Sort it

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4679 Accepted Submission(s): 3250

Problem Description

You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.
For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.

Output

For each case, output the minimum times need to sort it in ascending order on a single line.

Sample Input

1 2 3

4 3 2 1

Sample Output

题意很好理解，直接就想到冒泡排序了。

然而智障，在写的时候wa了一次。。。

代码:

#include<stdio.h>

int main()

{

    int a[];

    int i,j,t,n,ans;

    while(~scanf("%d\n",&n)){

        for(i=;i<n;i++)

            scanf("%d",&a[i]);

            ans=;

        for(j=;j<n-;j++){                  //是n-1,不是n。。。

            for(i=;i<n-j-;i++)             //是n-j-1,不是n-j,智障

            if(a[i]>a[i+]){

                t=a[i];

                a[i]=a[i+];

                a[i+]=t;

                ans++;

            }

        }

       printf("%d\n",ans);

    }

     return ;

}

巴特西

HDU 2689.Sort it-冒泡排序

Sort it

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