HDU——5427 A problem of sorting

A problem of sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1021    Accepted Submission(s): 431

Problem Description

There are many people's name and birth in a list.Your task is to print the name from young to old.(There is no pair of two has the same age.)

 

Input

First line contains a single integer  T≤100  which denotes the number of test cases. 

For each test case, there is an positive integer  n(1≤n≤100)  which denotes the number of people,and next  n  lines,each line has a name and a birth's year(1900-2015) separated by one space.

The length of name is positive and not larger than  100 .Notice name only contain letter(s),digit(s) and space(s).

 

Output

For each case, output  n  lines.

 

Sample Input

   
   
   
   

    
    
    
    2
1
FancyCoder 1996
2
FancyCoder 1996
xyz111 1997
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    FancyCoder
xyz111
FancyCoder
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    没想到名字可以是多个单词组成，结果一直wrong了！
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    #include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<map>
#include<numeric>
#include<stack>
#include<list>
const int INF=1<<30;
const int inf=-(1<<30);
const int MAX=100010;


using namespace std;


struct data
{
    char x[110];
    int y;
} a[110];


bool comp(data x,data y)
{
    return x.y>y.y;
}


int main()
{
    int T,n;
    cin>>T;
    while(T--)
    {
        cin>>n;
        getchar();
        char str[110];
        for(int i=0; i<n; i++)
        {
            cin.getline(str,110);
            int d=strlen(str);
            a[i].y=str[d-4]*1000+str[d-3]*100+str[d-2]*10+str[d-1];
            d=d-5;
            str[d]='\0';
            strcpy(a[i].x,str);
        }
        sort(a,a+n,comp);
        for(int i=0; i<n; i++)
            cout<<a[i].x<<endl;
    }
}