POJ 2299 Ultra-QuickSort

        今晚学习了逆序数，基本上照着别人的程序写的。利用归并排序来统计逆序数。

        当归并两条有序链时，可以同时计算逆序数。

        当left<=i<=mid,mid+1<=j<=right,如果有a[i]>a[j]，这样的说明在a的前段中i...mid的元素都比a[j]大，于是逆序数+=mid-i +1；如果a[i]<a[j]，这样的发生说明属于正常排序.

        这题如果是用O(n^2)的方法统计会超时，注意要用long long保存，用long会WA

#include <iostream>
#include <cstring>
#include <cstdio>
#define MAX 500005
using namespace std;

long long tmp[MAX];
long long array[MAX];
long long ans;

void merge_sort(int left, int right);
void merge(int left, int mid, int right);


int main()
{
	int n;
	while (cin >> n && n)
	{
		for (int i = 0; i < n; i++)
			scanf("%lld", &array[i]);
		
		ans = 0;		
		merge_sort(0, n-1);
		cout << ans << endl;
	}
	return 0;
}

void merge_sort(int left, int right)
{
	if (left < right)
	{
		int mid = (left + right)/2;
		merge_sort(left, mid);
		merge_sort(mid+1, right);
		merge(left, mid, right);
	}
}

void merge(int left, int mid, int right)
{
	int begin1, begin2;
	int end1, end2;
	int i = 0;

	begin1 = left;
	end1 = mid;
	begin2 = mid+1;
	end2 = right;

	while (begin1 <= end1 && begin2 <= end2)
	{
		if (array[begin1] <= array[begin2])	
			tmp[i++] = array[begin1++];
		else
		{
			tmp[i++] = array[begin2];
			ans += mid - begin1 + 1;   //计算逆序数
			begin2++;
		}	
	}

	while (begin1 <= end1)
		tmp[i++] = array[begin1++];
	
	while (begin2 <= end2)
		tmp[i++] = array[begin2++];
	
	for (int j = 0; j < i; j++)
		array[left+j] = tmp[j];
}