Marvolo Gaunt's Ring CodeForces - 855B

Posted 2020-10-12 啦啦啦

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·a_i?+?q·a_j?+?r·a_k for given p,?q,?r and array a₁,?a₂,?... a_n 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,?... a_n (?-?10⁹?≤?a_i?≤?10⁹).

Output

Output a single integer the maximum value of p·a_i?+?q·a_j?+?r·a_k that can be obtained provided 1?≤?i?≤?j?≤?k?≤?n.

Example

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.

题解：注意顺序是一定的，i<=j<=k，所以可以枚举中间值j，然后用两个数组分别存第一项的前缀最大值，第三项的后缀最大值，最后加起来遍历一遍就行了。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 
 5 const int maxn=1e5+5;
 6 
 7 ll n,p,q,r;
 8 ll a[maxn],b[maxn],c[maxn];
 9 
10 int main()
11 {   cin>>n>>p>>q>>r;
12     ll temp;
13     for(int i=1;i<=n;i++){
14         cin>>temp;
15         a[i]=p*temp;
16         b[i]=q*temp;
17         c[i]=r*temp;
18     }    
19     temp=a[1];
20     for(int i=1;i<=n;i++){
21         temp=max(temp,a[i]);
22         a[i]=temp;
23     }
24     temp=c[n];
25     for(int i=n;i>=1;i--){
26         temp=max(temp,c[i]);
27         c[i]=temp;
28     }
29     temp=a[1]+b[1]+c[1];
30     for(int i=1;i<=n;i++) temp=max(temp,a[i]+b[i]+c[i]);
31     cout<<temp<<endl;
32 }