Smallest number n, having k trailing zeros in n! (LightOJ – 1138)

Problem Link: LightOJ 1138

We can easily calculate trailing zeros in n!. But, this problem refers to find the smallest n where n! holds exactly k trailing zeros.

If you don’t know how to calculate trailing zeros of n!. Read here – Trailing Zeros.

The max range of k is 10^8. So, 5*10^8! can easily hold this amount of trailing zero ( as the number of total 5 will be= 5*10^8/ 5 + 5*10^8/ (5*5) + 5*10^8/(5*5*5) +…. ). As, the number of trailing zero increases with the increment of n. Now, search for n and check if that contains k trailing zero or not. Factorial of multiple numbers can contain same numbers of trailing zero, in this problem you just have to find the smallest one, which is obviously a multiple of 5. Now, code it and have a green verdict…

Code:

	#include<bits/stdc++.h>
	using namespace std;
	#define ll long long
	int main ()
	{
	ll tc,idd=0,b,n,temp,r,i,p,q,c,ans;
	scanf("%lld",&tc);
	while(tc--)
	{
	scanf("%lld",&n);
	printf("Case %lld: ",++idd);
	ans=0;
	ll lo=1,hi=5000000000,mid;
	while(lo<hi)
	{
	mid=(lo+hi)/2;
	ll cnt=0,tmp=mid;
	while(tmp/5)
	{
	cnt+=tmp/5;
	tmp/=5;
	}
	if(cnt==n)
	{
	ans=mid;
	break;
	}
	if(cnt<n)
	lo=mid+1;
	else
	hi=mid-1;
	}
	if(ans)
	printf("%lld\n",ans-ans%5);
	else
	printf("impossible\n");
	}
	return 0;
	}

view raw lightoj1138.cpp hosted with ❤ by GitHub

“For every complex problem there is an answer that is clear, simple, and wrong“

— H. L. Mencken