(Hdu step 1.3.7) As Easy As A + B (SORT), hdu1.3.7

(Hdu step 1.3.7) As Easy As A + B (SORT), hdu1.3.7

Question:

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 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

Authorlcy

Question Analysis:

Sort.

The Code is as follows:

/** H. cpp ** Created on: January 29, 2015 * 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]); // there is no space behind the last number} return 0 ;}