B. 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.
Examples
input
5 1 2 31 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
给出一个有n个数的序列a[1],a[2]……a[n],使得在i<=j<=k的条件下,令p*a[i]+q*a[j]+r*a[k]的值最大并输出这个值
思路
维护一个位置的前后缀left和right,left[i]表示位置i之前的元素与p相乘的最大值,right表示位置i之后的元素与q相乘的最大值,然后枚举每个位置,计算max(left[i]+q*a[i]+right([i])
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e6+10; 8 const int mod=1e9+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 ll Left[maxn]; 12 ll Right[maxn]; 13 ll a[maxn]; 14 int main(int argc, char const *argv[]) 15 { 16 #ifndef ONLINE_JUDGE 17 freopen("/home/wzy/in.txt", "r", stdin); 18 freopen("/home/wzy/out.txt", "w", stdout); 19 srand((unsigned int)time(NULL)); 20 #endif 21 ios::sync_with_stdio(false); 22 cin.tie(0); 23 int n; 24 ll p,q,r; 25 cin>>n>>p>>q>>r; 26 for(int i=0;i<n;i++) 27 cin>>a[i]; 28 Left[0]=a[0]*p; 29 Right[n-1]=a[n-1]*r; 30 for(int i=1;i<n;i++) 31 Left[i]=max(Left[i-1],p*a[i]); 32 for(int i=n-2;i>=0;i--) 33 Right[i]=max(Right[i+1],r*a[i]); 34 ll ans=-INF; 35 for(int i=0;i<n;i++) 36 ans=max(ans,Left[i]+q*a[i]+Right[i]); 37 cout<<ans<<endl; 38 #ifndef ONLINE_JUDGE 39 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 40 #endif 41 return 0; 42 }
Codeforces 855B:Marvolo Gaunt's Ring(枚举,前后缀)
原文地址:https://www.cnblogs.com/Friends-A/p/11372944.html