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HDU2689-Sort it-冒泡排序

2024-10-18 20:26:39   211人阅读

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
3
1 2 3
4
4 3 2 1
 

 

Sample Output
0
6
 
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[10000];
    int i,j,t,n,ans;
    while(~scanf("%d\n",&n)){
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
            ans=0;
        for(j=0;j<n-1;j++){                  //是n-1,不是n。。。
            for(i=0;i<n-j-1;i++)             //是n-j-1,不是n-j,智障
            if(a[i]>a[i+1]){
                t=a[i];
                a[i]=a[i+1];
                a[i+1]=t;
                ans++;
            }
        }
       printf("%d\n",ans);
    }
     return 0;
}

。。。

HDU2689-Sort it-冒泡排序


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