Find maximum value of x such that n! % (k^x) = 0 in C++

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Suppose we have two integers n and k. We have to find the maximum value of x, such that n! mod (k^x) = 0. So when n = 5, and k = 2, then output will be 3. As n! = 120, now for different values of x, it will be −

120 mod 2^0 = 0, 120 mod 2^1 = 0, 120 mod 2^2 = 0, 120 mod 2^3 = 0, 120 mod 2^4 = 8, 120 mod 2^5 = 24, 120 mod 2^6 = 56, 120 mod 2^7 = 120. As the max value of x = 3, the result is 0, so the output is 3.

To solve this, we have to follow these steps −

Take the square root of k, and store it into m
For i := 2 to m, do the following steps:
- When i = m, then set i := k
- if k is divisible by i, then, divide k by i
- Run a loop to n, and add quotient to a variable called u.
- Store the min value of r, after each loop

Example

Live Demo

#include <iostream>
#include <cmath>
using namespace std;
int calculateMaxX(int n, int k) {
   int result = n, v, u;
   int m = sqrt(k) + 1;
   for (int i = 2; i <= m && k > 1; i++) {
      if (i == m) {
         i = k;
      }
      for (u = v = 0; k % i == 0; v++) {
         k /= i;
      }
      if (v > 0) {
         int t = n;
         while (t > 0) {
            t /= i;
            u += t;
         }
         result = min(result, u / v);
      }
   }
   return result;
}
int main() {
   int n = 5;
   int k = 2;
   cout<<"Maximum value of x is: " << calculateMaxX(n, k);
}