(hdu step 1.3.7)As Easy As A+B(排序)

題目：

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2678 Accepted Submission(s): 1280 Problem DescriptionThese 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!
InputInput 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.
OutputFor 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

Authorlcy

題目分析：

排序。

代碼如下：

/*
 * h.cpp
 *
 *  Created on: 2015年1月29日
 *      Author: Administrator
 */

#include 
#include 
#include 


using namespace std;

const int maxn = 1005;
int a[maxn];


int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n;
		scanf("%d",&n);
		int i;
		for(i = 0 ; i < n ; ++i){
			scanf("%d",&a[i]);
		}

		sort(a,a+n);

		for(i = 0 ; i < n-1 ; ++i){
			printf("%d ",a[i]);
		}


		printf("%d\n",a[i]);//最後一個數的後面沒有空格
	}

	return 0;
}