兩種演算法的時間複雜度都是：O(nlogn)

但是，有可能樹狀陣列需要離散化！

所以，由許多元素共同影響下，歸併排序求逆序對   比   樹狀陣列求逆序對  

快

歸併排序：

#include <bits/stdc++.h>
#define ll long long
#define N 100005
using namespace std;
int a[N],tmp[N];
ll ans;
int n;
inline void merge(int l,int m,int r)
{
    int i = l;
    int j = m + 1;
    int k = l;
    while(i <= m && j <= r)
    {
        if(a[i] > a[j])
        {
            tmp[k++] = a[j++];
            ans += m - i + 1;
        }
        else
        {
            tmp[k++] = a[i++];
        }
    }
    while(i <= m) tmp[k++] = a[i++];
    while(j <= r) tmp[k++] = a[j++];
    for(int i=l;i<=r;i++) a[i]=tmp[i];
}
void merge_sort(int l,int r)
{
	int mid=(l+r)/2;
	if(l<r)
	{
		merge_sort(l,mid);
		merge_sort(mid+1,r);
		merge(l,mid,r);
	}
}
int main()
{
	scanf("%d",&n);
	for(int i = 1;i <= n;++i)
	{
		scanf("%d",&a[i]);
	}
	merge_sort(1,n);
	printf("%lld",ans);
	return 0;
}
 

樹狀陣列：

//樹狀陣列 求逆序對 
//找了好久，發現多取了一個mod 
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cstdlib>
#define ll long long
#define N 500005
using namespace std;
const int mod=99999997;

int n;
ll c[N];
int lowbit(int x)
{
	return x&(-x);
}
void add(int u,int v)
{
	if(!u)return ;
	while(u<=n)
	{
		c[u]+=v;
		u+=lowbit(u);
	}
}
ll query(int y)
{
	if(!y)return 0;
	ll ans=0;
	while(y>0)
	{
		ans+=c[y];
		y-=lowbit(y);
	}
	return ans;
}
struct node
{
	int num;
	int val;
}a[N];
ll seq[N];
int cmp(node x,node y)
{
	return x.val<y.val;
}
int main()
{
	scanf("%d",&n);
	for (int i = 1;i <= n;i++)
	{
        scanf("%d",&a[i].val);
        a[i].num=i;
    }
    sort(a+1,a+1+n,cmp);
    int tt=0;
	for (int i = 1;i <= n;i++) //離散化 
	{
		if(a[i].val==a[i-1].val) //注意 相等 
		{
			seq[a[i].num]=tt;	
			continue;
		}
		seq[a[i].num]=++tt;
	}
	ll cnt=0;
	for(int i = 1;i <= n;i++)
	{
		add(seq[i],1);
		cnt+=(i-query(seq[i]));
	}
	printf("%lld\n",cnt);
	return 0 ;
}