题目链接： http://cpp.zjut.edu.cn/ShowProblem.aspx?ShowID=1373
题面：
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!
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。
代码：
#include <stdio.h>
int main()
{
int temp,t,n;
int store[1005];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf ("%d,",&store[i]);
for (int i = 0; i < n; i++)
{
for (int j = i; j < n; j++)
{
if (store[i] > store[j])
{
temp = store[i];
store[i] = store[j];
store[j] = temp;
}
}
}
printf("%d",store[0]);
for(int i=1;i<n;i++)
printf(" %d",store[i] );
printf("\n");
}
return 0;
}