 # MAXPOW-Editorial

PRACTICE
Editorialist: Indrajit Kabiraj

Medium

# PREREQUISITES:

Array, Number Theory

# PROBLEM:

Sage has two weapons p and q.
She has an array A of attacks. She wants to choose two indices i and j (1<=i<=j and j<=n) such that her power pA[i]+qA[j] is maximum.

### Input:-

First line contains three integers n,p and q(1<=n<=10^5 and -10^9<=p,q<=10^9).
Second line contains n integers i.e the array of attacks (-10^9<=A[i]<=10^9).

### Output:-

Maximum possible value for pA[i]+qA[j].

## Example 1:-

5 4 6
1 2 3 4 5

50

### Explanation:-

Sage can choose index i=5 and j=5. Then the result will be :- 45 + 65 = 20+30 = 50.

2 -2 3
1 2

4

# EXPLANATION:

As we need to find the maximum value of pa[i] + qa[j] and we know that i<=j.
So, If we iterate over the array and choose an index j such that qa[j] is maximum and i can be any index less than or equals to j.
If we store maximum and minimum value until the jth index and if p is less than 0 then we can multiply the minimum value with p otherwise we can multiply p with the maximum one.
So, If we find the maximum of a[j]q + maximum of(minimum so farp, maximum so far
p) we will get the answer.Note: Minimum and Maximum so far means minimum and maximum value in the array upto the Jth index.

# SOLUTION:

``````#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define FASTANDFURIOUS ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define For(i,n) for(ll i=0;i<n;i++)
#define pll pair <ll,ll>
#define pii pair <int,int>
#define fi first
#define se second
#define INF 3e18
signed main(){
FASTANDFURIOUS;
int tc=1;
// cin>>tc;
For(ti,tc){
ll n,p,q;
cin>>n>>p>>q;
vl v(n);
For(i,n){
cin>>v[i];
}
vector<pll> pref(n);
pref.fi = v;
pref.se = v;
rep(i,1,n){
pref[i].fi = max(pref[i-1].fi,v[i]);
pref[i].se = min(pref[i-1].se,v[i]);
}
ll mx = 0LL,res=-INF;
rep(i,0,n){
ll a1 = p*1LL*pref[i].fi;
ll a2 = p*1LL*pref[i].se;
ll b1 = q*1LL*v[i];
mx = max(a1,a2)+b1;
res = max(res,mx);
}
cout<<res<<endl;
}
}``````