區間dp，dp[i][j] = min ( 左右兩邊去掉的數計算的和 + 當前數a[k]*a[i]*a[j] )

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<vector>
#include<map>
#include<algorithm>
#define lson l,m,level*2
#define rson m+1,r,level*2+1
using namespace std;
int a[110];
int dp[110][110];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	memset(dp,0,sizeof(dp));
	for(int l=3;l<=n;l++){
		for(int i=0;i<n;i++){
			int j = i+l-1;
			dp[i][j] = 0x3f3f3f3f;
			for(int k=i+1;k<j;k++){
				dp[i][j] = min(dp[i][j],dp[i][k] + dp[k][j] + a[i]*a[j]*a[k]);
			}
		}
	}
	printf("%d\n",dp[0][n-1]);
	return 0;
}