As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 58496    Accepted Submission(s): 25030 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   Recommend We have carefully selected several similar problems for you:   1096  1004  1093  1089  1092    水题，就是排一下序，我直接调用的c++里面的sort函数，进行快排 ac代码 #include <stdio.h> #include <iostream> #include <algorithm> using namespace std; int main() { int i,num,n,a[1005]; scanf("%d",&num); while(num--) { scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&a[i]); } sort(a,a+n); for(i=0;i<n;i++) { if(i==0) printf("%d",a[i]); else printf(" %d",a[i]); } printf("\n"); } return 0; }