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≤ which denotes the number of test cases. 

For each test case, there is an positive integer n(≤n≤) 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 .Notice name only contain letter(s),digit(s) and space(s).

Output

For each case, output n lines.

 

Sample Input

FancyCoder  

FancyCoder
xyz111 

 

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[];
 }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[];
         getchar();
         for(int i=;i<n;i++){
             person[i].year=;
             gets(s);
             //puts(s);
             int len=strlen(s);
             for(int j=;j<len-;j++){
                 person[i].ch[j]=s[j];
             }
             person[i].ch[len-]='\0';//记得加，否则wa

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

     return ;
 }