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 $T\le 100$ which denotes the number of test cases.

For each test case, there is an positive integer $n(1\le n\le 100)$ 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

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 : ITAKINGMotto :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;}