hdu 5427 A problem of sorting

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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
\leq 100T≤100 which
denotes the number of test cases.

For each test case, there is an positive integer n
(1 \leq n \leq 100)n(1≤n≤100) which
denotes the number of people,and next nn 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 100100.Notice
name only contain letter(s),digit(s) and space(s).

Output

For each case, output nn lines.

Sample Input

2
1
FancyCoder 1996
2
FancyCoder 1996
xyz111 1997

题目大意：就是让你排序，按照从大到小排序，然后输出姓名就行了

解题思路：注意这里有一个坑，就是可能有空格作为当前的名字比如说：  ITAK，输出的时候也需要输出空格，这就需要用到cin.getline了，这个就是能读单字符的。。。

读入啥，输出啥，可以手动实现一下：

上代码：

</pre><pre name="code" class="cpp"><pre name="code" class="cpp">/*
Date : 2015-09-05 晚上

Author : ITAKING

Motto :

今日的我要超越昨日的我，明日的我要胜过今日的我；
以创作出更好的代码为目标，不断地超越自己。
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
struct sa
{
    char name[105];
    int date;
} arr[105];
int cmp(sa a, sa b)
{
    return a.date > b.date;
}
char str[200];
int main()
{
    int t,m;
    scanf("%d",&t);
    while(t--)
    {
        cin>>m;
        getchar();
        for(int i=0; i<m; i++)
        {
            cin.getline(str,200);//能够正常读入单字符
            int len = strlen(str);
            arr[i].date = str[len-4]*1000+str[len-3]*100+str[len-2]*10+str[len-1];
            len -= 5;
            str[len] = '\0';
            strcpy(arr[i].name,str);
        }
        sort(arr, arr+m, cmp);
        for(int i=0; i<m; i++)
            cout<<arr[i].name<<endl;
    }
    return 0;
}