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Reach A and B by multiplying them with K and K^2 at every step

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We are given two numbers A and B, we need to write a program to determine if A and B can be reached starting from (1, 1) following the given steps. Start from (1, 1) and at every step choose a random number K and multiply K to any one of the two numbers obtained in the previous step and K2 to the other number. 
Examples: 

Input: A = 3,   B = 9 
Output: yes
Explanation: Starting from A = 1 and B = 1. 
We choose k=3 and multiply 3 with the first number to get A=3 and multiply k2=9 to the second-number to get B=9. 

Input: A = 60,   B = 450 
Output: yes
Explanation : Starting from A = 1 and B = 1,
Step 1: multiply k=3 and k2 to get 3 and 9 
Step 2: Multiply k=5 and k2 = 25 to get to 15 and 225 
Step 3: Multiply k2=4 and k=2 to get to A=60 and B=450 


 


The idea to solve this problem is to observe closely that at each step we are multiplying k and k2 to the numbers. So if A and B can be reached, it will have k^3 at every step as factors in A*B. In simple words, if the number A*B is a perfect cube and it divides A and B both, only then the number can be reached starting from 1 and 1 by performing given steps. 

Below is the implementation of above idea: 

C++

// CPP program to determine if
// A and B can be reached starting
// from 1, 1 following the given steps.
#include <bits/stdc++.h>
using namespace std;
  
// function to check is it is possible to reach
// A and B starting from 1 and 1
bool possibleToReach(int a, int b)
{
    // find the cuberoot of the number
    int c = cbrt(a * b);
  
    // divide the number by cuberoot
    int re1 = a / c;
    int re2 = b / c;
  
    // if it is a perfect cuberoot and divides a and b
    if ((re1 * re1 * re2 == a) && (re2 * re2 * re1 == b))
        return true;
    else
        return false;
}
  
int main()
{
    int A = 60, B = 450;
  
    if (possibleToReach(A, B))
        cout << "yes";
    else
        cout << "no";
  
    return 0;
}

                    

Java

// Java program to determine if
// A and B can be reached starting
// from 1, 1 following the given 
// steps.
class GFG {
      
    // function to check is it is 
    // possible to reach A and B 
    // starting from 1 and 1
    static boolean possibleToReach(int a, int b)
    {
          
        // find the cuberoot of the number
        int c = (int)Math.cbrt(a * b);
  
        // divide the number by cuberoot
        int re1 = a / c;
        int re2 = b / c;
  
        // if it is a perfect cuberoot and 
        // divides a and b
        if ((re1 * re1 * re2 == a) && 
                         (re2 * re2 * re1 == b))
            return true;
        else
            return false;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int A = 60, B = 450;
  
        if (possibleToReach(A, B))
            System.out.println("yes");
        else
            System.out.println("no");
    }
}
  
// This code is contributed by
// Smitha Dinesh Semwal

                    

Python 3

# Python 3 program to determine if
# A and B can be reached starting
# from 1, 1 following the given steps.
import numpy as np
  
# function to check is it is possible to 
# reach A and B starting from 1 and 1
def possibleToReach(a, b):
  
    # find the cuberoot of the number
    c = np.cbrt(a * b)
  
    # divide the number by cuberoot
    re1 = a // c
    re2 = b // c
  
    # if it is a perfect cuberoot and
    # divides a and b
    if ((re1 * re1 * re2 == a) and 
        (re2 * re2 * re1 == b)):
        return True
    else:
        return False
  
# Driver Code
if __name__ == "__main__":
      
    A = 60
    B = 450
  
    if (possibleToReach(A, B)):
        print("yes")
    else:
        print("no")
  
# This code is contributed by ita_c

                    

C#

// C# program to determine if
// A and B can be reached starting
// from 1, 1 following the given 
// steps.
using System;
  
public class GFG{
      
    // function to check is it is 
    // possible to reach A and B 
    // starting from 1 and 1
    public static bool possibleToReach(int a, int b)
    {
          
        // find the cuberoot of the number
        int c = (int)Math.Pow(a * b, (double) 1 / 3);
        // divide the number by cuberoot
        int re1 = a / c;
        int re2 = b / c;
  
        // if it is a perfect cuberoot  
        // and divides a and b
        if ((re1 * re1 * re2 == a) && 
              (re2 * re2 * re1 == b))
            return true;
        else
            return false;
    }
  
// Driver Code 
    static public void Main (String []args)
    {
          
        int A = 60, B = 450;
  
        if (possibleToReach(A, B))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code is contributed by Ajit.

                    

PHP

<?php
// PHP program to determine if
// A and B can be reached starting
// from 1, 1 following the given steps.
  
// function to check is it is
// possible to reach A and B 
// starting from 1 and 1
function possibleToReach($a, $b)
{
      
    // find the cuberoot
    // of the number
    $c =($a * $b);
  
    // divide the number
    // by cuberoot
    $re1 = $a / $c;
    $re2 = $b / $c;
  
    // if it is a perfect cuberoot
    // and divides a and b
    if (($re1 * $re1 * $re2 == $a) && 
        ($re2 * $re2 * $re1 == $b))
        return 1;
    else
        return -1;
}
  
    // Driver Code
    $A = 60; 
    $B = 450;
    if (possibleToReach($A, $B))
        echo "yes";
    else
        echo "no";
          
// This code is contributed by aj_36
?>

                    

Javascript

<script>
  
// JavaScript program to determine if
// A and B can be reached starting
// from 1, 1 following the given steps.
  
    // function to check is it is 
    // possible to reach A and B 
    // starting from 1 and 1
    function possibleToReach(a, b)
    {
            
        // find the cuberoot of the number
        let c = Math.cbrt(a * b);
    
        // divide the number by cuberoot
        let re1 = a / c;
        let re2 = b / c;
    
        // if it is a perfect cuberoot and 
        // divides a and b
        if ((re1 * re1 * re2 == a) && 
                         (re2 * re2 * re1 == b))
            return true;
        else
            return false;
    }
  
// Driver code
  
        let A = 60, B = 450;
    
        if (possibleToReach(A, B))
            document.write("Yes");
        else
            document.write("No");
  
// This code is contributed by splevel62.
</script>

                    

Output
yes

Time complexity: O(log(A*B)) for given A and B, as it is using cbrt function
Auxiliary space: O(1)



Last Updated : 20 Feb, 2023
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