Manthan, Codefest 17: B. Marvolo Gaunt's Ring

B. Marvolo Gaunt's Ring

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 31 2 3 4 5

output

30

input

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

output

12

题意：

给出三个数p, q, r，再给出n个数(a[1], a[1], … , a[n])，求p*a[i]+q*a[j]+r*a[k]的最大值，要求i<=j<=k

求出p*a[i]的前缀最大值b[i]，再求出r*a[k]的后缀最大值c[i]

最后答案就是max(b[i]+c[i]+q*a[i])  (1<=i<=n)

注意ans初始值一定要小于-1e18

#include<stdio.h>#include<algorithm>using namespace std;#define LL long longLL a[100005], b[100005], c[100005];int main(void){LL n, i, p, q, r, ans;scanf("%I64d%I64d%I64d%I64d", &n, &p, &q, &r);for(i=1;i<=n;i++)scanf("%I64d", &a[i]);for(i=1;i<=n;i++)b[i] = p*a[i];for(i=2;i<=n;i++)b[i] = max(b[i], b[i-1]);for(i=n;i>=1;i--)c[i] = r*a[i];for(i=n-1;i>=1;i--)c[i] = max(c[i], c[i+1]);ans = -(LL)1e18*4;for(i=1;i<=n;i++)ans = max(q*a[i]+b[i]+c[i], ans);printf("%I64d\n", ans);return 0;}/*3 1000000000 1000000000 1000000000-1000000000 -1000000000 -1000000000*/