Description
Give a set of numbers, output them after sort. You may use any algorithm you like to solve it.
Input
Each input file contains only one case.
Each test case begins with an integer N(0<N<=1000), the size of the set.
The Next line contains N numbers, represent the elements of the set. Each number range in [0..65535]
Output
Output the set in one line after sort.
Each pair of adjacent numbers is separated by one space. There is no space but '\n' at the end of the line.
Sample Input
4
4 15 8 5
Sample Output
4 5 8 15

我的解法 - -
#include<stdio.h>   
int main()   
{   
    int a[50];   
    int l,i,j,t;   
    scanf("%d",&l);   
    for(i=0;i<l;i++)   
        scanf("%d",&a[i]);   
    printf("\n");   
    for(j=0;j<l-1;j++)   
        for(i=0;i<l-j-1;i++)   
            if(a[i]>a[i+1])   
            {   
                t=a[i];   
                a[i]=a[i+1];   
                a[i+1]=t;   
            }   
            for(i=0;i<l;i++)   
                printf("%d ",a[i]);   
            printf("\n");   
  
}  
oj表示【Presentation Error】···
请问是哪里出问题了呢 看了老半天了

看缩进，大概是你最后一行的printf()出嫁了

- - 不懂啊

终于解决了 - - 菜鸟好开心
#include<stdio.h>      
int main()      
{      
    int a[1001];      
    int l,i,j,t;      
    scanf("%d",&l);      
    for(i=0;i<l;i++)      
        scanf("%d",&a[i]);      
      
    for(j=0;j<l-1;j++)      
        for(i=0;i<l-1;i++)      
            if(a[i]>a[i+1])      
            {      
                t=a[i];      
                a[i]=a[i+1];      
                a[i+1]=t;      
            }      
               
                for(i=0;i<l;i++)     
                   if(i<l-1)   
                   {   
                       printf("%d ",a[i]);   
                   }   
             else printf("%d\n",a[i]);   
     return 0;   
               
}  
 改正：1.去掉了 输出时的前一个换行  2.输出l-1个后带空格的数之后再输出最后一个数且带换行 3.开大数组范围（由50改为1001）

[ 本帖最后由 titus 于 2013-2-18 00:20 编辑 ]

这就是了嘛

不过打回车完全可以不用条件判断 试着改改 加油