Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
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 inversion number out of the above sequences.

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

Output

For each case, output the minimum inversion number on a single line.

 

Sample Input

 

Sample Output

 

#include<iostream>

#include<string.h>

#include<stdio.h>

using namespace std;

const int N=;

struct node

{

    int l,r;

    int num;

}tree[N*];

void build(int k,int l,int r)//建空树

{

    int mid=(l+r)/;

    tree[k].l=l;

    tree[k].r=r;

    tree[k].num=;

    if(l==r) return ;

    build(k*,l,mid);

    build(k*+,mid+,r);

}

void updata(int k,int c)//插入

{

    if(tree[k].l==c&&tree[k].r==c)

    {

        tree[k].num=;

        return ;

    }

    int mid=(tree[k].l+tree[k].r)/;

    if(c<=mid)

        updata(k*,c);

    else updata(k*+,c);

    tree[k].num=tree[k*].num+tree[k*+].num;

}

int get_sum(int k,int c,int n)//统计

{

    if(c<=tree[k].l&&tree[k].r<=n) return tree[k].num;

    else

    {

        int mid=(tree[k].l+tree[k].r)/;

        int sum1=,sum2=;

        if(c<=mid)

            sum1=get_sum(k*,c,n);

        if(n>mid)

            sum2=get_sum(k*+,c,n);

        return sum1+sum2;

    }

}

int main()

{

    int n;

    while(scanf("%d",&n)>)

    {

        int a[N];

        build(,,n-);

        int ans=;

        for(int i=;i<n;i++)

        {

            scanf("%d",&a[i]);

            ans+=get_sum(,a[i]+,n-);
//每输入一个数时，检验一下先他输入的并比他大的数的个数，对于前面比他大的数来说，他是他们的逆序数

            updata(,a[i]);

        }

        int minx=ans;

        for(int i=;i<n;i++)

        {

            ans=ans+n-*a[i]-;           //当a[i]由第一个变为最后一个时，要加上a[i]后面大于a[i]的数的个数，有n-1-a[i]个，要

            if(ans<minx)                    //减去a[i]后面小于a[i]的数的个数，有a[i]个（注意i是从0开始的）

                minx=ans;

        }

        printf("%d\n",minx);

    }

    return ;

}

#include<iostream>

#include<stdio.h>

#include<string.h>

#include<algorithm>

using namespace std;

const int N=+;

int a[N],b[N],c[N],ans[N];

int n,cnt,sum;

int lowbit(int x)

{

    return x&(-x);

}

int query(int x)

{

    int res=;

    x++;

    while(x<=n)

    {

        res+=c[x];

        x+=lowbit(x);

    }

    return res;

}

void update(int x)

{

    while(x)//对在他前且比他小的数，逆序数加1；

    {

        c[x]++;

        x-=lowbit(x);

    }

}

int main()

{

    while(~scanf("%d",&n))

    {

        memset(a,,sizeof(a));

        memset(c,,sizeof(c));

        int sum=;

        for(int i=;i<=n;i++)

        {

            scanf("%d",&a[i]);

            update(a[i]+);

            sum+=query(a[i]+);

        }

        int tmp=,s=sum;

        for(int i=;i<=n;i++)

        {

            tmp=sum-a[i]+(n-a[i]-);

            sum=tmp;

            if(tmp<s)

                s=tmp;

        }

        printf("%d\n",s);

    }

    return ;

}