Ultra-QuickSort - poj 2299 (归并排序+统计逆序数)

利用归并排序统计逆序数，利用归并求逆序在对子序列s1和s2在归并时（s1,s2已经排好序），若s1[i]>s2[j](逆序状况)，则逆序数加上s1.length-i,因为s1中i后面的数字对于s2[j]都是逆序的。

 #include <stdio.h>

 #include <stdlib.h>

 int N;

 int num[];

 int tmp[];

 __int64 count;

 void Merge(int l,int mid,int r){

     int i=l,j=mid+,k=;

     while(i<=mid&&j<=r){

      if(num[i]>num[j]) {

         tmp[k++]=num[j++];

         count+=mid-i+;

      } else

      {

          tmp[k++]=num[i++];

      }

     }

     while(i<=mid) tmp[k++]=num[i++];

      while(j<=r)  tmp[k++]=num[j++];

         for(i=; i<k; i++)

         {

             num[l+i]=tmp[i];

         }

 }

 void Mergesort(int l,int r){

     int mid=(l+r)/;

     if(l<r){

         Mergesort(l,mid);

         Mergesort(mid+,r);

         Merge(l,mid,r);

     }

 }

 int main(void) {

     scanf("%d",&N);

     int i=;

     while(N){

         for(i=;i<N;i++){

             scanf("%d",&num[i]);

         }

         count=;

         Mergesort(,N-);

         printf("%I64d \n",count);

         scanf("%d",&N);

     }

     return EXIT_SUCCESS;

 }

附：

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 48082		Accepted: 17536

Description

Ultra-QuickSort - poj 2299 (归并排序+统计逆序数)-LMLPHP

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

Sample Output

6

0