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Maximize profit when divisibility by two numbers have associated profits in C++
We are given with five integers N, A, B, X and Y. The goal is to maximize the profit by checking that between numbers in range [ 1 to N ] , if
A number is divisible by A, then profit increases by X.
A number is divisible by B then profit increases by Y.
A profit can be added once only, for a particular number in range.
Let’s understand with examples.
Input − N=4, A=2, B=3, X=2, Y=3
Output − Maximized profit is − 7
Explanation −
2,4 are divisible by A ( 2 ). Profit increases from 0 to 2, then 2 to 4 ( by X=2 )
3 is divisible by B ( 3 ). Profit increases from 4 to 7. (by Y=3)
Input − N=5, A=2, B=4, X=1, Y=3
Output − Maximized profit is: 4
Explanation −
2,4 are divisible by A ( 2 ).
4 is also divisible by B ( 4 ).
For 2 profit increases from 0 to 1 (by X). For 4 we chose divisibility by B. So profit increases by Y not X as Y is more. So it rises from 1 to 4 ( by Y=3 ).
Approach used in the below program is as follows
We have integers N, A, B, X, Y.
Function maximizeProfit(int n, int a, int b, int x, int y) calculates the profit and returns the value. It takes all variables as parameters and return maximized profit.
Variable profit will have an amount of profit, initially 0.
Starting from 1 to n, check divisibility of i by a and b, using for loop
If i is divisible by both a and b then increase profit by x or y whichever is more.
Else if i is divisible by a only, increase profit by x
Else if i is divisible by b only, increase profit by y
At the end return value present in profit as result.
Example
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum profit
int maximizeProfit(int n, int a, int b, int x, int y){
int profit=0;
for(int i=1;i<=n;i++){
if(i%a==0 && i%b==0){
int maxx=x>=y?x:y;
profit+=maxx;
}
else if(i%a==0){
profit+=x;
}
else if(i%b==0){
profit+=y;
}
}
return profit;
}
int main(){
int N = 6, A = 2, B =4, X = 6, Y = 3;
cout <<"Maximized profit is: "<<maximizeProfit(N,A,B,X,Y);
return 0;
}Output
If we run the above code it will generate the following output −
Maximized profit is: 2
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