牛客竞赛：冰冰的电子邮件

冰冰的电子邮件

先将字符串从 '@'  分割为s1，s2 ， 再分别判断。

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin>>t;
    
    while(t--)
    {
        string s;
        cin>>s;
        
        string s1;
        string s2;
        
        int pos=s.find('@');
        if(pos == string::npos)
        {
            cout<<"No"<<endl;
            continue;
        }
        s1=s.substr(0,pos);
        s2=s.substr(pos+1);
        int n=s1.size();
        int m=s2.size();
        
        if(s1[0] == '.' || s1[n-1] == '.' || s2[0] == '.' || 
           s2[m-1] == '.' || s2[0] == '-' || s2[m-1] == '-')
        {
            cout<<"No"<<endl;
            continue;
        }
        if(n>64 || n<1 || m>255 || m<1)
        {
            cout<<"No"<<endl;
            continue;
        }
        bool flag=true;
        for(int i=0;i<n;i++)
        {
            if(s1[i] == '.')continue;
            if(!((s1[i] >= 'A' && s1[i] <= 'Z') || (s1[i] >= 'a' && s1[i] <= 'z')
                 || (s1[i] >= '0' && s1[i] <= '9')))
            {
                flag=false;
                break;
            }
        }
        for(int i=0;i<m;i++)
        {
            if(s2[i] == '.' || s2[i] == '-')continue;
            if(!((s2[i] >= 'A' && s2[i] <= 'Z') || (s2[i] >= 'a' && s2[i] <= 'z')
                 || (s2[i] >= '0' && s2[i] <= '9')))
            {
                flag=false;
                break;
            }
        }
        if(flag)
        cout<<"Yes"<<endl;
        else
        cout<<"No"<<endl;
    }
    return 0;
}