#include<iostream> #include<cstring> using namespace std; char a[105][105],b[105][105]; int main(){ int n,m,g; cin >> n >> m; for(int i = 0;i < n;i ++){ for(int j = 0;j < m;j ++){//输入大地图 cin >> a[i][j]; } } for(int i =0;i < n;i ++){ for(int j = 0;j < m;j ++){ g = 0; if(a[i][j] == ''){ b[i][j] = ''; } else { if(i - 1 >= 0 and a[i - 1][j] == ''){//对于每一个方位进行判断 g += 1; } if(i + 1 < n and a[i + 1][j] == ''){ g += 1; } if(j - 1 >= 0 and a[i][j - 1] == ''){ g += 1; } if(j + 1 < m and a[i][j + 1] == ''){ g += 1; } if(i + 1 < n and j + 1 < m and a[i + 1][j + 1] == ''){ g += 1; } if(i - 1 >= 0 and j - 1 >= 0 and a[i - 1][j - 1] == ''){ g += 1; } if(i + 1 < n and j - 1 >= 0 and a[i + 1][j - 1] == ''){ g += 1; } if(i - 1 >= 0 and j + 1 < m and a[i - 1][j + 1] == ''){ g += 1; } b[i][j] = 48 + g; } } } for(int i = 0;i < n;i ++){ for(int j = 0;j < m;j ++){//将带有数据的数组输出 cout << b[i][j]; } cout << endl; } return 0; }

本人菜，代码太丑请见谅

题解 以下非题解

#include<iostream> #include<cmath> using namespace std; char a[110][110]; int main(){ int n,m; cin>>n>>m; for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ cin>>a[i][j]; } } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ int d=0; if(a[i][j]'?'){ if(a[i-1][j]'')d++; if(a[i-1][j-1]=='')d++; if(a[i-1][j+1]'*')d++; if(a[i][j-1]'')d++; if(a[i][j+1]=='')d++; if(a[i+1][j]'*')d++; if(a[i+1][j-1]'')d++; if(a[i+1][j+1]=='')d++; a[i][j]=d+'0'; } } } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ cout<<a[i][j]; } cout<<endl; } return 0; }

写了个check函数

写法有部分搜索的思想学搜索学的

#include<stdio.h> int main(void) { char a[101][101]; int sum[100][100]={0}; int j,i,n,m; scanf("%d%d",&n,&m); for(i=0;i<n;i++) { scanf("%s",a[i]);getchar(); } for(i=0;i<n;i++) { for(j=0;j<m;j++) { if(a[i][j]'*') {sum[i][j]=-1; continue; } if(a[i-1][j-1]'') sum[i][j]++; if(a[i-1][j]=='') sum[i][j]; if(a[i-1][j+1]=='*') sum[i][j]; if(a[i][j-1]'*') sum[i][j]++; if(a[i][j+1]'') sum[i][j]++; if(a[i+1][j-1]=='') sum[i][j]; if(a[i+1][j]=='*') sum[i][j]; if(a[i+1][j+1]=='') sum[i][j]; } } for(i=0;i<n;i) { for(j=0;j<m;j++) { if(sum[i][j]!=-1) printf("%d",sum[i][j]); else putchar(''); } putchar('\n'); } return 0; }