HDOJ 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): 1215    Accepted Submission(s): 485

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

21FancyCoder 19962FancyCoder 1996xyz111 1997

 

Sample Output

FancyCoderxyz111FancyCoder题意：给出一张许多人的年龄和生日表。你需要从年轻到年老输出人们的名字。（没有人年龄相同）注意：姓名可以含有空格ac代码：
#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<queue>#include<iostream>#include<algorithm>#define MAXN 10001#define LL long long#define INF 0x7fffffff#define mem(x) memset(x,0,sizeof(x))using namespace std;struct s{char name[110];int num;}a[110];bool cmp(s a,s b){return a.num>b.num;}int main(){int t,i,j;int n;char ss[1010];char str[1010];scanf("%d",&t);while(t--){scanf("%d",&n);getchar();for(i=0;i<n;i++){gets(ss);//printf("yes\n");int len=strlen(ss);int q=0;int k=1;int w=0;int bz=0;for(j=len-1;j>=0;j--){if(ss[j]==' '&&bz==0){bz=1;continue;}if(bz){str[w++]=ss[j];}else{q=q+(ss[j]-'0')*k;k*=10;}}a[i].num=q;k=0;for(j=w-1;j>=0;j--)a[i].name[k++]=str[j];a[i].name[k]='\0';//getchar();}sort(a,a+n,cmp);for(i=0;i<n;i++)printf("%s\n",a[i].name);}return 0;}