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): 12502    Accepted Submission(s): 5006

 

 

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

23 2 1 39 1 4 7 2 5 8 3 6 9

 

 

Sample Output

1 2 31 2 3 4 5 6 7 8 9

 

 

Author

lcy

 

 

 

 

 

/**
 * Presentation:
 *
 * Author: Vincent
 * Time:Sep 20, 2010 4:30:10 PM
 * ZheJiang University
 */
package HDOJ;

import java.io.BufferedInputStream;
import java.util.*;

public class hdoj1040_1 {
	//public class Main {
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		// TODO Auto-generated method stub
		Scanner cin = new Scanner(new BufferedInputStream(System.in));
		int cas = cin.nextInt();
		while(0 != cas)
		{
			int cursize = cin.nextInt();
			cas--;
			List list=new ArrayList();
			for(int i=0;i< cursize ; i++)
			{
				list.add( (Integer)cin.nextInt());
			}
			Collections.sort(list);
			Iterator irs = list.iterator(); //Iterator接口的枚举遍历   
			int isfirst = (Integer)irs.next();    
		    System.out.print(""+isfirst); 
			while(irs.hasNext())
			{   
			    int is = (Integer)irs.next();    
			    System.out.print(" "+is);   
			}   
			System.out.println();  
		}
	}
}