Count numbers up to N whose GCD with N is less than that number
Last Updated : 01 Feb, 2023
Given an integer N, the task is to count the values of K ( where 1 ? K? N ), such that 1< GCD(K, N) < K.
Examples:
Input: N = 10
Output: 3
Explanation: The values of K which satisfies the given conditions are:
- K = 4, gcd(4, 10) = 2
- K = 6, gcd(6, 10) = 2
- K = 8, gcd(8, 10) = 2
Input: N = 15
Output: 4
Explanation: The values of K which satisfies the given conditions are:
- K = 6, gcd(6, 15) = 3
- K = 9, gcd(9, 15) = 3
- K = 10, gcd(10, 15) = 5
- K = 12, gcd(12, 15) = 3
Naive Approach: The simplest approach is to iterate over the range [1, N], and check for each number, whether it satisfies the given condition or not. Finally, print the count of such numbers.
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
Efficient Approach: The idea is to find all the numbers in the range [1, N], for which gcd(K, N) = 1 and gcd(K, N) = K and then finally remove all these numbers from N to get the final answer. Follow the steps below to solve the problem:
- Store all the prime factors of N using Sieve of Eratosthenes.
- Store the count of all the numbers in the range [1, N] for which gcd(N, K) = K in a variable count1.
- Store the count of all the numbers in the range [1, N] for which gcd(N, K) = 1 using Euler's Totient Function in a variable count2.
- Remove these numbers from N, by updating ans to N - (count1 + count2).
- Print the value of ans as the result.
Below is the implementation of the above approach:
C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Store the all prime numbers // using Sieve of Eratosthenes const int MAXN = 100001; vector<int> spf; // Function to fill the prime numbers // using Sieve of Eratosthenes void sieve() { // Create a boolean array and // initialize all entries it as 0 int p[MAXN + 1] = { 0 }; p[2] = 1; for (long long int i = 3; i < MAXN; i += 2) { p[i] = 1; } // Push the first prime number spf.push_back(2); for (long long i = 3; i < MAXN; i += 2) { // If the current number is prime if (p[i]) { // Store the prime numbers // in spf which is prime spf.push_back(i); // Mark all its multiples as false for (long long int j = i * i; j < MAXN; j += 2 * i) p[j] = 0; } } } // Function to count the values of K where // 1?K?N such that 1< gcd(K, N) < K void countK(int N) { // Precalculate prime numbers sieve(); int N1 = N; // Store the smallest prime number int div = spf[0]; int index = 1, C = 0; // Store the count of those numbers in the // range [1, N] for which gcd(N, K) = K int count1 = 1; // Store the count of those numbers from // the range [1, N] such that gcd(N, K) = 1 float count2 = N; // Iterate through all prime factors of N while (div * div <= N) { if (N % div == 0) { C = 0; while (N % div == 0) { N /= div; C++; } count1 = count1 * (C + 1); // count2 is determined by // Euler's Totient Function count2 *= (1.0 - (1.0 / (float)div)); } div = spf[index]; index++; } if (N != 1) { count1 *= 2; count2 = count2 * (1.0 - (1.0 / (float)N)); } int ans = N1 - count1 - count2; // Add 1 to result as 1 is contributed // twice due to count1 and count2 ans = ans + 1; // Print the result cout << ans; return; } // Driver Code int main() { // Given Input int N = 10; // Function Call countK(N); return 0; } // Java code to implement the approach import java.util.ArrayList; class GFG { // Store the all prime numbers // using Sieve of Eratosthenes private static final int MAXN = 100001; private static ArrayList<Integer> spf = new ArrayList<>(); // Function to fill the prime numbers // using Sieve of Eratosthenes private static void sieve() { // Create a boolean array and // initialize all entries it as 0 int[] p = new int[MAXN + 1]; p[2] = 1; for (long i = 3; i < MAXN; i += 2) { p[(int)i] = 1; } // Push the first prime number spf.add(2); for (long i = 3; i < MAXN; i += 2) { // If the current number is prime if (p[(int)i] == 1) { // Store the prime numbers // in spf which is prime spf.add((int)i); // Mark all its multiples as false for (long j = i * i; j < MAXN; j += 2 * i) p[(int)j] = 0; } } } // Function to count the values of K where // 1 <= K <= N such that 1< gcd(K, N) < K private static void countK(int N) { // Precalculate prime numbers sieve(); int N1 = N; // Store the smallest prime number int div = spf.get(0); int index = 1, C = 0; // Store the count of those numbers in the // range [1, N] for which gcd(N, K) = K int count1 = 1; // Store the count of those numbers from // the range [1, N] such that gcd(N, K) = 1 double count2 = N; // Iterate through all prime factors of N while (div * div <= N) { if (N % div == 0) { C = 0; while (N % div == 0) { N /= div; C++; } count1 = count1 * (C + 1); // count2 is determined by // Euler's Totient Function count2 *= (1.0 - (1.0 / (double)div)); } div = spf.get(index); index++; } if (N != 1) { count1 *= 2; count2 = count2 * (1.0 - (1.0 / (double)N)); } int ans = N1 - count1 - (int)count2; // Add 1 to result as 1 is contributed // twice due to count1 and count2 ans = ans + 1; // Print the result System.out.println(ans); return; } // Driver Code public static void main(String[] args) { int N = 10; // Function call countK(N); } } // This code is contributed by phasing17 # Python3 program for the above approach # Store the all prime numbers # using Sieve of Eratosthenes MAXN = 100001 spf = [] # Function to fill the prime numbers # using Sieve of Eratosthenes def sieve(): global MAXN # Create a boolean array and # initialize all entries it as 0 p = [0] * (MAXN + 1) p[2] = 1 for i in range(3, MAXN, 2): p[i] = 1 # Push the first prime number spf.append(2) for i in range(3, MAXN, 2): # If the current number is prime if (p[i]): # Store the prime numbers # in spf which is prime spf.append(i) # Mark all its multiples as false for j in range(i * i, MAXN, 2 * i): p[j] = 0 # Function to count the values of K where # 1?K?N such that 1< gcd(K, N) < K def countK(N): # Precalculate prime numbers sieve() N1 = N # Store the smallest prime number div = spf[0] index, C = 1, 0 # Store the count of those numbers in the # range [1, N] for which gcd(N, K) = K count1 = 1 # Store the count of those numbers from # the range [1, N] such that gcd(N, K) = 1 count2 = N # Iterate through all prime factors of N while (div * div <= N): if (N % div == 0): C = 0 while (N % div == 0): N //= div C += 1 count1 = count1 * (C + 1) # count2 is determined by # Euler's Totient Function count2 *= (1.0 - (1.0 / div)) div = spf[index] index += 1 if (N != 1): count1 *= 2 count2 = count2 * (1.0 - (1.0 / N)) ans = N1 - count1 - count2 # Add 1 to result as 1 is contributed # twice due to count1 and count2 ans = ans + 1 # Print the result print(int(ans)) # Driver Code if __name__ == '__main__': # Given Input N = 10 # Function Call countK(N) # This code is contributed by mohit kumar 29
// C# code to implement the approach using System; using System.Collections.Generic; class GFG { // Store the all prime numbers // using Sieve of Eratosthenes private const int MAXN = 100001; private static List<int> spf = new List<int>(); // Function to fill the prime numbers // using Sieve of Eratosthenes private static void sieve() { // Create a boolean array and // initialize all entries it as 0 int[] p = new int[MAXN + 1]; p[2] = 1; for (long i = 3; i < MAXN; i += 2) { p[i] = 1; } // Push the first prime number spf.Add(2); for (long i = 3; i < MAXN; i += 2) { // If the current number is prime if (p[i] == 1) { // Store the prime numbers // in spf which is prime spf.Add((int)i); // Mark all its multiples as false for (long j = i * i; j < MAXN; j += 2 * i) p[j] = 0; } } } // Function to count the values of K where // 1?K?N such that 1< gcd(K, N) < K private static void countK(int N) { // Precalculate prime numbers sieve(); int N1 = N; // Store the smallest prime number int div = spf[0]; int index = 1, C = 0; // Store the count of those numbers in the // range [1, N] for which gcd(N, K) = K int count1 = 1; // Store the count of those numbers from // the range [1, N] such that gcd(N, K) = 1 double count2 = N; // Iterate through all prime factors of N while (div * div <= N) { if (N % div == 0) { C = 0; while (N % div == 0) { N /= div; C++; } count1 = count1 * (C + 1); // count2 is determined by // Euler's Totient Function count2 *= (1.0 - (1.0 / (double)div)); } div = spf[index]; index++; } if (N != 1) { count1 *= 2; count2 = count2 * (1.0 - (1.0 / (double)N)); } int ans = N1 - count1 - (int)count2; // Add 1 to result as 1 is contributed // twice due to count1 and count2 ans = ans + 1; // Print the result Console.WriteLine(ans); return; } // Driver Code static void Main(string[] args) { int N = 10; // Function call countK(N); } } // This code is contributed by phasing17 <script> // Javascript program for the above approach // Store the all prime numbers // using Sieve of Eratosthenes var MAXN = 100001; var spf = []; // Function to fill the prime numbers // using Sieve of Eratosthenes function sieve() { // Create a boolean array and // initialize all entries it as 0 var p = Array(MAXN + 1).fill(0); p[2] = 1; for (var i = 3; i < MAXN; i += 2) { p[i] = 1; } // Push the first prime number spf.push(2); for (var i = 3; i < MAXN; i += 2) { // If the current number is prime if (p[i]) { // Store the prime numbers // in spf which is prime spf.push(i); // Mark all its multiples as false for (var j = i * i; j < MAXN; j += 2 * i) p[j] = 0; } } } // Function to count the values of K where // 1?K?N such that 1< gcd(K, N) < K function countK(N) { // Precalculate prime numbers sieve(); var N1 = N; // Store the smallest prime number var div = spf[0]; var index = 1, C = 0; // Store the count of those numbers in the // range [1, N] for which gcd(N, K) = K var count1 = 1; // Store the count of those numbers from // the range [1, N] such that gcd(N, K) = 1 var count2 = N; // Iterate through all prime factors of N while (div * div <= N) { if (N % div == 0) { C = 0; while (N % div == 0) { N /= div; C++; } count1 = count1 * (C + 1); // count2 is determined by // Euler's Totient Function count2 *= (1.0 - (1.0 / div)); } div = spf[index]; index++; } if (N != 1) { count1 *= 2; count2 = count2 * (1.0 - (1.0 / N)); } var ans = N1 - count1 - count2; // Add 1 to result as 1 is contributed // twice due to count1 and count2 ans = ans + 1; // Print the result document.write(ans); return; } // Driver Code // Given Input var N = 10; // Function Call countK(N); // This code is contributed by rutvik_56. </script> Time Complexity: O(N*log(log(N)))
Auxiliary Space: O(N)
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