Given two positive integer x and y. we have to find the value of y mod 2x. That is remainder when y is divided by 2x.
Examples:
Input : x = 3, y = 14 Output : 6 Explanation : 14 % 23 = 14 % 8 = 6. Input : x = 4, y = 14 Output : 14 Explanation : 14 % 24 = 14 % 16 = 14.
To solve this question we can use pow() and modulo operator and can easily find the remainder.
But there are some points we should care about:
- For higher value of x such that 2x is greater than long long int range, we can not obtain proper result.
- Whenever y < 2x the remainder will always be y. So, in that case we can restrict ourselves to calculate value of 2x as:
y < 2x log y < x // means if log y is less than x, then // we can print y as remainder.
- The maximum value of 2x for which we can store 2x in a variable is 263. This means for x > 63, y is always going to be smaller than 2x and in that case remainder of y mod 2x is y itself.
keeping in mind the above points we can approach this problem as :
if (log y < x)
return y;
else if (x < 63)
return y;
else
return (y % (pow(2, x)))
Note: As python is limit free we can directly use mod and pow() function
C++
// C++ Program to find the // value of y mod 2^x #include <bits/stdc++.h> using namespace std; // function to calculate y mod 2^x long long int yMod(long long int y, long long int x) { // case 1 when y < 2^x if (log2(y) < x) return y; // case 2 when 2^x is out of // range means again y < 2^x if (x > 63) return y; // if y > 2^x return (y % (1 << x)); } // driver program int main() { long long int y = 12345; long long int x = 11; cout << yMod(y, x); return 0; } |
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Python
# Program to find the value # of y mod 2 ^ x function to # return y mod 2 ^ x def yMod(y, x) : return (y % pow(2, x)) # Driver code y = 12345x = 11print(yMod(y, x)) |
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Java
// Java Program to find // the value of y mod 2^x import java.io.*; class GFG { // function to calculate // y mod 2^x static long yMod(long y, long x) { // case 1 when y < 2^x if ((Math.log(y) / Math.log(2)) < x) return y; // case 2 when 2^x is // out of range means // again y < 2^x if (x > 63) return y; // if y > 2^x return (y % (1 << (int)x)); } // Driver Code public static void main(String args[]) { long y = 12345; long x = 11; System.out.print(yMod(y, x)); } } // This code is contributed by // Manish Shaw(manishshaw1) |
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C#
// C# Program to find the // value of y mod 2^x using System; class GFG { // function to calculate // y mod 2^x static long yMod(long y, long x) { // case 1 when y < 2^x if (Math.Log(y, 2) < x) return y; // case 2 when 2^x is // out of range means // again y < 2^x if (x > 63) return y; // if y > 2^x return (y % (1 << (int)x)); } // Driver Code static void Main() { long y = 12345; long x = 11; Console.Write(yMod(y, x)); } } // This code is contributed by // Manish Shaw(manishshaw1) |
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PHP
<?php // PHP Program to find the // value of y mod 2^x // function to // calculate y mod 2^x function yMod($y, $x) { // case 1 when y < 2^x if ((log($y) / log(2)) < $x) return $y; // case 2 when 2^x is // out of range means // again y < 2^x if ($x > 63) return $y; // if y > 2^x return ($y % (1 << $x)); } // Driver Code $y = 12345; $x = 11; echo (yMod($y, $x)); // This code is contributed by // Manish Shaw(manishshaw1) ?> |
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Output:
57
Compute modulus division by a power-of-2-number
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Improved By : manishshaw1
