题目描述

no

样例

i:
3 3
??
???
??
o:
10
221
11

算法1

(暴力枚举) $O(n^2)$

本题分析：使用char型二维数组，然后循环嵌套判断：
if(j+1<=m-1&&a[i][j+1]==’‘&&a[i][j]==’?’){
t;
}
if(j-1>=0&&a[i][j-1]==’*’&&a[i][j]==’?’){
t;
}
if(i+1<=n-1&&a[i+1][j]==’‘&&a[i][j]==’?’){
t;
}
if(i-1>=0&&a[i-1][j]==’*’&&a[i][j]==’?’){
t;
}
if(j-1>=0&&i-1>=0&&a[i-1][j-1]==’‘&&a[i][j]==’?’){
t;
}
if(j+1<=m-1&&i+1<=n-1&&a[i+1][j+1]==’*’&&a[i][j]==’?’){
t;
}
if(i+1<=n-1&&j-1>=0&&a[i+1][j-1]==’‘&&a[i][j]==’?’){
t;
}
if(i-1>=0&&j+1<=m-1&&a[i-1][j+1]==’*’&&a[i][j]==’?’){
t;
}
有些人把这道题做麻烦了，其实没有这么麻烦，该？上的数其实不用转换成字符型，直接输出就好了。
：：
if(a[i][j]==’?’){
cout<<t;
}else{
cout<<a[i][j];
}
时间复杂度分析：n*m

C++ 代码

include[HTML_REMOVED]

include[HTML_REMOVED]

include[HTML_REMOVED]

include[HTML_REMOVED]

include[HTML_REMOVED]

include[HTML_REMOVED]

include[HTML_REMOVED]

using namespace std;

int main(){
int n,m;
cin>>n>>m;
char a[n+1][m+1];
for(int i=0;i[HTML_REMOVED]>a[i][j];
}
}

for(int i=0;i<n;i++){
    for(int j=0;j<m;j++){
        int t=0;
        if(j+1<=m-1&&a[i][j+1]=='*'&&a[i][j]=='?'){
            t++;
        }
        if(j-1>=0&&a[i][j-1]=='*'&&a[i][j]=='?'){
            t++;
        }
        if(i+1<=n-1&&a[i+1][j]=='*'&&a[i][j]=='?'){
            t++;
        }
        if(i-1>=0&&a[i-1][j]=='*'&&a[i][j]=='?'){
            t++;
        }
        if(j-1>=0&&i-1>=0&&a[i-1][j-1]=='*'&&a[i][j]=='?'){
            t++;
        }
        if(j+1<=m-1&&i+1<=n-1&&a[i+1][j+1]=='*'&&a[i][j]=='?'){
            t++;
        }
        if(i+1<=n-1&&j-1>=0&&a[i+1][j-1]=='*'&&a[i][j]=='?'){
            t++;
        }
        if(i-1>=0&&j+1<=m-1&&a[i-1][j+1]=='*'&&a[i][j]=='?'){
            t++;
        }
        if(a[i][j]=='?'){
            cout<<t;
        }else{
            cout<<a[i][j];
        }
    }
    cout<<endl;
}
return 0;

}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度分析：blablabla

C++ 代码

blablabla