poj 2299 Ultra-QuickSort

                                              http://poj.org/problem?id=2299                     

                                                                 Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 40977		Accepted: 14817

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

题意就是求逆对数。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
__int64 a[5000005],c[5000005];
__int64 k;
void px(int low,int mid,int hight)
{
    int i=low,cnt=0,j=mid+1;
    while(i<=mid&&j<=hight)
     {
        if(a[i]>=a[j])
              {
                 c[cnt++]=a[j++];
                  k+=mid-i+1;
              }
         else
           c[cnt++]=a[i++];
     }
     while(i<=mid)
     {
         c[cnt++]=a[i++];
     }
     while(j<=hight)
     {
         c[cnt++]=a[j++];
     }
      i=low;cnt=0;
     while(i<=hight)
     {
         a[i++]=c[cnt++];
     }
}
void gb(int low,int hight)
{
    int mid=(low+hight)/2;
    if(low<hight)
    {
    gb(low,mid);
    gb(mid+1,hight);
    px(low,mid,hight);
    }
}
int main()
{
     int i,n;
    while(~scanf("%d",&n)&&n!=0)
    {
        k=0;
      for(i=0;i<n;i++)
       scanf("%I64d",&a[i]);
       gb(0,n-1);
       printf("%I64d\n",k);
    }
    return 0;
}