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题面：
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Easy as A+B
Time Limit:1000MS Memory Limit:32768K
Description:
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. You should know how easy the problem is now! Good luck!
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Input:
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 32bit-int.
Output:
For each case, print the sorting result, and one line one case.
Sample Input:
2
3 2 1 3
9 1 4 7 2 5 8 3 6 9
Sample Output:
1 2 3
1 2 3 4 5 6 7 8 9
Source:
lcy
题解：
冒泡排序一下，套了下别人写的，居然是错的，以后还是自己写吧。
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代码：
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#include
int main()
{
int temp,t,n;
int store[1005];
scanf(%d,&t);
while(t--)
{
scanf(%d,&n);
for(int i=0;i
store[j]) { temp = store[i]; store[i] = store[j]; store[j] = temp; } } } printf(%d,store[0]); for(int i=1;i
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