(Hdu step 1.3.7) As easy as a+b (sort)

Topic:

As easy as A+b

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)

Total submission (s): 2678 Accepted Submission (s): 1280

Problem Descriptionthese days, I am thinking on 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 was to sort these number ascending (ascending). Should know how easy the problem is now! Good luck!

Inputinput contains multiple test cases. The first line of the input was a single integer T which is the number of test cases. T test Cases follow. Each test case contains a integer n (1<=n<=1000 the number of integers to is sorted) and then N integers follow in The same line. It is guarantied, and all integers am in the range of 32-int.

Outputfor, 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

Authorlcy

Topic Analysis:

Sort.

The code is as follows:

/* * h.cpp * *  Created on:2015 January 29 *      author:administrator * * #include <iostream> #include <cstdio># Include <algorithm>using namespace Std;const int maxn = 1005;int A[maxn];int main () {int t;scanf ("%d", &t); while (t--) {int n;scanf ("%d", &n), int i;for (i = 0; i < n; ++i) {scanf ("%d", &a[i]);} Sort (a,a+n); for (i = 0; i < n-1; ++i) {printf ("%d", A[i]);} printf ("%d\n", A[i]);//After the last number there is no space}return 0;}

(Hdu step 1.3.7) As easy as a+b (sort)