hdoj-1040-As Easy As A+B（解题报告）

题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=1040

问题描述：

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 61329    Accepted Submission(s): 26366

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

题目分析：给出多组数据，每组数据有N个数，分别将每组数据从小到大排序。

解题思路：交换排序。若后一个数比前一个数小，则两数交换。

代码实现：

#include<stdio.h>
int main (void)
{
    int T,N,i,j,t;
    int a[1010];
    scanf("%d",&T);
    while(T--)
    {
       scanf("%d",&N);
       for(i=0;i<N;i++)
           scanf("%d",&a[i]);
       for(i=0;i<N-1;i++)
       {
           for(j=i+1;j<N;j++)
              if(a[i]>a[j])
              {
                t=a[i];
                a[i]=a[j];
                a[j]=t;
              }
       }
       for(i=0;i<N;i++)
       {
           if(i==0)
              printf("%d",a[i]);
           else
              printf(" %d",a[i]);
       }
      printf("\n");
    }
    return 0;
}