light oj 1138

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

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

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

3

1

2

5

Case 1: 5

Case 2: 10

Case 3: impossible

题意：寻找最小的自然数N，使其阶乘的末尾有Q个0.

思路：只有2和5相乘时末尾才会出现0，在阶乘过程中，因子2的个数足够多，所以每遇到一个5，末尾就会出现一个0，因此问题转化为求1到N这N个整数中包含了多少个因子5。

#include<stdio.h>
#include<string.h>
#define MAX 0x3f3f3f
#define LL long long
LL fun(LL x)
{
	LL ans=0;
	while(x)
	{
		ans+=x/5;
		x=x/5;
	}
	return ans;
}
int main()
{
	int n,k=1;
	LL m;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%lld",&m);
		LL left=0;
		LL right=400000010;
		LL mid;
		while(left<=right)
		{
			mid=(left+right)>>1;
			if(fun(mid)>=m)
			    right=mid-1;
			else 
			    left=mid+1;
		}
		printf("Case %d: ",k++);
		if(fun(left)!=m)
		printf("impossible\n");
		else
		printf("%lld\n",left);
	}
	return 0;
}