Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 32387 Accepted Submission(s): 13919

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

#include
int main()
{
 int amount,num[2000],a,t,i,cache,b,o;
 scanf("%d",&amount);
 for(t=0;tnum[i+1])
 {
 cache=num[i];
 num[i]=num[i+1];
 num[i+1]=cache;
 }
 }
 }
 for(i=0;i

 
水题一道，冒泡排序的应用。还是要注意一下冒泡排序中两个for()语句的用法，好久不写又忘了= =
还有就是数组不要开的太小，一开始num[100]，结果直接提示WA。本来还要欢乐的一次过呢……
今天下午考了马克思理论，考完感觉着实不爽，来A水题了……
	

绝对值排序

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35426    Accepted Submission(s): 17365

#include
#include
int main()
{
 int n,i,j,k,cache;
 double num[100];
 while(~scanf("%d",&n),n!=0)
 {
 for(i=0;ifabs(num[i]))
 {
 cache=num[i];
 num[i]=num[j];
 num[j]=cache;
 }
 }
 }
 for(i=0;i0)
 {
 printf(" ");
 }
 printf("%.0lf",num[i]);
 }
 printf("n");
 }
 return 0;
}