版权声明：本文为博主原创文章，未经博主同意不得转载。 https://blog.csdn.net/lx417147512/article/details/28855499

***************************************转载请注明出处：http://blog.csdn.net/lttree***************************************

SRE实战 互联网时代守护先锋，助力企业售后服务体系运筹帷幄！一键直达领取阿里云限量特价优惠。



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。就一个排序就能够了，并且数据范围就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;
}

 

 


    

        

            

                

                    

                        
                    

                    

                        

                            扫码关注我们
                        

                        

                            微信号：SRE实战
                        

                        

                            拒绝背锅 运筹帷幄