Program A-归并排序

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
 
此题就是要求逆序数，所以用归本排序较好。
 

#include<iostream>
 
using namespace std;
 
long long  cnt;
 
void merge(int array[],int left,int mid,int right)
{
    int* temp=new int[right-left+1];
    int i,j,p;
    for(i=left,j=mid+1,p=0;i<=mid&&j<=right;p++)
    {
        if(array[i]<=array[j])temp[p]=array[i++];
        else
        {
            temp[p]=array[j++];cnt+=(mid-i+1);
        }
    }
    while(i<=mid)temp[p++]=array[i++];
    while(j<=right)temp[p++]=array[j++];
    for(i=left,p=0;i<=right;i++)array[i]=temp[p++];
    delete temp;
}
 
void mergesort(int array[],int left,int right)
{
    if(left==right)array[left]=array[right];
    else
    {
        int mid=(left+right)/2;
        mergesort(array,left,mid);
        mergesort(array,mid+1,right);
        merge(array,left,mid,right);
    }
}
int main()
{
    int n,array[500005];
    while(cin>>n,n)
    {
        cnt=0;
        for(int i=0;i<n;i++)
            cin>>array[i];
        mergesort(array,0,n-1);
            cout<<cnt<<endl;
    }
    return 0;
}