As Easy As A+B

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 42164 Accepted Submission(s): 18001



[align=left]Problem Description[/align]
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!

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each case, print the sorting result, and one line one case.

[align=left]Sample Input[/align]

2
3 2 1 3
9 1 4 7 2 5 8 3 6 9

[align=left]Sample Output[/align]

1 2 3
1 2 3 4 5 6 7 8 9

[align=left]Author[/align]
lcy
确实是水题……

#include<iostream>//1040
#include<algorithm>
using namespace std;
int main()
{
int T,N,i,a[1000];
cin>>T;
while(T--)
{
cin>>N;
for(i=0;i<N;i++)
cin>>a[i];
sort(a,a+N);
for(i=0;i<N-1;i++)
cout<<a[i]<<" ";
cout<<a[i]<<endl;
}
return 0;
}