A problem of sorting（简单排序+读取一整行数据的用法）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5427

A problem of sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 91    Accepted Submission(s): 53

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

 

Source

BestCoder Round #54 (div.2)

 

中文题意：

A problem of sorting

 

 Accepts: 371

 

 Submissions: 1706

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)

问题描述

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）

输入描述

第一行包含一个正整数$T(T\le 5)$，表示数据组数。对于每组数据，第一行包含一个正整数$n(1\le n\le 100)$，表示人数，接下来n行，每行包含一个姓名和生日年份(1900-2015)，用一个空格隔开。姓名长度大于0且不大于100。注意姓名中只包含字母，数字和空格。

输出描述

对于每组数据，输出$n$行姓名。

输入样例

21FancyCoder 19962FancyCoder 1996xyz111 1997

输出样例

FancyCoderxyz111FancyCoder

官方题解：

A problem of sorting

选择任意喜欢的排序方法即可。注意名字中可能有空格。

AC code：

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>#include<queue>#define LL long long#define MAXN 1000010using namespace std;const int INF=0xffffff;struct node{int y;char name[111];}p[111];bool cmp(node a,node b){return a.y>b.y;}string name;int n,m,k,s,a,b;int ans;int main(){int T,i;scanf("%d",&T);while(T--){scanf("%d",&n);getchar();for(i=1;i<=n;i++){getline(cin,name);//读取一整行（包括空格用getline(cin,name);!!!） int len=name.length();int year=0;int j=0;for(j=len-4;j<=len-1;j++){year=year*10+(name[j]-'0');}p[i].y=year;for(j=0;j<len-5;j++)p[i].name[j]=name[j];p[i].name[j]='\0';}sort(p+1,p+n+1,cmp);for(i=1;i<=n;i++){puts(p[i].name);}}return 0;}