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题目：
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 |
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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 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!
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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.
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OutputFor each case, print the sorting result, and one line one case.
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Sample Input2
3 2 1 3
9 1 4 7 2 5 8 3 6 9 |
Sample Output1 2 3
1 2 3 4 5 6 7 8 9 |
| Authorlcy |
题目分析：
排序。
代码如下：
/*
* h.cpp
*
* Created on: 2015年1月29日
* Author: Administrator
*/
#include
#include
#include
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]);//最后一个数的后面没有空格 } return 0; }
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