Find a positive number M such that gcd(N^M, N&M) is maximum
Given a number N, the task is to find a positive number M such that gcd(N^M, N&M) is the maximum possible and M < N. The task is to print the maximum gcd thus obtained.
Examples:
Input: N = 5 Output: 7 gcd(2^5, 2&5) = 7 Input: N = 15 Output: 5
Approach: There are two cases which need to be solved to get the maximum gcd possible.
- If a minimum of one bit is not set in the number, then M will be a number whose bits are flipped at every position of N. And after that get the maximum gcd.
- If all bits are set, then the answer will the maximum factor of that number except the number itself.
Below is the implementation of the above approach:
C++
// C++ program to implement above approach #include <bits/stdc++.h>using namespace std; // Recursive function to count// the number of unset bitsint countBits(int n){ // Base case if (n == 0) return 0; // unset bit count else return !(n & 1) + countBits(n >> 1);} // Function to return the max gcdint maxGcd(int n){ // If no unset bits if (countBits(n) == 0) { // Find the maximum factor for (int i = 2; i * i <= n; i++) { // Highest factor if (n % i == 0) { return n / i; } } } else { int val = 0; int power = 1; int dupn = n; // Find the flipped bit number while (n) { // If bit is not set if (!(n & 1)) { val += power; } // Next power of 2 power = power * 2; // Right shift the number n = n >> 1; } // Return the answer return __gcd(val ^ dupn, val & dupn); } // If a prime number return 1;} // Driver Codeint main(){ // First example int n = 5; cout << maxGcd(n) << endl; // Second example n = 15; cout << maxGcd(n) << endl; return 0;} |
Java
// Java program to implement above approach class GFG{ static int __gcd(int a,int b){ if(b == 0) return a; return __gcd(b, a % b);} // Recursive function to count// the number of unset bitsstatic int countBits(int n){ // Base case if (n == 0) return 0; // unset bit count else return ((n & 1) == 0 ? 1 : 0) + countBits(n >> 1);} // Function to return the max gcdstatic int maxGcd(int n){ // If no unset bits if (countBits(n) == 0) { // Find the maximum factor for (int i = 2; i * i <= n; i++) { // Highest factor if (n % i == 0) { return n / i; } } } else { int val = 0; int power = 1; int dupn = n; // Find the flipped bit number while (n > 0) { // If bit is not set if ((n & 1) == 0) { val += power; } // Next power of 2 power = power * 2; // Right shift the number n = n >> 1; } // Return the answer return __gcd(val ^ dupn, val & dupn); } // If a prime number return 1;} // Driver Codepublic static void main(String[] args){ // First example int n = 5; System.out.println(maxGcd(n)); // Second example n = 15; System.out.println(maxGcd(n));}} // This code contributed by Rajput-Ji |
Python3
# Python 3 program to implement # above approachfrom math import gcd, sqrt # Recursive function to count# the number of unset bitsdef countBits(n): # Base case if (n == 0): return 0 # unset bit count else: return (((n & 1) == 0) + countBits(n >> 1)) # Function to return the max gcddef maxGcd(n): # If no unset bits if (countBits(n) == 0): # Find the maximum factor for i in range(2, int(sqrt(n)) + 1): # Highest factor if (n % i == 0): return int(n / i) else: val = 0 power = 1 dupn = n # Find the flipped bit number while (n): # If bit is not set if ((n & 1) == 0): val += power # Next power of 2 power = power * 2 # Right shift the number n = n >> 1 # Return the answer return gcd(val ^ dupn, val & dupn) # If a prime number return 1 # Driver Codeif __name__ == '__main__': # First example n = 5 print(maxGcd(n)) # Second example n = 15 print(maxGcd(n)) # This code is contributed by# Surendra_Gangwar |
C#
// C# program to implement above approachusing System;class GFG{ static int __gcd(int a,int b){ if(b == 0) return a; return __gcd(b, a % b);}// Recursive function to count// the number of unset bitsstatic int countBits(int n){ // Base case if (n == 0) return 0; // unset bit count else return ((n & 1) == 0 ? 1 : 0) + countBits(n >> 1);} // Function to return the max gcdstatic int maxGcd(int n){ // If no unset bits if (countBits(n) == 0) { // Find the maximum factor for (int i = 2; i * i <= n; i++) { // Highest factor if (n % i == 0) { return n / i; } } } else { int val = 0; int power = 1; int dupn = n; // Find the flipped bit number while (n > 0) { // If bit is not set if ((n & 1) == 0) { val += power; } // Next power of 2 power = power * 2; // Right shift the number n = n >> 1; } // Return the answer return __gcd(val ^ dupn, val & dupn); } // If a prime number return 1;} // Driver Codestatic void Main(){ // First example int n = 5; Console.WriteLine(maxGcd(n)); // Second example n = 15; Console.WriteLine(maxGcd(n));}} // This code is contributed by mits |
PHP
<?php// PHP program to implement above approach // Recursive function to count // the number of unset bits function countBits($n) { // Base case if ($n == 0) return 0; // unset bit count else return !($n & 1) + countBits($n >> 1); } // Recursive function to return// gcd of a and b function gcd($a, $b) { // Everything divides 0 if ($a == 0) return $b; if ($b == 0) return $a; // base case if($a == $b) return $a ; // a is greater if($a > $b) return gcd($a - $b , $b ); return gcd($a , $b - $a ); } // Function to return the max gcd function maxGcd($n) { // If no unset bits if (countBits($n) == 0) { // Find the maximum factor for ($i = 2; $i * $i <= $n; $i++) { // Highest factor if ($n % $i == 0) { return floor($n / $i); } } } else { $val = 0; $power = 1; $dupn = $n; // Find the flipped bit number while ($n) { // If bit is not set if (!($n & 1)) { $val += $power; } // Next power of 2 $power = $power * 2; // Right shift the number $n = $n >> 1; } // Return the answer return gcd($val ^ $dupn, $val & $dupn); } // If a prime number return 1; } // Driver Code // First example $n = 5; echo maxGcd($n), "\n"; // Second example $n = 15; echo maxGcd($n), "\n"; // This code is contributed by Ryuga?> |
Output:
7 5
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