Topic link: https://vjudge.net/problem/HDU-1394

The reverse number of a given number sequence a1, a2,..., an is a logarithm satisfying I < J and AI > AJ (ai, aj). For a given number sequence a1, a2,..., an, if we move the first m > = 0 number to the end of the seqence, we will get another sequence. In total, there are the following sequences:
a1, a2,..., an-1, an (where m = 0-initial sequence)
a2, a3,..., an, a1 (where m = 1)
a3, a4,..., an, a1, a2 (where m = 2)
... 
An, a1, a2,..., an-1 (where m = n-1)
You are asked to write a program to find the minimum number of inversions in the above sequence.

Analysis: This problem data is not big, direct violence can also be found, here, we use a more efficient method - line segment tree to solve. In fact, only one thing is clear: if the initial inverse logarithm has been found, then when the first item is put i n the last inverse logarithm k will become k-a[i]+n-1-a[i].

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#define L(u) (u<<1)
#define R(u) (u<<1|1)
using namespace std;
const int maxn=5005;
const int INF=0x3f3f3f3f;
int a[maxn];

struct Node {
    int l,r;
    int num;
} node[maxn<<2];

void Pushup(int u) {
    node[u].num=node[L(u)].num+node[R(u)].num;
    return;
}

void Build(int u,int left,int right) {
    node[u].l=left;
    node[u].r=right;
    if(left==right) {
        node[u].num=0;
        return;
    }
    int mid=(left+right)>>1;
    Build(L(u),left,mid);
    Build(R(u),mid+1,right);
    Pushup(u);
}

int Quary(int u,int left,int right) {
    if(left==node[u].l&&right==node[u].r) return node[u].num;
    int mid=(node[u].l+node[u].r)>>1;
    if(mid>=right) return Quary(L(u),left,right);
    else if(mid<left) return Quary(R(u),left,right);
    else return Quary(L(u),left,mid)+Quary(R(u),mid+1,right);
}

void Update(int u,int val) {
    if(node[u].l==node[u].r) {
        node[u].num++;
        return;
    }
    int mid=(node[u].l+node[u].r)>>1;
    if(mid>=val) Update(L(u),val);
    else Update(R(u),val);
    Pushup(u);
}

int main() {
    //freopen("in.txt","r",stdin);
    int n;
    while(~scanf("%d",&n)) {
        for(int i=0; i<n; i++)
            scanf("%d",a+i);
        Build(1,0,n-1);
        int sum=0,ans=INF;
        for(int i=0; i<n; i++) {
            sum+=Quary(1,a[i],n-1);
            Update(1,a[i]);
        }
        ans=min(ans,sum);
        for(int i=0; i<n; i++) {
            sum=sum-a[i]+n-1-a[i];
            ans=min(ans,sum);
        }
        printf("%d\n",ans);
    }

    return 0;
}