hdoj1040 As Easy As A+B

As Easy As A+B

[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

 

比较水的一个题，只需用一下快排就行。

代码如下：

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#define N 1010

int s
;

int cmp(const void *a,const void *b)

    {return *(int *)a-*(int *)b;

    }

int main()

{int T;

scanf("%d",&T);

while(T--)

    {int n,i;

    scanf("%d",&n);

    memset(s,0,sizeof(s));

    for(i=1;i<=n;i++)

        scanf("%d",&s[i]);

        qsort(s+1,n,sizeof(s[1]),cmp);

    for(i=1;i<n;i++)

    printf("%d ",s[i]);

    printf("%d\n",s[i]);

    }   

return 0;

}