Hduoj1040【水题】

/*As Easy As A+BTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 46241    Accepted Submission(s): 19757Problem 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 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!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 Input23 2 1 39 1 4 7 2 5 8 3 6 9Sample Output1 2 31 2 3 4 5 6 7 8 9AuthorlcyRecommendWe have carefully selected several similar problems for you:  1062 1391 2673 1236 1056 */#include<stdio.h>#include<stdlib.h>int a[1010];int cmp(const void *a, const void *b){return *(int *)a - *(int *)b;}int main(){int i, j, k, t, n;scanf("%d", &t);while(t--){scanf("%d", &n);for(i = 0; i < n; ++i)scanf("%d", &a[i]); qsort(a, n, sizeof(a[0]), cmp);for(i = 0; i < n-1; ++i)printf("%d ", a[i]);printf("%d\n",a[n-1]);}return 0;}