HDU 5427

A problem of sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 78    Accepted Submission(s): 44

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
   
   
   
   

 

Source

BestCoder Round #54 (div.2)

 

Recommend

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>

using namespace std;

struct node
{
    char s[110];
    int a;
}p[1000],t;
int T,n;
int main()
{
    while(~scanf("%d",&T))
    {
        while(T--)
        {
            scanf("%d",&n);
            getchar();
            for(int i=0;i<n;i++)
            {
                gets(p[i].s);
            }
            for(int i=0;i<n;i++)
            {
                int sum=0;
                int len=strlen(p[i].s);
                for(int j=len-4;j<len;j++)
                {
                    sum*=10;
                    sum+=p[i].s[j]-'0';
                }
                p[i].s[len-5]='\0';   //处理字符串后的空格<span id="transmark"></span>
                p[i].a=sum;
                //printf("%d\n",p[i].a);
            }

            for(int i=0;i<n-1;i++)
            {
                for(int j=0;j<n-1-i;j++)
                {
                    if(p[j].a<p[j+1].a)
                    {
                        t=p[j];
                        p[j]=p[j+1];
                        p[j+1]=t;
                    }
                }
            }

            for(int i=0;i<n;i++)
            printf("%s\n",p[i].s);
        }
    }
}