CareerCup

unsigned long CountZerosOfFactorial(unsigned long n)
{
unsigned long x2 = n, n2 = 0;

while (x2)
n2 += x2 /= 2;

unsigned long x5 = n, n5 = 0;

while (x5)
n5 += x5 /= 5;

return min(n2, n5);
}

r = 0
for i = 1, n
   k = i
   while k % 5 =0
      r = r + 1
      k = k/5

return r

=> O(.)=sum(log(i)) = O(n log n)

I don't think we have to use any while loop, simply divide the number by 5 and quotient is the number of zero's at end.
So,
int numZero( int n){

return n/5;

}

sum=n/power(5,0)+n/power(5,1)+n/power(5,2) and so on .untill ou get 0 so sum will be no of zero

this is the same problem as question id=2577

{{
The idea is to find the number of 5's in the n!.

size_t getTrailingZeroes(size_t n)
{
size_t count = 0;

for (size_t div = 5;; div *= 5)
{
size_t result = n/div;

if (result == 0)
break;

count += result;
}

return count;
}
}}

The idea is to find the number of 5's in the n!.

size_t getTrailingZeroes(size_t n)
{
	size_t count = 0;
	
	for (size_t div = 5;; div *= 5)
	{
		size_t result = n/div;

		if (result == 0)
			break;

		count += result;
	}

	return count;
}

def fun(n):
       if n == 1:
          return 1
       if n == 2:
          return 2
       return fun(n-1) * n
    def main():
        n = 10
        resultList = list(fun(10))
        zeroCounter = 0
        for element in resultList:
            if element == "0"
               zeroCounter += 1
        print "answer:",zeroCounter