标签：space find line bit ons def ted lower alt
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.
Example
5 1 2 3
1 2 3 4 5
30
5 1 2 -3
-1 -2 -3 -4 -5
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 }
Marvolo Gaunt's Ring CodeForces - 855B
标签：space find line bit ons def ted lower alt
原文地址：http://www.cnblogs.com/zgglj-com/p/7745426.html
