在每一步将A和B乘以K和K ^ 2来达到

我们给了两个数字A和B，我们需要编写一个程序来确定是否可以按照给定的步骤从(1，1)开始到达A和B。从(1，1)，并在每个步骤开始选择一个随机数K和乘法K至在先前步骤和K ²到其它数量获得的两个数字中的任一个。
例子：

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 

解决此问题的想法是密切观察，在每一步中，我们将k和k ²乘以数字。因此，如果可以达到A和B，则每一步的k ^ 3都将作为A * B中的因子。简而言之，如果数字A * B是一个完美的立方体，并且将A和B都除，那么只有执行指定的步骤才能从1和1开始获得数字。
下面是上述想法的实现：

C++

// CPP program to determine if
// A and B can be reached starting
// from 1, 1 following the given steps.
#include 
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.

Javascript

 
// 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.

输出：

Yes