Maximum Production

Author: akashmodak97

Code Chef/Maximum Production.cpp

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+/* Code Chef */
+/* Title - Maximum Production */
+/* Created By - Akash Modak */
+/* Date - 12/07/2021 */
+
+// Chefland has 7 days in a week. Chef is very conscious about his work done during the week.
+
+// There are two ways he can spend his energy during the week. The first way is to do x units of work every day and the second way is to do y (>x) units of work for the first d (<7) days and to do z (<x) units of work thereafter since he will get tired of working more in the initial few days.
+
+// Find the maximum amount of work he can do during the week if he is free to choose either of the two strategies.
+
+// Input
+// The first line contains an integer T, the number of test cases. Then the test cases follow.
+// Each test case contains a single line of input, four integers d, x, y, z.
+// Output
+// For each testcase, output in a single line the answer to the problem.
+
+// Sample Input
+// 3
+// 1 2 3 1
+// 6 2 3 1
+// 1 2 8 1
+// Sample Output
+// 14
+// 19
+// 14
+
+
+
+#include<bits/stdc++.h>
+#define fast                  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+#define F                     first
+#define S                     second
+#define pb                    push_back
+#define MP                    make_pair
+#define REP(i,a,b)            for (int i = a; i <= b; i++)
+#define FLSH                  fflush(stdout)
+#define count_1(n)            __builtin_popcountll(n) 
+#define max(x,y)              (x>y)?x:y 
+#define min(x,y)              (x<y)?x:y 
+#define mid(s,e)              (s+(e-s)/2) 
+#define mini                  INT_MIN 
+#define maxi		      INT_MAX
+
+const int MOD = 1000000007; 
+const int FMOD = 998244353; 
+using namespace std;
+
+typedef long long int ll;
+typedef vector<int> vi;
+typedef pair<int,int> pi;
+
+int main() {
+	// your code goes here
+	fast;
+	int t;
+	cin>>t;
+	while(t--){
+	    int d,x,y,z;
+	    cin>>d>>x>>y>>z;
+	    ll for_x = 7*x;
+	    ll for_other = (d*y)+((7-d)*z);
+	    if(for_x>for_other){
+	        cout<<for_x<<"\n";
+	    }
+	    else{
+	        cout<<for_other<<"\n";
+	    }
+	}
+	return 0;
+}