HDU1040

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47474 Accepted Submission(s): 20354

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!

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.

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

Author
lcy

用快排写了一下0ms 做个记号

#include "cstring"
#include "cstdio"
#include "string"
#include "iostream"
#include "cstdlib"

using namespace std;

int Compare(const void *elem1, const void *elem2)
{
    return *((int *)(elem1)) - *((int *)(elem2));
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
            int count;
            scanf("%d",&count);
            int a[1005];
            for(int i=0;i<=count-1;i++)
            {
                scanf("%d",&a[i]);
            }
            qsort(a, count, sizeof(int), Compare);
            for(int i=0;i<=count-1;i++)
            {
                printf("%d",a[i]);
                if(i!=count-1)
                    printf(" ");
                else
                    printf("\n");
            }
        }
    }
}