高效算法——A 归并排序

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

解题思路：这个题目求的就是一串数的逆序数，但是必须用到归并排序，归并：关键在于如何把两个有序表合成一个，每次只需要把两个序列的最小元素加以比较，删除其中的较小元素并加入合并后的新表即可。思路：先把序列分成元素个数尽量相等的两半，r,l在两边尽量控制逆序对的个数，对于右边的每个数，统计左边比它大的个数。
程序代码：

#include<iostream>
#include<cstdio>
using namespace std;
long long cnt;
int A[],T[];
void merge_sort(int l,int r)
{
    if(r-l>)
    {
        int m=l+(r-l)/;
        int p=l,q=m,i=l;
        merge_sort(l,m);
        merge_sort(m,r);
        while(p<m || q<r)
        {
            if(q>=r || (p<m && A[p]<=A[q]) ) T[i++]=A[p++];
            else {
                T[i++]=A[q++];
                cnt+=m-p;
                }
        }
        for(i=l;i<r;++i) A[i]=T[i];
    }
}
int main()
{
    int n;
    while(cin>>n&&n)
    {
        cnt=;
        for(int i=;i<n;++i)
           scanf("%d",&A[i]);
        merge_sort(,n);
        cout<<cnt<<endl;
    }
    return ;
}