Create Max power of sage.cpp

Max power of sage.cpp

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+/*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:-### Input:- 5 4 6
+1 2 3 4 5 ### Output:- 50 ### Explanation:- Sage can choose index i=5 and j=5. Then the result will be :- 45 + 65 = 20+30 = 50.
+
+Example 2:-
+2 -2 3
+1 2
+
+Output:-
+4
+*/
+#include<bits/stdc++.h>
+using namespace std; 
+#define ll long long
+#define vi vector<int>
+#define v3 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 t2=1;
+    // cin>>t2;
+    For(ti,t2){
+        ll n,p,q;
+        cin>>n>>p>>q;
+        v3 v(n);
+        For(i,n){
+            cin>>v[i];
+        }
+        vector<pll> pref(n);
+        pref[0].fi = v[0];
+        pref[0].se = v[0];
+        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;
+    }
+}