归并排序、求逆序对

归并排序

#include<iostream>
using namespace std;
int n;
int a[5000005];
int b[5000005];
long long ans = 0;
void Merge(int a[], int s, int m, int e, int tmp[])
{
	int pb = 0;
	int p1 = s, p2 = m + 1;
	while (p1 <= m && p2 <= e)
	{
		if (a[p1] < a[p2])
			tmp[pb++] = a[p1++];
		else
		{
			tmp[pb++] = a[p2++];
			//ans += (long long)m - p2 + 1;//与归并排序的唯一区别在这里
		}
			
	}
	while (p1 <= m)
		tmp[pb++] = a[p1++];
	while (p2 <= e)
		tmp[pb++] = a[p2++];
	for (int i = s; i < e + 1; i++)
		a[i] = tmp[i - s];
}
void MergeSort(int a[], int s, int e, int tmp[])
{
	if (s < e)
	{
		int m = s + (e - s) / 2;
		MergeSort(a, s, m, tmp);
		MergeSort(a, m + 1, e, tmp);
		Merge(a, s, m, e, tmp);
	}
}
int main()
{
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	MergeSort(a, 0, n - 1, b);
	cout << ans;
	return 0;
}

求逆序对

#include <iostream>
#include<cstdio>
using namespace std;
int n, a[5000005], b[5000005];
long long ans;
void MerageSortandCount(int left, int right)
{
	//主排序部分↓
	int mid = (left + right) / 2;
	if (left == right)
	{
		return;
	}
	else
	{
		MerageSortandCount(left, mid);
		MerageSortandCount(mid + 1, right);
	}
	//归并部分函数↓
	int i = left;
	int j = mid + 1;
	int m = left;
	while (i <= mid && j <= right)
	{
		if (a[i] > a[j])
		{
			ans += (long long)mid - i + 1;
			//因为要求的是逆序对，也就是左边大右边小的数对，因为1~mid之间的数也就是i
			//代表的数都相对j来着左边，因此如上统计，并且i右边的数都要计入。
			b[m++] = a[j++];
		}
		else
		{
			b[m++] = a[i++];
		}
	}
	while (i <= mid)
	{
		b[m++] = a[i++];
	}
	while (j <= right)
	{
		b[m++] = a[j++];
	}
	for (int i = left; i <= right; ++i)
	{
		a[i] = b[i];
	}
}
int main()
{
	scanf_s("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf_s("%d", &a[i]);
	MerageSortandCount(1, n);
	printf("%lld", ans);
	return 0;
}