Preimage Size of Factorial Zeroes Function in C++

Suppose we have a function f(x), this will return the number of zeroes at the end of factorial of x. So for f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has 2 zeroes at the end. Now when we have K, we have to find how many non-negative integers x have the property that f(x) = K.

So if the input is like K = 2, then the answer will be 5.

To solve this, we will follow these steps −

• Define a function ok(), this will take x,
• ret := 0
• for initialize i := 5, when i <= x, update i := i * 5, do −
  • ret := ret + x / i
• return ret
• From the main method, do the following −
• if K is same as 0, then −
  • return 5
• low := 1, high := K * 5
• while low < high, do −
  • mid := low + (high - low) /2
  • x := ok(mid)
  • if x < K, then −
    • low := mid + 1
  • Otherwise
    • high := mid
• return (if ok(low) is same as K, then 5, otherwise 0)

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; typedef long long int lli; class Solution { public:    lli ok(lli x){       int ret = 0;       for(lli i = 5; i <= x; i *= 5){          ret += x / i;       }       return ret;    }    int preimageSizeFZF(int K) {       if(K == 0) return 5;       lli low = 1;       lli high = (lli)K * 5;       while(low < high){          lli mid = low + (high - low) / 2;          lli x = ok(mid);          if(x < K){             low = mid + 1;          }else high = mid;       }       return ok(low) == K ? 5 : 0;    } }; main(){    Solution ob;    cout << (ob.preimageSizeFZF(2)); }

Input

2

Output

5