小米校园招聘 2017 编程题：号码分身

//根据特殊字符一个一个排除即可

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<string.h>
using namespace std; 

int has[26]; 
// one two three four five six seven eight nine
// six seven five four two eight three nine one
// 0 1 2 3 4 5 6 7 8 9
// 8 9 0 1 2 3 4 5 6 7
char ch[] = {'X' , 'S' , 'V' , 'F' , 'W' , 'G' , 'R' , 'I' , 'O' };
string str[] = {"SIX" ,"SEVEN","FIVE","FOUR","TWO","EIGHT","THREE","NINE","ONE"};
int num[] = { 8 , 9 , 7 , 6 ,4 ,0 ,5,1,3}; 
void solve(vector<int> &ans){
   
   for(int i = 0 ; i < 9 ; i ++){
       
          while(has[ch[i] - 'A'] > 0) {
                 ans.push_back(num[i]);
                 string tmp = str[i];
                 for(int j = 0 ; j < tmp.length() ; j ++) has[tmp[j] - 'A'] --;
       }
       
   }
    
}
int main(){
    
    freopen("data.in","r",stdin); 
    
    int T;
    while(cin>>T){
        string str;
        for(int i = 0 ; i < T ; i ++){
            vector<int> ans;
            memset(has, 0 , sizeof has);
            cin>>str;
            
            for(int i = 0 ; i < str.length() ; i ++){
                  has[str[i] - 'A'] ++;
            }            
            solve(ans);
            sort(ans.begin() , ans.end());
            for(int i = 0 ; i < ans.size() ; i ++){
                
                cout<<ans[i];
                
            }
            cout<<endl;    
        }    
        
    }
    return 0;
}