HDU1040 As Easy As A+B【排序】

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 69843    Accepted Submission(s): 29716


[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

问题链接：HDU1040 As Easy As A+B。
问题描述：参见上文。
问题分析：这是一个排序问题，关键在于处理输入的循环控制。套路要掌握好，输入有多个case。
程序说明：（略）

参考链接：（略）

AC的C++语言程序：
/* HDU1040 As Easy As A+B */

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1000;
int a
;

int main()
{
int t, n;

while(cin >> t) {
while(t--) {
cin >> n;
for(int i=0; i<n; i++)
cin >> a[i];

sort(a, a + n);

for(int i=0; i<n; i++) {
if(i != 0)
cout << " ";
cout << a[i];
}
cout << endl;
}
}

return 0;
}