杭电Sort it 2689树状数组

Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3510    Accepted Submission(s): 2546

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
   
   
   
   
   
   
   
   

    
    
    
    //虽然是水题，但是对于初学者来说很想通，自己好好推理推理
   
   
   
   
   
   
   
   

    
    
    
    #include <iostream>
#include <cstring>
using namespace std;
const int maxn=1010;
int C[maxn],n;
int lowbit(int x)
{
    return x&(-x);
}
void update(int i)
{
    while(i<=n)
    {
        C[i]+=1;
        i+=lowbit(i);
    }
}
int query(int i)
{
    int sum=0;
    while(i>0)
    {
        sum+=C[i];
        i-=lowbit(i);
    }
    return sum;
}
int main()
{
    while(cin>>n)
    {
        int sum=0;
        memset(C,0,sizeof(C));
        for(int i=1;i<=n;i++)
        {
            int a;
           cin>>a;
            update(a);
            sum+=i-query(a);//得出比i大的的元素个数，即要交换的次数 
        }
    cout<<sum<<endl;
    }
    return 0;
}