hdu 5427 A problem of sorting(字符排序)

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)

 

具体看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<cmath>
using namespace std;
#define N 106
int n;
struct Node{
    int year;
    char ch[106];
}person[N];
bool cmp(Node a,Node b){
    return a.year>b.year;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        char s[150];
        getchar();
        for(int i=0;i<n;i++){
            person[i].year=0;
            gets(s);
            //puts(s);
            int len=strlen(s);
            for(int j=0;j<len-5;j++){
                person[i].ch[j]=s[j];
            }
            person[i].ch[len-5]='';//记得加，否则wa

            for(int j=len-4;j<len;j++){
                person[i].year=person[i].year*10+s[j]-'0';
            }
            //printf("%s
",person[i].ch);
            //printf("%d
",person[i].year);
        }
        sort(person,person+n,cmp);
        for(int i=0;i<n;i++){
            printf("%s
",person[i].ch);
        }
    }

    return 0;
}