This is a C++ problem: It is a well-known mathematical fact that every integer g

Question

This is a C++ problem:

It is a well-known mathematical fact that every integer greater than or equal to 2 has a unique factorization into prime numbers. For example:

40 = 2 * 2 * 2 * 5

58 = 2 * 29

9 = 3 * 3

30 = 2 * 3 * 5

100 = 2 * 2 * 5 * 5

Write a recursive function:

void printPrimeFactorization(int n, bool firstcall);

that prints out the prime factorization of the number n. (You will need to figure out for yourself the purpose of the boolean argument firstcall.)

For example, if n == 40, it should print out:

The prime factorization of 40 is 2 * 2 * 2 * 5.

also, our professor ask us to write a loop to find the smallest integer greater than or equal to 2 that divides n,

then print out this integer and call the function recursively on n/k

Explanation / Answer

#include <iostream>
using namespace std;

void printPrimeFactorization(int n,bool firstCall)
{
   if(n<2)
   {
       return;
   }
   else
   {
       int i=0;
       for(i=2;i<=n;i++)
       {
           if(n%i==0)
           {
               if((n/i)<2)
                   cout << i;
               else
                   cout << i << "*";
               break;
           }
       }
       return printPrimeFactorization(n/i,false);
   }
}

int main() {
   // your code goes here
   cout << "40 = ";
   printPrimeFactorization(40,true);
   cout << endl;
  
   cout << "100 = ";
   printPrimeFactorization(100,true);
   cout << endl;
  
   cout << "58 = ";
   printPrimeFactorization(58,true);
   cout << endl;
   return 0;
}

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