Ultra-QuickSort

Ultra-QuickSort

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=85904#problem/A

题目：

     

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,
Ultra-QuickSort produces the output 
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0


题意：
    给出一个序列，求这个序列的逆序对个数。
方法：
   用归并排序（时间复杂度为O（nlongn））法进行排序，然后在排序过程中计数即可。

#include<iostream>
using namespace std;
const int maxn=500010;
long long a[maxn],b[maxn],count=0;
int n;
void merge(long long*A,int x,int y,long long*T)
{
    if(y-x>1)
    {
       int m=x+(y-x)/2;
        int p=x,q=m,i=x;
         merge(A,x,m,T);
         merge(A,m,y,T);
            while(p<m||q<y)
            {
              if(q>=y||(p<m&&A[p]<=A[q]))
                   T[i++]=A[p++];
                   else 
                   {
                       T[i++]=A[q++];
                       count+=m-p;
                   }
            }
            for(int j=x;j<y;j++)
                A[j]=T[j];
    }
}

int main()
{
    while(cin>>n&&n)
    {    
    count=0;
    for(int j=0;j<n;j++)
        cin>>a[j];
   merge(a,0,n,b);
    cout<<count<<endl;
    }
return 0;
}