C++ Exercises: Get the maximum product from a given integer after breaking the integer into the sum of at least two positive integers

C++ Math: Exercise-14 with Solution

Write a C++ programming to get the maximum product from a given integer after breaking the integer into the sum of at least two positive integers.

Example-1:
Input: 12
Output: 81
Explanation: 12 = 3 + 3 + 3 + 3, 3 x 3 × 3 × 3 = 81.
Example-2:
Input: 7
Output: 12
Explanation: 7 = 3 + 2 + 2, 3 x 2 x 2 = 12.

Sample Solution:

C++ Code :

#include <iostream>
#include <cmath>
using namespace std;

int integer_Break(int n) {
            if (n == 2) {
                return 1;
            } else if (n == 3) {
                return 2;
            } else if (n % 3 == 0) {
                return (int) pow(3, n / 3);
            } else if (n % 3 == 1) {
                return 2 * 2 * (int) pow(3, (n - 4) / 3);
            } else {
                return 2 * (int) pow(3, n / 3);
            }
        }
   
int main()
{
    int n = 7;
    cout << "\nMaximum product of " << n << " after breaking the integer is " << integer_Break(n) << endl;
    n = 9;
    cout << "\nMaximum product of " << n << " after breaking the integer is " << integer_Break(n) << endl;
    n = 12;
    cout << "\nMaximum product of " << n << " after breaking the integer is " << integer_Break(n) << endl; 
    return 0;
}

Sample Output:

Maximum product of 7 after breaking the integer is 12

Maximum product of 9 after breaking the integer is 27

Maximum product of 12 after breaking the integer is 81

Flowchart:

C++ Code Editor:

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