# Preimage Size of Factorial Zeroes Function in C++

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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