UVA 10327

转载于 2013-03-09 11:25:00 发布 · 124 阅读

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原文链接：http://www.cnblogs.com/huzhenbo113/archive/2013/03/09/2951270.html

文章标签：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=24&problem=1268&mosmsg=Submission+received+with+ID+11407977

输入一组数据，求出在排序中数据交换的次数

View Code

#include<stdio.h>
int main()
{
 int a[1010],i,j,sum,n,temp;
 while(scanf("%d",&n)!=EOF)
     {
      sum=0;
      for(i=0;i<n;i++)  
         scanf("%d",&a[i]);
      for(j=1;j<n;j++)
         for(i=0;i<n-j;i++)
            if(a[i]>a[i+1])
              {
               temp=a[i];
               a[i]=a[i+1];
               a[i+1]=temp;
               sum++;
              }
      printf("Minimum exchange operations : %d\n",sum);
     } 
 return 0;
}