hdu 2689 Sort it(树状数组)

http://acm.hdu.edu.cn/showproblem.php?pid=2689

Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2261    Accepted Submission(s): 1628

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

 

 

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看到这道题，说是树状数组，想了一下，八竿子打不到的都可以用树状数组？？？

然后还是乖乖的去看了题解，才知道……

如果你说，我不懂，怎么也不懂，那就看看这个吧：http://www.cnblogs.com/xiaohongmao/archive/2012/05/28/2520710.html 

 

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>

int n;
int c[1005];

int Lowbit(int x)
{
    return x&(-x);
}

void Update(int x,int i)
{
    while(x<=n)
    {
        c[x]+=i;
        x+=Lowbit(x);
    }
}

int Sum(int  x)
{
    int s=0;
    while(x>0)
    {
        s+=c[x];
        x-=Lowbit(x);
    }
    return s;
}

int main()
{
    int i,j,m;
    while(scanf("%d",&n)!=EOF)
    {
         memset(c,0,sizeof(c));
         int sum=0;
         for(i=1;i<=n;i++)
         {
             scanf("%d",&m);
             Update(m,1);
             sum+=(m-Sum(m));
         }
         printf("%d
",sum);
    }
    return 0;
}