去年用樹狀陣列做過這道題，今年思路一樣，多虧這是個n的全排列，每動一個數，你知道會有多少數發生改變。第五次手寫線段樹，寫過了高興。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;

int a[5010];

struct node
{
    int l,r;
    int sum;
    int lazy;
    void update(int v)
    {
        sum+=(r-l+1)*v;
        lazy+=v;
    }
};

node tree[5010*4];

inline void push_down(int rt)
{
    int lazyval=tree[rt].lazy;
    tree[rt*2].update(lazyval);
    tree[rt*2+1].update(lazyval);
    tree[rt].lazy=0;
}

inline void push_up(int rt)
{
    tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;
}

inline void build_tree(int l,int r,int rt)
{
    tree[rt].l=l;
    tree[rt].r=r;
    tree[rt].sum=tree[rt].lazy=0;
    if (l==r) tree[rt].sum=0;
    else
    {
        int mid=(r+l)/2;
        build_tree(l,mid,rt*2);
        build_tree(mid+1,r,rt*2+1);
        push_up(rt);
    }
    //cout<<"l="<<l<<" r="<<r<<" rt="<<rt<<" sum="<<tree[rt].sum<<endl;
}

inline void update(int l,int r,int v,int rt)
{
    int L=tree[rt].l;
    int R=tree[rt].r;
    if (l<=L && R<=r) tree[rt].update(v);
    else
    {
        push_down(rt);
        int mid=L+R>>1;
        if (l<=mid) update(l,r,v,rt*2);
        if (mid<r) update(l,r,v,rt*2+1);
        push_up(rt);
    }
}

inline int query(int l,int r,int rt)
{
    int L=tree[rt].l;
    int R=tree[rt].r;
    if (l<=L && r>=R)
    {
        return tree[rt].sum;
    }
    else
    {
        push_down(rt);
        int mid=L+R>>1;
        int res=0;
        if (l<=mid) res+=query(l,r,rt*2);
        if (r>mid) res+=query(l,r,rt*2+1);
        push_up(rt);
        //cout<<"l="<<l<<" r="<<r<<" rt="<<rt<<" L="<<L<<" R="<<R<<endl;
        //cout<<"res="<<res<<endl;
        return res;
    }
}

int n;

int main()
{
	while (~scanf("%d",&n))
	{
		for (int i=1; i<=n; i++) scanf("%d",&a[i]);
		build_tree(1,n,1);
		int tm=0; int ans;
		for (int i=1; i<=n; i++)
		{
			tm+=query(1,n-a[i],1);
			update(n-a[i],n-a[i],1,1);
		}
		ans=tm;
		for (int i=1; i<=n-1; i++)
	    {
	        update(n-a[i],n-a[i],-1,1);
	        tm+=query(1,n-a[i],1)-a[i];
	        update(n-a[i],n-a[i],1,1);
	    	if (ans>tm) ans=tm;
		}
		printf("%d\n",ans);
	}
	return 0;
 }