Hdoj 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): 36141 accepted submission (s): 15642


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 waking night.
Give you some integers, your task is to sort these number ascending (ascending ).
You shoshould 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

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

 

You can use the quick sort order.

#include<stdio.h>#include<stdlib.h>int a[1002];int cmp (const void *a,const void *b){return *(int *)a-*(int *)b;}int main(){int T;scanf("%d",&T);while(T--){int N,i;scanf("%d ",&N);for(i=0;i<N;i++)scanf("%d",&a[i]);qsort(a,N,sizeof(a[0]),cmp);printf("%d",a[0]);for(i=1;i<N;i++)printf(" %d",a[i]);printf("\n");}return 0;}