Description

给定 8 个数，如果将它们排成一列，每个数的权值是它与相邻的数之积，求一个排列方式，所有数的权值之和最大，输出该权值和.

例如 13242315 的权值和为 1∗3+3∗1∗2+2∗3∗4+4∗2∗2+2∗4∗3+3∗2∗1+1∗3∗5+5∗1=99

Input

每组数据一行 8 个空格隔开的数 1≤��≤100.

Output

求最大的权值和.

Sample

#0

Input

Copy

1 3 2 4 2 3 1 5
3 5 6 8 7 1 2 6
1 2 3 4 5 6 7 8
23 99 63 67 20 64 44 51

Output

Copy

190
1129
987
1483816

回溯法秒了

#include<iostream>
#include<cmath>
#include"stdio.h"
#include "string"
using namespace std;
#include "cstring"
#include "vector"
#include "algorithm"
void calAns(vector<int>&path,int &Max){int max_temp=0;for(int i=0;i<8;i++){if(i==0){max_temp+=path[0]*path[1];}else if(i==7){max_temp+=path[7]*path[6];}else{max_temp+=path[i]*path[i-1]*path[i+1];}}
//    cout<<"max="<<max_temp<<endl;Max=max(max_temp,Max);
}
void backtrack(int a[],int startIndex,vector<int>&path,vector<int>&contain,int &Max){if(path.size()==8){calAns(path,Max);return;}for(int i=0;i<8;i++){if(find(contain.begin(), contain.end(),i)==contain.end()){path.push_back(a[i]);contain.push_back(i);backtrack(a,startIndex+1,path,contain,Max);path.pop_back();contain.pop_back();}}
}
int b[10];
int main()
{while(cin>>b[0]){for(int i=1;i<8;i++){cin>>b[i];}int Max=0;vector<int>path,contain;backtrack(b,0,path,contain,Max);cout<<Max<<endl;}return 0;
}