poj 2299 归并排序求逆序数 （可做模板）

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 48077		Accepted: 17533

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

Source

 

#include<string.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;

long long  sum;
int temp[500005];

void sort2(int a[],int l,int mid,int r){
 // memset(temp,0,sizeof(temp));
   int i=l,j=mid+1,k=0;
   while(i<=mid&&j<=r){
      if(a[i]<a[j]){
        temp[k++]=a[i++];
      }
      else{
        temp[k++]=a[j++];
        sum+=mid-i+1;
      }
   }
   while(i<=mid)  temp[k++]=a[i++];
   while(j<=r)    temp[k++]=a[j++];

   for(int i=l,k=0;i<=r;k++,i++)
    a[i]=temp[k];

}

void sort1(int a[],int l,int r){
   int mid;
   if(l<r){
       mid=(l+r)/2;
       sort1(a,l,mid);
       sort1(a,mid+1,r);
       sort2(a,l,mid,r);
   }
}
int main(){
   int n;
int a[500005];
   while(scanf("%d",&n)!=EOF){

      if(n==0)
        break;
       
        sum=0;
      for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
      sort1(a,0,n-1);
      printf("%lld
",sum);


   }
   return 0;
}