杭电HDOJ 1040 解题报告

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27708    Accepted Submission(s): 11774

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

23 2 1 39 1 4 7 2 5 8 3 6 9

 

Sample Output

1 2 31 2 3 4 5 6 7 8 9

 

Author

lcy

基础题。（排序）

解题思路：

使用快速排序对数组进行排序。

注意输出格式，最后一个数字后是没有空格的。

Qsort函数可参考：qsort函数

代码：

#include<stdio.h>int comp(const void *a,const void *b){return *(int *)a-*(int *)b;}main(){    int i,n,t;    int s[1001];    scanf("%d",&n);    while(n--)    {        scanf("%d",&t);        for(i=0;i<t;i++)        {            scanf("%d",&s[i]);        }        qsort(s,t,sizeof(int),comp);        for(i=0;i<t;i++){        printf("%d",s[i]);        if(i!=t-1) printf(" ");        }        printf("\n");    }}