Ultra-QuickSort 求逆序对 归并排序

Problem 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

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求逆序对   归并排序

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#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
using namespace std;
int a[500001],c[500001];
int n,i,j,k;
long long sum;
void merge(int le,int m,int ri)//归并
{
    int it,jt;
    int p=0;
    it=le,jt=m+1;
    while(it<=m&&jt<=ri)
    {
        if(a[it]>a[jt])
        {
            c[p++]=a[jt++];
            sum+=m-it+1;
        }
        else
        {
            c[p++]=a[it++];
        }
    }
    while(it<=m) c[p++]=a[it++];
    while(jt<=ri) c[p++]=a[jt++];
    for(it=0;it<p;it++)
    {
        a[le+it]=c[it];
    }
}
void mergesort(int le,int ri)//二分
{
    if(le<ri)
   {
    int mid=(le+ri)>>1;
    mergesort(le,mid);
    mergesort(mid+1,ri);
    merge(le,mid,ri);
   }
}
int main()
{
    while(scanf("%d",&n)&&n)
    {
        for(i=0;i<n;i++)
         scanf("%d",&a[i]);
        sum=0;
        mergesort(0,n-1);
        printf("%lld
",sum);
    }
    return 0;
}