Number of trailing zeroes in base B representation of N!

Given two positive integers B and N. The task is to find the number of trailing zeroes in b-ary (base B) representation of N! (factorial of N)

Examples:

Input: N = 5, B = 2
Output: 3
5! = 120 which is represented as 1111000 in base 2. 

Input: N = 6, B = 9
Output: 1


A naive solution is to find the factorial of the given number and convert it into given base B. Then, count the number of trailing zeroes but that would be a costly operation. Also, it will not be easy to find the factorial of large numbers and store it in integer.

Efficient Approach: Suppose, the base is 10 i.e., decimal then we’ll have to calculate the highest power of 10 that divides N! using Legendre’s formula. Thus, number B is represented as 10 when converted into base B. Let’s say base B = 13, then 13 in base 13 will be represented as 10, i.e., 1310 = 1013. Hence, problem reduces to finding the highest power of B in N!. (Largest power of k in n!)

Below is the implementation of the above approach.

C++

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// CPP program to find the number of trailing
// zeroes in base B representation of 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> > primeFactorsofB(int B)
{
    // vector to store all the prime factors
    // along with their number of occurrence
    // in factorization of B
    vector<pair<int, int> > ans;
  
    for (int i = 2; B != 1; i++) {
        if (B % i == 0) {
            int count = 0;
            while (B % i == 0) {
                B = B / i;
                count++;
            }
  
            ans.push_back(make_pair(i, count));
        }
    }
    return ans;
}
  
// Returns largest power of B that
// divides N!
int largestPowerOfB(int N, int B)
{
    vector<pair<int, int> > vec;
    vec = primeFactorsofB(B);
    int ans = INT_MAX;
    for (int i = 0; i < vec.size(); i++)
  
        // calculating minimum power of all
        // the prime factors of B
        ans = min(ans, findPowerOfP(N,
                                    vec[i].first)
                           / vec[i].second);
  
    return ans;
}
  
// Driver code
int main()
{
    cout << largestPowerOfB(5, 2) << endl;
    cout << largestPowerOfB(6, 9) << endl;
    return 0;
}

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Python3

# Python 3 program to find the number of
# trailing zeroes in base B representation of N!
import sys

# To find the power of a prime
# p in factorial N
def findPowerOfP(N, p):
count = 0
r = p
while (r <= N): # calculating floor(n/r) # and adding to the count count += int(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 def primeFactorsofB(B): # vector to store all the prime factors # along with their number of occurrence # in factorization of B' ans = [] i = 2 while(B!= 1): if (B % i == 0): count = 0 while (B % i == 0): B = int(B / i) count += 1 ans.append((i, count)) i += 1 return ans # Returns largest power of B that # divides N! def largestPowerOfB(N, B): vec = [] vec = primeFactorsofB(B) ans = sys.maxsize # calculating minimum power of all # the prime factors of B ans = min(ans, int(findPowerOfP(N, vec[0][0]) / vec[0][1])) return ans # Driver code if __name__ == '__main__': print(largestPowerOfB(5, 2)) print(largestPowerOfB(6, 9)) # This code is contributed by # Surendra_Gangwar [tabbyending]

Output:

3
1


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Improved By : SURENDRA_GANGWAR