Marvolo Gaunt's Ring（巧妙利用前后缀进行模拟）

Description

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.

Sample Input

Input

5 1 2 3
1 2 3 4 5
Output

30
Input

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

12
Hint

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：找出最大的ai*p+aj*q+ak*r。保证i<=j<=k

　　　　   解析：目前我在各个博客中找到了三种解法（其实思想差不多），首先是前后缀数组模拟的：用L[]来记录 i 左边的最大a[]*p值，R[]来记录 i 右边的最大a[]*r最大值，q为中间，最后遍历的时候直接q*a[]来更新最大值就好了：

#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const ll inf=0x3f3f3f3f3f3f3f3f;　　//这注意了，1e18会WA，目前不知道原因
const int maxn = 1e5+10;
ll a[maxn],L[maxn],R[maxn];
int main()
{
    ll n,p,q,r;
    cin>>n>>p>>q>>r;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    L[1]=a[1]*p;
    for(int i=2;i<=n;i++)
        L[i]=max(L[i-1],a[i]*p);
    R[n]=a[n]*r;
    for(int i=n-1;i>=1;i--)
        R[i]=max(R[i+1],a[i]*r);
    ll sum=-inf;
    for(int i=1;i<=n;i++)
        sum=max(sum,L[i]+R[i]+q*a[i]);
    cout<<sum<<endl;
}

　　　　第二种：也是模拟，代码最为简单

#include <iostream>
using namespace std;
typedef long long ll;
const ll inf=0x3f3f3f3f3f3f3f3f;
int main()
{
    int n,p,q,r;
    cin>>n>>p>>q>>r;
    long long x;long long a=-inf,aa=-inf,aaa=-inf;
    while(n--){
        cin>>x;
        a=max(a,p*x);
        aa=max(aa,a+q*x);
        aaa=max(aaa,aa+r*x);
    }
    cout<<aaa<<endl;
}

　　　　第三种：按背包问题来做

#include<iostream>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
typedef long long ll;
const int maxn=1e5+10;
ll a[maxn],b[4],dp[maxn][4];
int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1;i<=3;i++)
          cin>>b[i];
        for(int i=1;i<=n;i++)
          cin>>a[i];
        for(int i=0;i<=n;i++)
        {
            dp[i][0]=0;
            for(int j=1;j<=3;j++)
              dp[i][j]=-INF;
        }
        for(int i=1;i<=n;i++)
          for(int j=1;j<=3;j++)
            dp[i][j]=max(dp[i][j],max(dp[i-1][j],dp[i][j-1]+(ll)(a[i]*b[j])));
        cout<<dp[n][3]<<endl;
    }
    return 0;
}

Marvolo Gaunt's Ring（巧妙利用前后缀进行模拟）

原文地址：https://www.cnblogs.com/liyexin/p/12339734.html