Maximize the product of four factors of a Number in C++

C++Server Side ProgrammingProgramming

Given the task is to calculate the maximum product that can be obtained from four factors A, B, C, D of a given number N, given the condition −

The sum of the four factors should be equal to the number N, that is, N=A+B+C+D.

Input − N=10

Output − 20

Explanation − The factors of 10 are: 1, 2, 5, 10.

Maximum product can be obtained by multiplying 5*2*2*1=20 and also it satisfies the given condition, that is, 5+2+2+1=10.

Input − N=16

Output − 256

Explanation − The factors of 16 are: 1, 2, 4, 8, 16.

Maximum product can be obtained by multiplying 4*4*4*4=256 and also it satisfies the given condition, that is, 4+4+4+4=16.

Approach used in the below program as follows

  • Create an array Factors[] of type int to store the factors of given number and a variable K=0 of type int to keep track of the size of array occupied .

  • Create a function FindFactors() to find the factors of given numbers.

  • Loop from i=1; i*i<=N; i++

  • Inside the loop set if (N%i == 0) to check if I is a factor or not.

  • If i is a factor, then check if (N/I == i). If yes then insert i into the Factors[] else pass both N/i and i into Factors[].

  • Create function Product() to find the maximum product out of the factors.

  • Initialize int product=0; and size=K+1;

  • Initialize four new nested loops and run them till ‘size’.

  • Inside the loops, initialize int sum= Factors[i] + Factors[] + Factors[k] + Factors[l];

  • Check if (sum == N) and if so then initialize pro=Factors[i] * Factors[j] * Factors[k] * Factors[l];

  • Then check if (pro > product), if so then put product=pro;

  • return product

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
//Array to store the factors
int Factors[30];
int K=0;
//Function to find out the factors
int FindFactors(int N){
   //Looping until i reaches the sqrt(N)
   for (int i = 1; i * i <= N; i++){
      if (N % i == 0){
         /* if both the factors are same then only one will be inserted*/
         if ((N / i) == i){
            Factors[K]=i;
            K++;
         }
         else{
            //Inserting 1st factor in array
            Factors[K]=N/i;
            K++;
            //Inserting 2st factor in array
            Factors[K]=i;
            K++;
         }
      }
   }
}
// Function to find the maximum product
int Product(int N){
   int product = 0;
   int size = K+1;
   for (int i = 0; i < size; i++)
      for (int j = 0; j < size; j++)
         for (int k = 0; k < size; k++)
            for (int l = 0; l < size; l++){
               //Adding each set of factors
               int sum = Factors[i] + Factors[j] + Factors[k] + Factors[l];
               //Checking if the sum is equal to N
               if (sum == N){
                  //Multiplying the factors
                  int pro = Factors[i] * Factors[j] * Factors[k] * Factors[l];
                  //Replacing the value of product if a larger value is found
                  if(pro > product)
                     product = pro;
               }
            }
   return product;
}
//Main function
int main(){
   int N = 10;
   //Calling function to find factors of N
   FindFactors(N);
   //Calling function to find the maximum product
   cout<<Product(N);
   return 0;
}

Output

If we run the above code we will get the following output −

Maximum Profit: 20
raja
Published on 14-Aug-2020 07:23:49
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