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

?

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#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();  }