CF855B Marvolo Gaunt's Ring

给了三个数: p,q,r(−1e9<=p,q,r<=1e9)p,q,r(−1e9<=p,q,r<=1e9)

然后给了n个数a1,a2...an(−1e9<=ai<=1e9)a1​,a2​...an​(−1e9<=ai​<=1e9)

求找出三个数ai,aj,ak(1<=i<=j<=k<=n)ai​,aj​,ak​(1<=i<=j<=k<=n)使得p×ai+q×aj+r×akp×ai​+q×aj​+r×ak​最大。

  直接dp了

  f[ i ][ 3] = f[i-1][3] + f[i-1][2]+a3*

 f[i][2] = f[i-1][2]+ f[i-1][1]+a2*

 f[i][1] =f[i-1][1]+ a1

#include <iostream>
#include<queue> 
#include <cstring>
#define IOS std::ios::sync_with_stdio(0)
using namespace std;
 const int N =1e5+3;
 
 #define int long long
 int f[N][4],a[N],n,a1,a2,a3;
 
 void solve(){
 	int i,j;
 	cin>>n>>a1>>a2>>a3;
 	for(i=1;i<=n;i++)cin>>a[i];
 	
 	f[1][1]=a[1]*a1;
 	
 	for(i=2;i<=n;i++){
 		f[i][1]=max(f[i-1][1],a1*a[i]);
 	}
 	
 	f[1][2]=f[1][1]+a[1]*a2;
 	for(i=2;i<=n;i++){
 		f[i][2]=max(f[i-1][2],f[i][1]+a2*a[i]);
 	}
 	
 	f[1][3]=f[1][2]+a[1]*a3;
 	for(i=2;i<=n;i++){
 		f[i][3]=max(f[i-1][3],f[i][2]+a3*a[i]);
 	}
 	cout<<f[n][3]<<endl;
 }
 signed main(){
     IOS;
     solve();
 }