标签：acm
As Easy As A+B |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 2882 Accepted Submission(s): 1357 |
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Problem 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 (升序). 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 32-int. |
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Output
For each case, print the sorting result, and one line one case.
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Sample Input
2
3 2 1 3
9 1 4 7 2 5 8 3 6 9
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Sample Output
1 2 3
1 2 3 4 5 6 7 8 9
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Author
lcy
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#include<stdio.h>
#include<stdlib.h>
int cmp(const void *a,const void *b)
{
return *(int *)a - *(int *)b;//由小到大排序
}
int main()
{
int n,m;
int i;
int data[1001]={0};
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
for(i=0;i<m;i++)
{
scanf("%d",&data[i]);
}
qsort(data,m,sizeof(int),cmp);
for(i=0;i<m-1;i++)
{
printf("%d ",data[i]);
}
printf("%d\n",data[i]);
}
return 0;
}
标签：acm
原文地址：http://blog.csdn.net/u010275850/article/details/45623729
