i <= n; i++) 
    {
        for (int j = 1; j <= m; j++)
        {
            if (arr[i][j]) cout << "*";
            else cout << arr1[i][j];
        } 
        cout << endl;
    }
    return 0;
}
// 64 位输出请用 printf("%lld")