zoukankan      html  css  js  c++  java
  • CodeForces

     Marvolo Gaunt's Ring

     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
    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.

    题意:找到a[i]*p+a[j]*q+a[k]*r的最大值,i<=j<=k;

    #include<stdio.h>
    #include<iostream>
    using namespace std;
    int main()
    {
        long long n,a1,a2,a3,a=-8e18,b=-8e18,c=-8e18;
        cin>>n>>a1>>a2>>a3; int x;
        for(int i=0;i<n;i++)
        {
            cin>>x;
            a=max(a,a1*x);
            b=max(b,a+a2*x);
            c=max(c,b+a3*x);
        }
        cout<<c<<endl;
    }




  • 相关阅读:
    NodeJS学习之3:express和Utility的配合使用
    NodeJS学习之2:express版的Hello World
    NodeJS学习之1:express安装
    9:Node.js GET/POST请求
    8:Node.js 文件系统
    7:Node.js 全局对象
    PowerShell工作流学习-4-工作流中重启计算机
    PowerShell工作流学习-3-挂起工作流
    PowerShell工作流学习-2-工作流运行Powershell命令
    PowerShell工作流学习-1-嵌套工作流和嵌套函数
  • 原文地址:https://www.cnblogs.com/da-mei/p/9053311.html
Copyright © 2011-2022 走看看