ZOJ-2386

问题可转化为求逆序对数，感觉用O(N^2)的排序算法会超时，没试过，我是用的归并排序，在排序的同时统计逆序对，注意merge的时候那个统计的逻辑，还有结果用long long 存，否则溢出，这题暂时排名第一，感觉良好，哈哈

#include<stdio.h>
#include<string.h>

static int array[500000];
static long long res;

static void merge(int *a, int begin, int mid, int end)
{
	int i = begin, j = mid + 1, index = 0;
	while (i <= mid && j <= end)
		if (a[i] <= a[j])
			array[index++] = a[i++];
		else
		{
			array[index++] = a[j++];
			res += mid - i + 1;//main point!!!
		}
	if (i > mid)
		while (j <= end)
			array[index++] = a[j++];
	if (j > end)
		while (i <= mid)
			array[index++] = a[i++];
	memcpy(a + begin, array, index * sizeof(int));
}

static void merge_sort(int *a, int begin, int end)
{
	if (begin == end)
		return;
	else if (begin + 1 == end)
	{
		if (a[begin] > a[end])
		{
			int temp = a[begin];
			a[begin] = a[end];
			a[end] = temp;
			res++;
		}
	}
	else
	{
		int mid = (begin + end) / 2;
		merge_sort(a, begin, mid);
		merge_sort(a, mid + 1, end);
		merge(a, begin, mid, end);
	}
}

int main()
{
	int n, a[500000];
	while (scanf("%d", &n), n)
	{
		int i;
		for (i = 0; i < n; i++)
			scanf("%d", &a[i]);
		res = 0;
		merge_sort(a, 0, n - 1);
		printf("%lld\n", res);
	}
	return 0;
}