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

10

1 3 6 9 0 8 5 7 4 2

Sample Output

16
题目大意，从初始的数组開始，每次把第一个加到最后一个，求逆序数，共求出n个逆序数，找出最小值
对每种情况都求逆序数，能够每次都归并排序，或树状数组，或线段树，也能够由上一个的逆序数推出；
a[1] , a[2] , a[3] 。。。a[n]，将a[1]放到最后，然后将a[2]放到最后，能够找到规律，将首个a[i]放到最后时，逆序数添加了 a[i]之前比a[i]大的，添加a[i]之后比a[i]大的，减小了a[i]之前比a[i]小的，减小了a[i]之后比a[i]小的，又由于每次给出的数n个数在0到n，且都不同，最后得出 逆序数会添加 n-a[i]个，降低a[i]-1个



#include <cstdio>

#include <cstring>

#define INF 0x3f3f3f3f

#include <algorithm>

using namespace std;

int tree[100000] , p[6000] , q[6000];

void update(int o,int x,int y,int u)

{

    if( x == y && x == u )

        tree[o]++ ;

    else

    {

        int mid = (x + y)/ 2;

        if( u <= mid )

            update(o*2,x,mid,u);

        else

            update(o*2+1,mid+1,y,u);

        tree[o] = tree[o*2] + tree[o*2+1];

    }

}

int sum(int o,int x,int y,int i,int j)

{

    int ans = 0 ;

    if( i <= x && y <= j )

        return tree[o] ;

    else

    {

        int mid = (x + y) /2 ;

        if( i <= mid )

            ans += sum(o*2,x,mid,i,j);

        if( mid+1 <= j )

            ans += sum(o*2+1,mid+1,y,i,j);

    }

    return ans ;

}

int main()

{

    int i , j , n , min1 , num ;

    while(scanf("%d", &n)!=EOF)

    {

        min1 = 0 ;

        memset(q,0,sizeof(q));

        for(i = 1 ; i <= n ; i++)

        {

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

            p[i]++ ;

        }

        memset(tree,0,sizeof(tree));

        for(i = 1 ; i <= n ; i++)

        {

            min1 += sum(1,1,n,p[i],n);

            update(1,1,n,p[i]);

        }

        num = min1 ;

        for(i = 1 ; i < n ; i++)

        {

            num = num + ( n - p[i] ) - (p[i] - 1) ;

            if( num < min1 )

                min1 = num ;

        }

        printf("%d\n", min1);

    }

    return 0;

}