题解 | #电话号码分身#C++暴力遍历字符排除解法

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() 
{
    int i=0;
    int n=0;
    cin>>n;
    for(i=0;i<n;i++)
    {
        string a;
        string ans;
        cin>>a;
        vector <int> z(26,0);
        int j=0;
        int len=a.size();
        for(j=0;j<len;j++)
        {
            z[a[j]-'A']++;
        }
        if(z[25])
        {
            for(j=0;j<z[25];j++)
            {
                ans+='2';
            }
            z['O'-'A']-=z[25];
            z['R'-'A']-=z[25];
        }
        if(z['G'-'A'])
        {
            for(j=0;j<z['G'-'A'];j++)
            {
                ans+='0';
            }
            z['I'-'A']-=z['G'-'A'];
        }
        if(z['X'-'A'])
        {
            for(j=0;j<z['X'-'A'];j++)
            {
                ans+='8';
            }
            z['S'-'A']-=z['X'-'A'];
            z['I'-'A']-=z['X'-'A'];
        }
        if(z['W'-'A'])
        {
            for(j=0;j<z['W'-'A'];j++)
            {
                ans+='4';
            }
            z['O'-'A']-=z['W'-'A'];
        }
        if(z['U'-'A'])
        {
            for(j=0;j<z['U'-'A'];j++)
            {
                ans+='6';
            }
            z['R'-'A']-=z['U'-'A'];
            z['O'-'A']-=z['U'-'A'];
        }
        if(z['O'-'A'])
        {
            for(j=0;j<z['O'-'A'];j++)
            {
                ans+='3';
            }
        }
        if(z['S'-'A'])
        {
            for(j=0;j<z['S'-'A'];j++)
            {
                ans+='9';
            }
            z['V'-'A']-=z['S'-'A'];
        }
        if(z['V'-'A'])
        {
            for(j=0;j<z['V'-'A'];j++)
            {
                ans+='7';
            }
            z['I'-'A']-=z['V'-'A'];
        }
        if(z['R'-'A'])
        {
            for(j=0;j<z['R'-'A'];j++)
            {
                ans+='5';
            }
        }
        if(z['I'-'A'])
        {
            for(j=0;j<z['I'-'A'];j++)
            {
                ans+='1';
            }
        }
        sort(ans.begin(),ans.end());
        cout<<ans;
        if(i!=n-1)
        {
            cout<<endl;
        }
    }
    return 0;
}