ACM-简单题之As Easy As A+B——hdu1040

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As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35503    Accepted Submission(s): 15331

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
   
   
   
   

 

Author

lcy

 

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1040

做一道简单题吧，题目也说了，和A+B一样简单。

求一些数的升序，就一个排序就可以了，而且数据范围就1000.

/******************************************* 
******************************************** 
*           Author:Tree                    * 
*    From :  blog.csdn.net/lttree          * 
*      Title : As Easy As A+B              * 
*    Source: hdu 1040                      * 
*      Hint :  简单题                      * 
******************************************** 
*******************************************/  

#include <stdio.h>
#include <algorithm>
using namespace std;
int a[1005];
int main()
{
	int i,n,t;
	scanf("%d",&t);
	while( t-- )
	{
		scanf("%d",&n);
		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[n-1]);
	}
	return 0;
}