2299 Poj Ultra-QuickSort（归并排序求逆序数）

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 46926 Accepted: 17131

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

59105431230

Sample Output

60

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int a[500010],b[500010];long long ans;void mer(int s,int mid,int e){    int i=s,j=mid+1,k=0;    while(i<=mid&&j<=e)    {        if(a[i]>a[j])  //一定是看小的a[i]是否>a[j]，成立的话从i到mid 的a[i]也是一定大于a[j].        {            ans+=mid-i+1;            b[k++]=a[j++];        }        else            b[k++]=a[i++];    }    while(i<=mid)        b[k++]=a[i++];    while(j<=e)        b[k++]=a[j++];    for(i=s,k=0;i<=e;i++,k++)        a[i]=b[k];}void fen(int s,int e){    if(s<e)    {        int mid=(s+e)/2;        fen(s,mid);//分两部分不断递归        fen(mid+1,e);        mer(s,mid,e);    }}int main(){    int n,m,i,j;    ios::sync_with_stdio(false);    while(cin>>n&&n)    {        ans=0;        for(i=0;i<n;i++)        {            cin>>a[i];        }        fen(0,n-1);        cout<<ans<<endl;    }    return 0;}