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:
// 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 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 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# 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 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 ?> |
<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> |
yes
Time complexity: O(log(A*B)) for given A and B, as it is using cbrt function
Auxiliary space: O(1)