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Count ways to represent an integer as an exponent
  • Difficulty Level : Basic
  • Last Updated : 25 Feb, 2021

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.


Output: 

4

 

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

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