poj 2299 Ultra-QuickSort(归并排序，树状数组，线段树)

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

求逆序对数,具体思路是，在归并的过程中计算每个小区间的逆序对数，进而计算出大区间的逆序对数.
归并过程为：比较a[i]和a[j]的大小，若a[i]≤a[j]，则将第一个有序表中的元素a[i]复制到r[k]中，并令i和k分别加上1；否则将第二个有序表中的元素a[j]复制到r[k]中，并令j和k分别加上1，如此循环下去，直到其中一个有序表取完，然后再将另一个有序表中剩余的元素复制到r中从下标k到下标t的单元。归并排序的算法我们通常用递归实现，先把待排序区间[s,t]以中点二分，接着把左边子区间排序，再把右边子区间排序，最后把左区间和右区间用一次归并操作合并成有序的区间[s,t]。

归并实现：直接套模板

#include <iostream>
#include <cstdio>
using namespace std;
int n,a[500005],b[500005];
long long count;
void Merge(int *a,int left,int mid,int right)
{
    int i=left,j=mid+1,k=left;
    while(i!=mid+1&&j!=right+1)
    {
        if(a[i]>a[j])
        {
            //count+=j-k;
            b[k++]=a[j++];
            count+=mid-i+1;
        }
        else
        {
            b[k++]=a[i++];
        }
    }
    while(i!=mid+1)
        b[k++]=a[i++];
    while(j!=right+1)
        b[k++]=a[j++];
    for(i=left; i<=right; i++)
        a[i]=b[i];
}
void Merge_sort(int *a,int left,int right)
{
    if(left==right)
        return ;
    int mid=(left+right)>>1;
    Merge_sort(a,left,mid);
    Merge_sort(a,mid+1,right);
    Merge(a,left,mid,right);
}
int main()
{
    while(scanf("%d",&n),n)
    {
        count=0;
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        Merge_sort(a,0,n-1);
        printf("%lld
",count);
    }
    return 0;
}