D

The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row. 

The goal is to take cards in such order as to minimize the total number of scored points. 

For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring 

10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000

If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be 

1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150.

Input

The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces.

Output

Output must contain a single integer - the minimal score.

Sample Input

6
10 1 50 50 20 5

Sample Output

3650

区间dp

对于每个区间其内部 的 去掉顺序 对于 下一步 没有 影响

而对于每个更长的序列 都是有更短一些的区间序列 推得

因此 我们很容易的推出了状态转移方程

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int dp[105][105];
int  a[105];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(dp,0x3f,sizeof(dp));
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        for(int i=1; i<n; i++)
            dp[i][i+1]=0;
        //dp[j][k]
        for(int i=2; i<n; i++) //i: lenth of a[i]
        {
            for(int j=1; i+j<=n; j++) // j: begin of a[i];
            {
                int k=i+j;// k : end of a[i]
                for(int l=j+1;l<k;l++)//for each l  cal
                {
                    dp[j][k]=min(dp[j][k],dp[j][l]+dp[l][k]+a[l]*a[j]*a[k]);
                }
            }
        }
        printf("%d
",dp[1][n]);
    }
}