HDOJ-1040 As Easy As A+B

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 80266 Accepted Submission(s): 34137

Problem Description
These days, I am thinking about a question, how can I get a problem as easy as A+B? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights.
Give you some integers, your task is to sort these number ascending (升序).
You should know how easy the problem is now!
Good luck!

Input
Input contains multiple test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case contains an integer N (1<=N<=1000 the number of integers to be sorted) and then N integers follow in the same line.
It is guarantied that all integers are in the range of 32-int.

Output
For each case, print the sorting result, and one line one case.

Sample Input
2
3 2 1 3
9 1 4 7 2 5 8 3 6 9

Sample Output
1 2 3
1 2 3 4 5 6 7 8 9

解题思路：数据量大，适合使用快速排序算法，降低时间复杂度

#include
#include
using namespace std;

int a[1010]; //数组中存放整数 
void quicksort(int left,int right)//快速排序算法，可详见《啊哈！算法》快速排序章节 
{
    int i,j,t,temp;//整数按从小到大排序 
    if(left>right)
       return;  
    
    temp=a[left];//把数组的第一个数设为基准数 
    i=left;
    j=right;
    while(i!=j)  
    {
        while(a[j]>=temp&&i>T;  //有T个样例 
     while(T--)
     {
         cin>>N; 
         for(i=1;i<=N;i++)
            cin>>a[i];
         quicksort(1,N);  //数组元素排序 
         for(i=1;i