light oj 1138

1138 - Trailing Zeroes (III)

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Time Limit: 2 second(s)

Memory Limit: 32 MB

You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero on the trail.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer Q (1 ≤ Q ≤ 10⁸) in a line.

Output

For each case, print the case number and N. If no solution is found then print 'impossible'.

Sample Input	Output for Sample Input
3 1 2 5	Case 1: 5 Case 2: 10 Case 3: impossible

以前在杭电上做过跟这个类似的题，主要考察的知识点一样，只是那个是给出n求0的个数，这个给出0的个数求n

题解：主要还是求数中含有5的个数

#include<stdio.h>
#include<string.h>
#define LL long long
LL fun(LL x)
{
	LL ans=0;
	while(x)
	{
		ans+=x/5;
		x/=5;
	}
	return ans;
}
int main()
{
	int t,k=1;
	LL n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld",&n);
		LL l=0,r=500000000,mid,ans;
		while(r>l)
		{
			mid=(l+r)/2;
			if(fun(mid)>=n)
			{
				ans=mid;
				r=mid;
			}	
			else
			    l=mid+1;
		} 
		printf("Case %d: ",k++);
		if(fun(ans)!=n)                           
		printf("impossible
");
		else
		printf("%lld
",ans);
	}
	return 0;
}