Largest power of k in n! (factorial) where k may not be prime

Given two numbers k and n, find the largest power of k that divides n!

Constraints:

 K > 1 

Examples:



Input : n = 7, k = 2
Output : 4
Explanation : 7! = 5040
The largest power of 2 that
divides 5040 is 24.

Input : n = 10, k = 9
Output :  2
The largest power of 9 that
divides 10! is 92.

We have discussed a solution in below post when k is always prime.

Legendre’s formula (Given p and n, find the largest x such that p^x divides n!)

Now to find the power of any non-prime number k in n!, we first find all the prime factors of the number k along with the count of number of their occurrences. Then for each prime factor, we count occurrences using Legendre’s formula which states that the largest possible power of a prime number p in n is ⌊n/p⌋ + ⌊n/(p2)⌋ + ⌊n/(p3)⌋ + ……

Over all the prime factors p of K, the one with the minimum value of findPowerOfK(n, p)/count will be our answer where count is number of occurrences of p in k.

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// CPP program to find the largest power
// of k that divides n!
#include <bits/stdc++.h>
using namespace std;
  
// To find the power of a prime p in
// factorial N
int findPowerOfP(int n, int p)
{
    int count = 0;
    int r=p;
    while (r <= n) {
  
        // calculating floor(n/r)
        // and adding to the count
        count += (n / r);
  
        // increasing the power of p
        // from 1 to 2 to 3 and so on
        r = r * p;
    }
    return count;
}
  
// returns all the prime factors of k
vector<pair<int, int> > primeFactorsofK(int k)
{
    // vector to store all the prime factors
    // along with their number of occurrence
    // in factorization of k
    vector<pair<int, int> > ans;
  
    for (int i = 2; k != 1; i++) {
        if (k % i == 0) {
            int count = 0;
            while (k % i == 0) {
                k = k / i;
                count++;
            }
  
            ans.push_back(make_pair(i, count));
        }
    }
    return ans;
}
  
// Returns largest power of k that
// divides n!
int largestPowerOfK(int n, int k)
{
    vector<pair<int, int> > vec;
    vec = primeFactorsofK(k);
    int ans = INT_MAX;
    for (int i = 0; i < vec.size(); i++)
  
        // calculating minimum power of all
        // the prime factors of k
        ans = min(ans, findPowerOfP(n, 
              vec[i].first) / vec[i].second);
  
    return ans;
}
  
// Driver code
int main()
{
    cout << largestPowerOfK(7, 2) << endl;
    cout << largestPowerOfK(10, 9) << endl;
    return 0;
}

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Output:

4
2

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