CodeForces - 855B Marvolo Gaunt's Ring（dp）

Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly x drops of the potion he made.

Value of x is calculated as maximum of p·ai + q·aj + r·ak for given p, q, r and array a1, a2, ... an such that 1 ≤ i ≤ j ≤ k ≤ n. Help Snape find the value of x. Do note that the value of x may be negative.

Input

First line of input contains 4 integers n, p, q, r ( - 109 ≤ p, q, r ≤ 109, 1 ≤ n ≤ 105).

Next line of input contains n space separated integers a1, a2, ... an ( - 109 ≤ ai ≤ 109).

Output

Output a single integer the maximum value of p·ai + q·aj + r·ak that can be obtained provided 1 ≤ i ≤ j ≤ k ≤ n.

Examples

Input

5 1 2 3
1 2 3 4 5

Output

30

Input

5 1 2 -3
-1 -2 -3 -4 -5

Output

12

Note

In the first sample case, we can take i = j = k = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.

In second sample case, selecting i = j = 1 and k = 5 gives the answer 12.

求max(p*a[i]+q*a[j]+r*a[k]),满足i<=j<=k

首先可以用dp做  dp[j][i]表示前i个数中已经取了j个数的最大值.把p q r存到x[]数组里。

dp[j][i]=max(dp[j-1][i],dp[j][i-1]+x[i]*a[j])    

#include<bits/stdc++.h>///vj B
using namespace std;
const int N=1e5+5;
long long dp[N][4],a[N],x[4];
int main()
{
    long long n,m;
    while(scanf("%lld%lld%lld%lld",&n,&x[1],&x[2],&x[3])!=EOF)
    {
        long long inf=-8e18;
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=3;j++)
                dp[i][j]=inf;
        }
        dp[1][1]=a[1]*x[1];
        for(int i=1;i<=3;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==1)
                    dp[j][i]=max(dp[j-1][i],a[j]*x[1]);
                else
                    dp[j][i]=max(dp[j-1][i],dp[j][i-1]+a[j]*x[i]);
            }
        }
        cout<<dp[n][3]<<'\n';
    }
    return 0;
}

还有一种比较简单的办法  orz

#include<bits/stdc++.h>
using namespace std;
const long long N=-8e18;
int main()
{
    long long n,m,p,q,r;
    while(scanf("%lld%lld%lld%lld",&n,&p,&q,&r)!=EOF)
    {
        long long a=N,b=N,c=N;
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&m);
            a=max(a,m*p);
            b=max(b,a+m*q);
            c=max(c,b+m*r);
        }
        cout<<c<<'\n';
    }
    return 0;
}