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Count ways to represent an integer as an exponent
  • Difficulty Level : Basic
  • Last Updated : 05 Mar, 2021
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Given an integer N, the task is to count the number of ways in which N can be expressed as an exponent, i.e., xy, where x and y are positive integers.

Examples:

 
 

Input: N = 64
Output: 4
Explanation: 64 can be expressed as 26, 43, 82 and 641

Input: N = 27
Output: 2



 

 

Approach: The idea to solve the given problem is to find the prime factorization of the number N and then, find the number of prime factors of the GCD of exponents of the prime factors of the given number N.

 

Below is the implementation of the above approach:

 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate GCD of a
// and b using Euclidean Algorithm
long long int gcd(long long int a,
                  long long int b)
{
    // Iterate until b is non-zero
    while (b > 0) {
 
        long long int rem = a % b;
        a = b;
        b = rem;
    }
 
    // Return the GCD
    return a;
}
 
// Function to count the number of
// ways N can be expressed as x^y
int countNumberOfWays(long long int n)
{
    // Base Case
    if (n == 1)
        return -1;
 
    // Stores the gcd of powers
    long long int g = 0;
 
    int power = 0;
 
    // Calculate the degree of 2 in N
    while (n % 2 == 0) {
        power++;
        n /= 2;
    }
 
    g = gcd(g, power);
 
    // Calculate the degree of prime numbers in N
    for (int i = 3; i <= sqrt(n); i += 2) {
        power = 0;
 
        // Calculate the degree of
        // prime 'i' in N
        while (n % i == 0) {
 
            power++;
            n /= i;
        }
        g = gcd(g, power);
    }
 
    // If N is a prime, g becomes 1.
    if (n > 2)
        g = gcd(g, 1);
 
    // Stores the number of ways
    // to represent N as x^y
    int ways = 1;
 
    // Find the number of Factors of g
    power = 0;
 
    while (g % 2 == 0) {
        g /= 2;
        power++;
    }
 
    // Update the count of ways
    ways *= (power + 1);
 
    // Iterate to find rest of the prime numbers
    for (int i = 3; i <= sqrt(g); i += 2) {
        power = 0;
 
        // Find the power of i
        while (g % i == 0) {
            power++;
            g /= i;
        }
 
        // Update the count of ways
        ways *= (power + 1);
    }
 
    // If g is prime
    if (g > 2)
        ways *= 2;
 
    // Return the total number of ways
    return ways;
}
 
// Driver Code
int main()
{
    int N = 64;
    cout << countNumberOfWays(N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to calculate GCD of a
// and b using Euclidean Algorithm
static int gcd(int a,
                  int b)
{
    // Iterate until b is non-zero
    while (b > 0) {
 
        int rem = a % b;
        a = b;
        b = rem;
    }
 
    // Return the GCD
    return a;
}
 
// Function to count the number of
// ways N can be expressed as x^y
static int countNumberOfWays(int n)
{
   
    // Base Case
    if (n == 1)
        return -1;
 
    // Stores the gcd of powers
    int g = 0;
 
    int power = 0;
 
    // Calculate the degree of 2 in N
    while (n % 2 == 0) {
        power++;
        n /= 2;
    }
 
    g = gcd(g, power);
 
    // Calculate the degree of prime numbers in N
    for (int i = 3; i <= (int)Math.sqrt(n); i += 2) {
        power = 0;
 
        // Calculate the degree of
        // prime 'i' in N
        while (n % i == 0) {
 
            power++;
            n /= i;
        }
        g = gcd(g, power);
    }
 
    // If N is a prime, g becomes 1.
    if (n > 2)
        g = gcd(g, 1);
 
    // Stores the number of ways
    // to represent N as x^y
    int ways = 1;
 
    // Find the number of Factors of g
    power = 0;
 
    while (g % 2 == 0) {
        g /= 2;
        power++;
    }
 
    // Update the count of ways
    ways *= (power + 1);
 
    // Iterate to find rest of the prime numbers
    for (int i = 3; i <= (int)Math.sqrt(g); i += 2) {
        power = 0;
 
        // Find the power of i
        while (g % i == 0) {
            power++;
            g /= i;
        }
 
        // Update the count of ways
        ways *= (power + 1);
    }
 
    // If g is prime
    if (g > 2)
        ways *= 2;
 
    // Return the total number of ways
    return ways;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 64;
    System.out.print(countNumberOfWays(N));
}
}
 
// This code is contributed by code_hunt.

Python3




# Python3 program for the above approach
import math
 
# Function to calculate GCD of a
# and b using Euclidean Algorithm
def gcd(a, b) :
 
    # Iterate until b is non-zero
    while (b > 0) :
        rem = a % b
        a = b
        b = rem
     
    # Return the GCD
    return a
 
# Function to count the number of
# ways N can be expressed as x^y
def countNumberOfWays(n) :
     
    # Base Case
    if (n == 1) :
        return -1
 
    # Stores the gcd of powers
    g = 0
    power = 0
 
    # Calculate the degree of 2 in N
    while (n % 2 == 0) :
        power += 1
        n //= 2
    g = gcd(g, power)
 
    # Calculate the degree of prime numbers in N
    for i in range(3, int(math. sqrt(g)) + 1, 2):
        power = 0
 
        # Calculate the degree of
        # prime 'i' in N
        while (n % i == 0) :
            power += 1
            n //= i      
        g = gcd(g, power)
     
    # If N is a prime, g becomes 1.
    if (n > 2) :
        g = gcd(g, 1)
 
    # Stores the number of ways
    # to represent N as x^y
    ways = 1
 
    # Find the number of Factors of g
    power = 0
    while (g % 2 == 0) :
        g //= 2
        power += 1
     
    # Update the count of ways
    ways *= (power + 1)
 
    # Iterate to find rest of the prime numbers
    for i in range(3, int(math. sqrt(g)) + 1, 2):
        power = 0
 
        # Find the power of i
        while (g % i == 0) :
            power += 1
            g /= i
         
        # Update the count of ways
        ways *= (power + 1)
     
    # If g is prime
    if (g > 2) :
        ways *= 2
 
    # Return the total number of ways
    return ways
 
# Driver Code
N = 64
print(countNumberOfWays(N))
 
# This code is contributed by sanjoy_62.

C#




// C# program for the above approach
using System;
public class GFG
{
 
  // Function to calculate GCD of a
  // and b using Euclidean Algorithm
  static int gcd(int a,
                 int b)
  {
 
    // Iterate until b is non-zero
    while (b > 0)
    {
      int rem = a % b;
      a = b;
      b = rem;
    }
 
    // Return the GCD
    return a;
  }
 
  // Function to count the number of
  // ways N can be expressed as x^y
  static int countNumberOfWays(int n)
  {
 
    // Base Case
    if (n == 1)
      return -1;
 
    // Stores the gcd of powers
    int g = 0;
    int power = 0;
 
    // Calculate the degree of 2 in N
    while (n % 2 == 0)
    {
      power++;
      n /= 2;
    }
    g = gcd(g, power);
 
    // Calculate the degree of prime numbers in N
    for (int i = 3; i <= (int)Math.Sqrt(n); i += 2)
    {
      power = 0;
 
      // Calculate the degree of
      // prime 'i' in N
      while (n % i == 0)
      {
        power++;
        n /= i;
      }
      g = gcd(g, power);
    }
 
    // If N is a prime, g becomes 1.
    if (n > 2)
      g = gcd(g, 1);
 
    // Stores the number of ways
    // to represent N as x^y
    int ways = 1;
 
    // Find the number of Factors of g
    power = 0;
    while (g % 2 == 0)
    {
      g /= 2;
      power++;
    }
 
    // Update the count of ways
    ways *= (power + 1);
 
    // Iterate to find rest of the prime numbers
    for (int i = 3; i <= (int)Math.Sqrt(g); i += 2)
    {
      power = 0;
 
      // Find the power of i
      while (g % i == 0)
      {
        power++;
        g /= i;
      }
 
      // Update the count of ways
      ways *= (power + 1);
    }
 
    // If g is prime
    if (g > 2)
      ways *= 2;
 
    // Return the total number of ways
    return ways;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int N = 64;
    Console.Write(countNumberOfWays(N));
  }
}
 
// This code is contributed by shikhasingrajput
Output: 
4

 

Time Complexity: O(√N)
Auxiliary Space: O(1)

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