LightOJ - 1138 - Trailing Zeroes (III) （二分）

最新推荐文章于 2020-03-05 23:18:07 发布

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本文介绍了一种使用二分查找法求解最小自然数N，使得N!在十进制表示中恰好含有Q个尾随0的方法。通过具体实例展示了输入输出样例及解析过程。

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题目链接---LightOJ-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

Case 1: 5

Case 2: 10

Case 3: impossible

PROBLEM SETTER: JANE ALAM JAN

题意：本题主要讲给出一个数n的阶乘的末尾连续0的个数，输出最小n;

题解：典型的二分思想，难点就是求该数阶乘末尾的0的个数，可以用下面的代码求的，读者可以自行体会一下;

long long sum(long long n)
{
	long long res=0;
	while(n)
	{
	    res=res+n/5;
	    n=n/5;
	}
	return res;
}

下面是该题代码的实现：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
long long sum(long long n)//求n的阶乘末尾连续0的个数 
{
	long long res=0;
	while(n)
	{
	    res=res+n/5;
	    n=n/5;
	}
	return res;
}
int main()
{
	int t;
	cin>>t;
	int k=1;
	while(t--)
	{
		long long n,ans=0,head,tail,mid;
		cin>>n;
		head=1;
		tail=INF;
		while(tail>=head)
		{
			mid=(head+tail)/2;
			if(sum(mid)==n)
			{
				ans=mid;
				tail=mid-1; 
			}
			else if(sum(mid)>n)
			{
				tail=mid-1;
			}
			else
			{
				head=mid+1;
			}
		}
		printf("Case %d: ",k++);
		if(ans)
		cout<<ans<<endl;
		else
		cout<<"impossible"<<endl; 
	}
return 0;	
}