描述

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.

输入

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.

输出

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.

样例输入

59105431230

样例输出

60

题目来源

 

求最小的交换次数，即求逆序对数问题，并归排序。

 

【并归排序】

如下演示并归排序的整个过程：

并归排序主要是深搜实现的。

{9,1,0,4,5,8,7,4,3}=>{9,1,0,4,5} {8,7,4,3}

{9,1,0,4,5}=>{9,1,0} {4,5}=>{9}{1}{0} {4}{5}

{8,7,4,3}=>{8,7} {4,3}=>{8}{7} {4}{3}

合并子表

{0,1,9} {4,5}  {7,8} {3,4}

{0,1,4,5,9} {3,4,7,8}

{0,1,3,4,4,7,8,9}

 

#include 
     
      __int64 sum;void mergeSort(int* a,int low,int mid,int high){	int* p=new int[high+1];	int i=low;	int j=low;//左侧表的起始位置 	int h=mid+1;//右侧表的起始位置 	while(h<=high&&j<=mid){		if(a[j]<=a[h]){			p[i]=a[j];			j++;			i++;		}else{			//求逆序数 			sum+=h-i;			p[i]=a[h];			h++;			i++;		}	}	for(;j<=mid;j++,i++){		p[i]=a[j];	}	for(;h<=high;h++,i++){		p[i]=a[h];	}	for(i=low;i<=high;i++){		a[i]=p[i];	}	delete[] p;}void merge(int* a,int low,int high){	if(low
      
       >1;		//划分子表 		merge(a,low,mid);		merge(a,mid+1,high);		//合并子表 		mergeSort(a,low,mid,high); 	}}int main(){	int n;	int arr[500010];	while(scanf("%d",&n)!=EOF && n){		for(int i=0; i