nlogn求逆序數 POJ 2299解題報告

前幾天自己想出了利用歸併排序求逆序數的方法,找了一個求逆序數的題2299 交了300++MS水過...

 

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 19686		Accepted: 6959

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<iostream>using namespace std;__int64 sum=0;int *guibing(int *data,int n){int i,j,k,L,R,s;int *now;for(i=2;i<n*2;i=i<<1){now=new int[n];for(j=0;j<=n/i;j++){k=L=i*j;R=L+i/2;s=0;while(L<n&&R<n&&L<i*j+i/2&&R<i*(j+1)){if(data[L]<=data[R]){now[k++]=data[L++];sum+=s;}else{now[k++]=data[R++];s++;}}while(L<n&&L<i*j+i/2){now[k++]=data[L++];sum+=s;}while(R<n&&R<(j+1)*i)now[k++]=data[R++];}delete data;data=now;}return data;}int main(){int n,i;int *data;while(cin>>n&&n){sum=0;data=new int[n];for(i=0;i<n;i++)scanf("%d",data+i);data=guibing(data,n);delete data;printf("%I64d/n",sum);}}