Given two integers 
and . The task is to find the maximum value of x, such that, n! % (k^x) = 0.
Examples:
Input : n = 5, k = 2 Output : 3 Explanation : Given n = 5 and k = 2. So, n! = 120. Now for different values of x: n! % 2^0 = 0, n! % 2^1 = 0, n! % 2^2 = 0, n! % 2^3 = 0, n! % 2^4 = 8, n! % 2^5 = 24, n! % 2^6 = 56, n! % 2^7 = 120. So, the answer should be 3. Input : n = 1000, x = 2 Output : 994
Approach:
- First take the squareroot of
and store it in a variable say, 
. - Run the loop from i=2 to m.
- If i = m then copy k to i.
- If k is divisible by i then divide k by i.
- Run a loop to n and add the quotient to a variable say,
. - Store the minimum value of r after every loop.
Below is the implementation of the above approach:
C++
// C++ program to maximize the value // of x such that n! % (k^x) = 0 #include<iostream>#include<math.h>using namespace std;class GfG{ // Function to maximize the value // of x such that n! % (k^x) = 0 public: int findX(int n, int k) { int r = n, v, u; // Find square root of k and add 1 to it int m = sqrt(k) + 1; // Run the loop from 2 to m and k // should be greater than 1 for (int i = 2; i <= m && k > 1; i++) { if (i == m) { i = k; } // optimize the value of k for (u = v = 0; k % i == 0; v++) { k /= i; } if (v > 0) { int t = n; while (t > 0) { t /= i; u += t; } // Minimum store r = min(r, u / v); } } return r; } }; // Driver Code int main() { GfG g; int n = 5; int k = 2; cout<<g.findX(n, k); } //This code is contributed by Soumik |
Java
// Java program to maximize the value// of x such that n! % (k^x) = 0import java.util.*;public class GfG { // Function to maximize the value // of x such that n! % (k^x) = 0 private static int findX(int n, int k) { int r = n, v, u; // Find square root of k and add 1 to it int m = (int)Math.sqrt(k) + 1; // Run the loop from 2 to m and k // should be greater than 1 for (int i = 2; i <= m && k > 1; i++) { if (i == m) { i = k; } // optimize the value of k for (u = v = 0; k % i == 0; v++) { k /= i; } if (v > 0) { int t = n; while (t > 0) { t /= i; u += t; } // Minimum store r = Math.min(r, u / v); } } return r; } // Driver Code public static void main(String args[]) { int n = 5; int k = 2; System.out.println(findX(n, k)); }} |
Python 3
# Python 3 program to maximize the value# of x such that n! % (k^x) = 0import math# Function to maximize the value# of x such that n! % (k^x) = 0def findX(n, k): r = n # Find square root of k # and add 1 to it m = int(math.sqrt(k)) + 1 # Run the loop from 2 to m # and k should be greater than 1 i = 2 while i <= m and k > 1 : if (i == m) : i = k # optimize the value of k u = 0 v = 0 while k % i == 0 : k //= i v += 1 if (v > 0) : t = n while (t > 0) : t //= i u += t # Minimum store r = min(r, u // v) i += 1 return r# Driver Codeif __name__ == "__main__": n = 5 k = 2 print(findX(n, k))# This code is contributed# by ChitraNayal |
C#
// C# program to maximize the value // of x such that n! % (k^x) = 0 using System;class GfG{ // Function to maximize the value // of x such that n! % (k^x) = 0 public int findX(int n, int k) { int r = n, v, u; // Find square root of k and add 1 to it int m = (int)Math.Sqrt(k) + 1; // Run the loop from 2 to m and k // should be greater than 1 for (int i = 2; i <= m && k > 1; i++) { if (i == m) { i = k; } // optimize the value of k for (u = v = 0; k % i == 0; v++) { k /= i; } if (v > 0) { int t = n; while (t > 0) { t /= i; u += t; } // Minimum store r = Math.Min(r, u / v); } } return r; } } // Driver Code class geek{ public static void Main() { GfG g = new GfG(); int n = 5; int k = 2; Console.WriteLine(g.findX(n, k)); } } |
PHP
<?php// PHP program to maximize the value // of x such that n! % (k^x) = 0 // Function to maximize the value // of x such that n! % (k^x) = 0 function findX($n, $k) { $r = $n; // Find square root of k and add 1 to it $m = (int)sqrt($k) + 1; // Run the loop from 2 to m and k // should be greater than 1 for ($i = 2; $i <= $m && $k > 1; $i++) { if ($i == $m) { $i = $k; } // optimize the value of k for ($u = $v = 0; $k % $i == 0; $v++) { $k = (int)($k / $i); } if ($v > 0) { $t = $n; while ($t > 0) { $t = (int)($t / $i); $u = $u + $t; } // Minimum store $r = min($r, (int)($u / $v)); } } return $r; } // Driver Code $n = 5; $k = 2; echo findX($n, $k); // This code is contributed by// Archana_kumari?> |
Javascript
<script> // Javascript program to maximize the value // of x such that n! % (k^x) = 0 // Function to maximize the value // of x such that n! % (k^x) = 0 function findX(n, k) { let r = n; let v = 0; let u = 0; // Find square root of k and add 1 to it let m = Math.floor(Math.sqrt(k) + 1); // Run the loop from 2 to m and k // should be greater than 1 for (let i = 2; i <= m && k > 1; i++) { if (i == m) { i = k; } // optimize the value of k for (let u = v = 0; k % i == 0; v++) { k = Math.floor(k / i); } if (v > 0) { let t = n; while (t > 0) { t = Math.floor(t / i); u += t; } // Minimum store r = Math.min(r, Math.floor(u / v)); } } return r; } // Driver Code let n = 5; let k = 2; document.write(findX(n, k)); //This code is contributed by gfgking</script> |
Output:
3
Time complexity: O(sqrt(k)log(n)), where n is the input parameter n and k is the input parameter k.
Auxiliary space: O(1)
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