Given a positive integer N, the task is to count the number of integers from the range [1, N] having exactly 5 divisors.
Examples:
Input: N = 18
Output: 1
Explanation:
From all the integers over the range [1, 18], 16 is the only integer that has exactly 5 divisors, i.e. 1, 2, 8, 4 and 16.
Therefore, the count of such integers is 1.Input: N = 100
Output: 2
Naive Approach: The simplest approach to solve the given problem is to iterate over the range [1, N] and count those integers in this range having the count of divisors as 5.
C++
// C++ code for the approach#include <iostream>#include <cmath>using namespace std;void SieveOfEratosthenes(int n, bool prime[], bool primesquare[], int a[]) { //For more details check out: https://www.neveropen.co.uk/sieve-of-eratosthenes/ // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // Not a prime, else true. for (int i = 2; i <= n; i++) prime[i] = true; // Create a boolean array "primesquare[0..n*n+1]" // and initialize all entries it as false. A value // in squareprime[i] will finally be true if i is // square of prime, else false. for (int i = 0; i <= (n * n + 1); i++) primesquare[i] = false; // 1 is not a prime number prime[1] = false; for (int p = 2; p * p <= n; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true) { // Update all multiples of p starting from p * p for (int i = p * p; i <= n; i += p) prime[i] = false; } } int j = 0; for (int p = 2; p <= n; p++) { if (prime[p]) { // Storing primes in an array a[j] = p; // Update value in primesquare[p*p], // if p is prime. primesquare[p * p] = true; j++; } }} // Function to count divisorsint countDivisors(int n) { // If number is 1, then it will have only 1 // as a factor. So, total factors will be 1. if (n == 1) return 1; bool prime[n + 1], primesquare[n * n + 1]; int a[n]; // for storing primes upto n // Calling SieveOfEratosthenes to store prime // factors of n and to store square of prime // factors of n SieveOfEratosthenes(n, prime, primesquare, a); // ans will contain total number of distinct // divisors int ans = 1; // Loop for counting factors of n for (int i = 0;; i++) { // a[i] is not less than cube root n if (a[i] * a[i] * a[i] > n) break; // Calculating power of a[i] in n. int cnt = 1; // cnt is power of prime a[i] in n. while (n % a[i] == 0) // if a[i] is a factor of n { n = n / a[i]; cnt = cnt + 1; // incrementing power } // Calculating the number of divisors // If n = a^p * b^q then total divisors of n // are (p+1)*(q+1) ans = ans * cnt; } // if a[i] is greater than cube root of n // First case if (prime[n]) ans = ans * 2; // Second case else if (primesquare[n]) ans = ans * 3; // Third case else if (n != 1) ans = ans * 4; return ans; // Total divisors}// Function to count the number of integers with exactly 5 divisorsint countIntegers(int n) { // Store count of numbers with exactly 5 divisors int count = 0; // loop from 1 to n to check its distinct count of divisors for (int i = 1; i <= n; i++) { // Function Call int divisors = countDivisors(i); // If the number of divisors is 5, check if it is a prime square if (divisors == 5 ) { count++; } } return count;}// Driver codeint main() { int n = 100; cout << countIntegers(n) << endl; return 0;} |
Java
// Java code for the approachimport java.util.Vector;public class GFG { static void SieveOfEratosthenes(int n, boolean prime[], boolean primesquare[], int a[]) { // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // Not a prime, else true. for (int i = 2; i <= n; i++) prime[i] = true; /* Create a boolean array "primesquare[0..n*n+1]" and initialize all entries it as false. A value in squareprime[i] will finally be true if i is square of prime, else false.*/ for (int i = 0; i < ((n * n) + 1); i++) primesquare[i] = false; // 1 is not a prime number prime[1] = false; for (int p = 2; p * p <= n; p++) { // If prime[p] is not changed, // then it is a prime if (prime[p] == true) { // Update all multiples of p for (int i = p * 2; i <= n; i += p) prime[i] = false; } } int j = 0; for (int p = 2; p <= n; p++) { if (prime[p]) { // Storing primes in an array a[j] = p; // Update value in // primesquare[p*p], // if p is prime. primesquare[p * p] = true; j++; } } } // Function to count divisors static int countDivisors(int n) { // If number is 1, then it will // have only 1 as a factor. So, // total factors will be 1. if (n == 1) return 1; boolean prime[] = new boolean[n + 1]; boolean primesquare[] = new boolean[(n * n) + 1]; // for storing primes upto n int a[] = new int[n]; // Calling SieveOfEratosthenes to // store prime factors of n and to // store square of prime factors of n SieveOfEratosthenes(n, prime, primesquare, a); // ans will contain total number // of distinct divisors int ans = 1; // Loop for counting factors of n for (int i = 0;; i++) { // a[i] is not less than cube root n if (a[i] * a[i] * a[i] > n) break; // Calculating power of a[i] in n. // cnt is power of prime a[i] in n. int cnt = 1; // if a[i] is a factor of n while (n % a[i] == 0) { n = n / a[i]; // incrementing power cnt = cnt + 1; } // Calculating the number of divisors // If n = a^p * b^q then total // divisors of n are (p+1)*(q+1) ans = ans * cnt; } // if a[i] is greater than cube root // of n // First case if (prime[n]) ans = ans * 2; // Second case else if (primesquare[n]) ans = ans * 3; // Third case else if (n != 1) ans = ans * 4; return ans; // Total divisors } // Function to count the number of integers with exactly // 5 divisors static int countIntegers(int n) { // Store count of numbers with exactly 5 divisors int count = 0; // loop from 1 to n to check its distinct count of // divisors for (int i = 1; i <= n; i++) { // Function Call int divisors = countDivisors(i); // If the number of divisors is 5, check if it // is a prime square if (divisors == 5) { count++; } } return count; } // Driver code public static void main(String[] args) { int N = 100; System.out.println(countIntegers(N)); }} |
Python3
# Python3 code for the approachimport mathdef sieveOfEratosthenes(n): # Create a boolean array "prime[0..n]" and initialize # all entries it as true. A value in prime[i] will # finally be false if i is Not a prime, else true. prime = [True for i in range(n+1)] p = 2 while(p * p <= n): # If prime[p] is not changed, then it is a prime if (prime[p] == True): # Update all multiples of p for i in range(p * p, n+1, p): prime[i] = False p += 1 # Store prime numbers primes = [] for p in range(2, n+1): if prime[p]: primes.append(p) return primes # Function to count divisorsdef countDivisors(n, primes): # If number is 1, then it will have only 1 as a factor. # So, total factors will be 1. if (n == 1): return 1 ans = 1 # Loop for counting factors of n i = 0 while (primes[i] <= math.sqrt(n)): # a[i] is not less than square root of n cnt = 1 # cnt is power of prime a[i] in n. while (n % primes[i] == 0): # if a[i] is a factor of n n = n // primes[i] cnt += 1 # incrementing power ans = ans * cnt # Calculating the number of divisors i += 1 # if a[i] is greater than square root of n if (n > 1): ans = ans * 2 return ans # Total divisors # Function to count the number of integers with exactly 5 divisorsdef countIntegers(n): # Store count of numbers with exactly 5 divisors count = 0 # Get all prime numbers up to n primes = sieveOfEratosthenes(n) # loop from 1 to n to check its distinct count of divisors for i in range(1, n+1): # Function Call divisors = countDivisors(i, primes) # If the number of divisors is 5, check if it is a prime square if (divisors == 5 and int(math.sqrt(i))**2 == i): count += 1 return count # Driver codeif __name__ == "__main__": # Input integer n = 100 # Function call print(countIntegers(n)) |
Javascript
function sieveOfEratosthenes(n){ // Create a boolean array "prime[0..n]" and initialize // all entries it as true. A value in prime[i] will // finally be false if i is Not a prime, else true. let prime = new Array(n+1).fill(true); let p = 2; while(p * p <= n) { // If prime[p] is not changed, then it is a prime if (prime[p] == true) { // Update all multiples of p for (let i = p * p; i <= n; i += p) { prime[i] = false; } } p += 1; } // Store prime numbers let primes = []; for (let p = 2; p <= n; p++) { if (prime[p] == true) { primes.push(p); } } return primes;}// Function to count divisorsfunction countDivisors(n, primes) { // If number is 1, then it will have only 1 as a factor. // So, total factors will be 1. if (n == 1) { return 1; } let ans = 1; // Loop for counting factors of n let i = 0; while (primes[i] <= Math.sqrt(n)) { // a[i] is not less than square root of n let cnt = 1; // cnt is power of prime a[i] in n. while (n % primes[i] == 0) { // if a[i] is a factor of n n = n / primes[i]; cnt += 1; // incrementing power } ans = ans * cnt; // Calculating the number of divisors i += 1; } // if a[i] is greater than square root of n if (n > 1) { ans = ans * 2; } return ans; // Total divisors}// Function to count the number of integers with exactly 5 divisorsfunction countIntegers(n) { // Store count of numbers with exactly 5 divisors let count = 0; // Get all prime numbers up to n let primes = sieveOfEratosthenes(n); // loop from 1 to n to check its distinct count of divisors for (let i = 1; i <= n; i++) { // Function Call let divisors = countDivisors(i, primes); // If the number of divisors is 5, check if it is a prime square if (divisors == 5 && Math.sqrt(i)**2 == i) { count += 1; } } return count;}// Driver codelet n = 100;console.log(countIntegers(n)); |
Output
2
Time Complexity: O(N4/3)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by observing a fact that the numbers that have exactly 5 divisors can be expressed in the form of p4, where p is a prime number as the count of divisors is exactly 5. Follow the below steps to solve the problem:
- Generate all primes such that their fourth power is less than 1018 by using Sieve of Eratosthenes and store it in vector, say A[].
- Initialize two variables, say low as 0 and high as A.size() – 1.
- For performing the Binary Search iterate until low is less than high and perform the following steps:
- Find the value of mid as the (low + high)/2.
- Find the value of fourth power of element at indices mid (mid – 1) and store it in a variable, say current and previous respectively.
- If the value of current is N, then print the value of A[mid] as the result.
- If the value of current is greater than N and previous is at most N, then print the value of A[mid] as the result.
- If the value of current is greater than N then update the value of high as (mid – 1). Otherwise, update the value of low as (mid + 1).
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>#define ll long long intconst int MAX = 1e5;using namespace std;// Function to calculate the value of// (x^y) using binary exponentiationll power(ll x, unsigned ll y){ // Stores the value of x^y ll res = 1; // Base Case if (x == 0) return 0; while (y > 0) { // If y is odd multiply // x with result if (y & 1) res = (res * x); // Otherwise, divide y by 2 y = y >> 1; x = (x * x); } return res;}// Function to perform the Sieve Of// Eratosthenes to find the prime// number over the range [1, 10^5]void SieveOfEratosthenes( vector<pair<ll, ll> >& v){ bool prime[MAX + 1]; memset(prime, true, sizeof(prime)); prime[1] = false; for (int p = 2; p * p <= MAX; p++) { // If prime[p] is not changed // then it is a prime if (prime[p] == true) { // Set all the multiples of // p to non-prime for (int i = p * 2; i <= MAX; i += p) prime[i] = false; } } int num = 1; // Iterate over the range [1, MAX] for (int i = 1; i <= MAX; i++) { // Store all the prime number if (prime[i]) { v.push_back({ i, num }); num++; } }}// Function to find the primes having// only 5 divisorsint countIntegers(ll n){ // Base Case if (n < 16) { return 0; } // First value of the pair has the // prime number and the second value // has the count of primes till that // prime numbers vector<pair<ll, ll> > v; // Precomputing all the primes SieveOfEratosthenes(v); int low = 0; int high = v.size() - 1; // Perform the Binary search while (low <= high) { int mid = (low + high) / 2; // Calculate the fourth power of // the curr and prev ll curr = power(v[mid].first, 4); ll prev = power(v[mid - 1].first, 4); if (curr == n) { // Return value of mid return v[mid].second; } else if (curr > n and prev <= n) { // Return value of mid-1 return v[mid - 1].second; } else if (curr > n) { // Update the value of high high = mid - 1; } else { // Update the value of low low = mid + 1; } } return 0;}// Driver Codeint main(){ ll N = 100; cout << countIntegers(N); return 0;} |
Java
// Java program for the above approachimport java.util.Vector;public class GFG { static int MAX = (int)1e5; public static class pair { long first; long second; pair(long first, long second) { this.first = first; this.second = second; } } // Function to calculate the value of // (x^y) using binary exponentiation static long power(long x, long y) { // Stores the value of x^y long res = 1; // Base Case if (x == 0) return 0; while (y > 0) { // If y is odd multiply // x with result if ((y & 1) == 1) res = (res * x); // Otherwise, divide y by 2 y = y >> 1; x = (x * x); } return res; } // Function to perform the Sieve Of // Eratosthenes to find the prime // number over the range [1, 10^5] static void SieveOfEratosthenes(Vector<pair> v) { boolean prime[] = new boolean[MAX + 1]; for (int i = 0; i < prime.length; i++) { prime[i] = true; } prime[1] = false; for (int p = 2; p * p <= MAX; p++) { // If prime[p] is not changed // then it is a prime if (prime[p] == true) { // Set all the multiples of // p to non-prime for (int i = p * 2; i <= MAX; i += p) prime[i] = false; } } int num = 1; // Iterate over the range [1, MAX] for (int i = 1; i <= MAX; i++) { // Store all the prime number if (prime[i]) { v.add(new pair(i, num)); num++; } } } // Function to find the primes having // only 5 divisors static long countIntegers(long n) { // Base Case if (n < 16) { return 0; } // First value of the pair has the // prime number and the second value // has the count of primes till that // prime numbers Vector<pair> v = new Vector<>(); // Precomputing all the primes SieveOfEratosthenes(v); int low = 0; int high = v.size() - 1; // Perform the Binary search while (low <= high) { int mid = (low + high) / 2; // Calculate the fourth power of // the curr and prev long curr = power(v.get(mid).first, 4); long prev = power(v.get(mid - 1).first, 4); if (curr == n) { // Return value of mid return v.get(mid).second; } else if (curr > n && prev <= n) { // Return value of mid-1 return v.get(mid - 1).second; } else if (curr > n) { // Update the value of high high = mid - 1; } else { // Update the value of low low = mid + 1; } } return 0; } // Driver code public static void main(String[] args) { long N = 100; System.out.println(countIntegers(N)); }}// This code is contributed by abhinavjain194 |
Python3
# Python program for the above approach# Function to calculate the value of# (x**y) using binary exponentiationdef power(x, y): # Stores the value of x**y res = 1 # Base Case if x == 0: return 0 while y > 0: # If y is odd multiply # x with result if y&1: res = (res*x) # otherwise, divide y by 2 y = y >> 1 x = (x*x) return res# Function to perform the Sieve of# Eratosthenes to find the prime# number over the range [1, 10^5]def sieveofeartoshenes(vec): prime = [] for i in range(pow(10, 5)+1): prime.append(True) prime[1] = False p = 2 while (p * p <= pow(10, 5)): # If prime[p] is not # changed, then it is a prime if (prime[p] == True): # Updating all multiples of # to non-prime for i in range(p * p, pow(10, 5) + 1, p): prime[i] = False p += 1 num = 1 # Iterate over the range [1, pow(10, 5)] for i in range(1, pow(10, 5)+1): # Stores all the prime number if prime[i]: vec.append([i, num]) num += 1def count_integer(n): # Base Case if n < 16: return 0 # First value of the pair has the # prime number and the second value # has the cont of primes till that # prime numbers vec = [[]] # precomputing all the primes sieveofeartoshenes(vec) low = 0 high = len(vec)-1 # perform the Binary Search while low <= high: mid = (low+high)//2 # Calculate the fourth power of # the curr and prev curr = power(vec[mid][0], 4) prev = power(vec[mid-1][0], 4) if curr == n: # Return value of mid return vec[mid][1] elif curr > n and prev <= n: # Return value of mid-1 return vec[mid-1][1] elif curr > n: # Update the value of low high = mid - 1 else: # Update the value of high low = mid + 1n = 100ans = count_integer(n)print(ans)# This code is contributed by Aditya Sharma |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;public class pair { public long first; public long second; public pair(long first, long second) { this.first = first; this.second = second; }}class GFG { static int MAX = (int)1e5; // Function to calculate the value of // (x^y) using binary exponentiation static long power(long x, long y) { // Stores the value of x^y long res = 1; // Base Case if (x == 0) return 0; while (y > 0) { // If y is odd multiply // x with result if ((y & 1) == 1) res = (res * x); // Otherwise, divide y by 2 y = y >> 1; x = (x * x); } return res; } // Function to perform the Sieve Of // Eratosthenes to find the prime // number over the range [1, 10^5] static void SieveOfEratosthenes(List<pair> v) { bool[] prime = new bool[MAX + 1]; for (int i = 0; i < prime.Length; i++) { prime[i] = true; } prime[1] = false; for (int p = 2; p * p <= MAX; p++) { // If prime[p] is not changed // then it is a prime if (prime[p] == true) { // Set all the multiples of // p to non-prime for (int i = p * 2; i <= MAX; i += p) prime[i] = false; } } int num = 1; // Iterate over the range [1, MAX] for (int i = 1; i <= MAX; i++) { // Store all the prime number if (prime[i]) { v.Add(new pair(i, num)); num++; } } } // Function to find the primes having // only 5 divisors static long countIntegers(long n) { // Base Case if (n < 16) { return 0; } // First value of the pair has the // prime number and the second value // has the count of primes till that // prime numbers List<pair> v = new List<pair>(); // Precomputing all the primes SieveOfEratosthenes(v); int low = 0; int high = v.Count - 1; // Perform the Binary search while (low <= high) { int mid = (low + high) / 2; // Calculate the fourth power of // the curr and prev long curr = power(v[mid].first, 4); long prev = power(v[mid - 1].first, 4); if (curr == n) { // Return value of mid return v[mid].second; } else if (curr > n && prev <= n) { // Return value of mid-1 return v[mid - 1].second; } else if (curr > n) { // Update the value of high high = mid - 1; } else { // Update the value of low low = mid + 1; } } return 0; } // Driver code public static void Main(string[] args) { long N = 100; Console.WriteLine(countIntegers(N)); }}// This code is contributed by phasing17 |
Javascript
// JavaScript program for the above approach// Function to calculate the value of// (x**y) using binary exponentiationfunction power(x, y){ // Stores the value of x**y let res = 1; // Base Case if (x === 0) { return 0; } while (y > 0) { // If y is odd multiply // x with result if (y&1) { res = (res*x); } // otherwise, divide y by 2 y = y >> 1; x = (x*x); } return res;}// Function to perform the Sieve of// Eratosthenes to find the prime// number over the range [1, 10^5]function sieveofeartoshenes(vec) { let prime = []; for (let i = 0; i <= Math.pow(10, 5)+1; i++) { prime.push(true); } prime[1] = false; let p = 2; while (p * p <= Math.pow(10, 5)) { // If prime[p] is not // changed, then it is a prime if (prime[p] === true) { // Updating all multiples of // to non-prime for (let i = p * p; i <= Math.pow(10, 5) + 1; i += p) { prime[i] = false; } } p += 1; } let num = 1; // Iterate over the range [1, pow(10, 5)] for (let i = 1; i <= Math.pow(10, 5)+1; i++) { // Stores all the prime number if (prime[i]) { vec.push([i, num]); num += 1; } }}function count_integer(n) { // Base Case if (n < 16) { return 0; } // First value of the pair has the // prime number and the second value // has the cont of primes till that // prime numbers let vec = [[]]; // precomputing all the primes sieveofeartoshenes(vec); let low = 0; let high = vec.length-1; // perform the Binary Search while (low <= high) { let mid = (low+high)//2; // Calculate the fourth power of // the curr and prev let curr = power(vec[mid][0], 4); let prev = power(vec[mid-1][0], 4); if (curr === n) { // Return value of mid return vec[mid][1]; } else if (curr > n && prev <= n) { // Return value of mid-1 return vec[mid-1][1]; } else if (curr > n) { // Update the value of low high = mid - 1; } else { // Update the value of high low = mid + 1 } }}let n = 100let ans = count_integer(n)console.log(ans)// This code is contributed by phasing17 |
Output:
2
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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