E - Trailing Zeroes (III)（二分法+一个数末尾有几个零）

E - Trailing Zeroes (III)

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.

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Input

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

Each case contains an integer Q (1 ≤ Q ≤ 10^8) in a line.

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Output

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

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Sample Input

3

1

2

5

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Sample Output

Case 1: 5

Case 2: 10

Case 3: impossible

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解决方法：二分法

这道题n!尾部有几个零可以用以下代码解决，解决这个问题，这道题就简单多了

int zero(int n)//此函数求n！末尾有几个0; 
{
	int ans=0;
	while(n)
	{
		ans+=n/5;
		n/=5;
	}
	return ans;
}

上面的代码原理参考博客https://blog.csdn.net/ee8736199/article/details/48105017

AC代码

#include<iostream>
#include<cstdio>
using namespace std;
int zero(int n)//此函数求n！末尾有几个0; 
{
	int ans=0;
	while(n)
	{
		ans+=n/5;
		n/=5;
	}
	return ans;
}
int main()
{
	int t,q,k=1;
	cin>>t;
	while(t--)
	{
		cin>>q;
		int left,right,ans=0,mid;//ans储存要求的结果 
		//使用二分法 
		left=1,right=(int)1e9;//因为题目中n!最多Q=1e8个0；所以将n最大设置为1e9足够了；
		while(left<=right) 
		{
			mid=(left+right)/2;
			if(zero(mid)==q)
			{
				ans=mid;//满足用ans记下，因为要求最小的n,所以下面将right=mid-1,继续循环向小值逼近； 
				right=mid-1;
			}
			else if(zero(mid)>q)
				right=mid-1;
			else
				left=mid+1; 
		}
		if(ans)
			 printf("Case %d: %d\n",k++,ans);
		else
			printf("Case %d: impossible\n",k++);
	}
	return 0;
}

写的第一个博客，代码有水的地方欢迎大佬留言指点