HDU 5427

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本文介绍了一个简单的年龄排序问题，任务是根据输入的人名和出生年份列表，将人名按年龄从小到大进行排序输出。文章提供了一段C++代码实现，通过读取测试用例数量、每组数据中人的数量及姓名和出生年份，然后提取出生年份并进行排序。

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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)

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#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);
        }
    }
}