Given two integers A and B, the task is to calculate 22A % B.
Examples:
Input: A = 3, B = 5
Output: 1
223 % 5 = 28 % 5 = 256 % 5 = 1.Input: A = 10, B = 13
Output: 3
Approach: The problem can be efficiently solved by breaking it into sub-problems without overflow of integer by using recursion.
Let F(A, B) = 22A % B.
Now, F(A, B) = 22A % B
= 22 * 2A – 1 % B
= (22A – 1 + 2A – 1) % B
= (22A – 1 * 22A – 1) % B
= (F(A – 1, B) * F(A – 1, B)) % B
Therefore, F(A, B) = (F(A – 1, B) * F(A – 1, B)) % B.
The base case is F(1, B) = 221 % B = 4 % B.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define ll long long// Function to return 2^(2^A) % Bll F(ll A, ll B){ // Base case, 2^(2^1) % B = 4 % B if (A == 1) return (4 % B); else { ll temp = F(A - 1, B); return (temp * temp) % B; }}// Driver codeint main(){ ll A = 25, B = 50; // Print 2^(2^A) % B cout << F(A, B); return 0;} |
Java
// Java implementation of the above approach class GFG { // Function to return 2^(2^A) % B static long F(long A, long B) { // Base case, 2^(2^1) % B = 4 % B if (A == 1) return (4 % B); else { long temp = F(A - 1, B); return (temp * temp) % B; } } // Driver code public static void main(String args[]) { long A = 25, B = 50; // Print 2^(2^A) % B System.out.println(F(A, B)); } } // This code is contributed by Ryuga |
Python3
# Python3 implementation of the approach# Function to return 2^(2^A) % Bdef F(A, B): # Base case, 2^(2^1) % B = 4 % B if (A == 1): return (4 % B); else: temp = F(A - 1, B); return (temp * temp) % B;# Driver codeA = 25;B = 50;# Print 2^(2^A) % Bprint(F(A, B));# This code is contributed by mits |
C#
// C# implementation of the approachclass GFG{ // Function to return 2^(2^A) % Bstatic long F(long A, long B){ // Base case, 2^(2^1) % B = 4 % B if (A == 1) return (4 % B); else { long temp = F(A - 1, B); return (temp * temp) % B; }}// Driver codestatic void Main(){ long A = 25, B = 50; // Print 2^(2^A) % B System.Console.WriteLine(F(A, B));}}// This code is contributed by mits |
PHP
<?php// PHP implementation of the approach // Function to return 2^(2^A) % Bfunction F($A, $B){ // $Base case, 2^(2^1) % B = 4 % B if ($A == 1) return (4 % $B); else { $temp = F($A - 1, $B); return ($temp * $temp) % $B; }} // Driver code$A = 25;$B = 50; // Print 2^(2^$A) % $Becho F($A, $B);// This code contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Function to return 2^(2^A) % Bfunction F(A, B){ // Base case, 2^(2^1) % B = 4 % B if (A == 1) return (4 % B); else { var temp = F(A - 1, B); return (temp * temp) % B; }}// Driver codevar A = 25, B = 50;// Print 2^(2^A) % Bdocument.write( F(A, B));// This code is contributed by noob2000</script> |
Output:
46
Time Complexity: O(A)
Auxiliary Space: O(A), due to recursive call stack
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