poj3696 The Luckiest number

最新推荐文章于 2019-11-09 11:19:14 发布

overload1997

最新推荐文章于 2019-11-09 11:19:14 发布

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分类专栏：数论欧拉函数

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本文介绍了一个有趣的数学问题，即如何找到最小的正整数，该整数仅由数字'8'组成，并且是给定幸运数字L的倍数。通过解析算法步骤，包括最大公约数、快速幂等技术的应用，展示了如何高效地解决这一问题。

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The Luckiest number

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5120		Accepted: 1370

Description

Chinese people think of '8' as the lucky digit. Bob also likes digit '8'. Moreover, Bob has his own lucky number L. Now he wants to construct his luckiest number which is the minimum among all positive integers that are a multiple of L and consist of only digit '8'.

Input

The input consists of multiple test cases. Each test case contains exactly one line containing L(1 ≤ L ≤ 2,000,000,000).

The last test case is followed by a line containing a zero.

Output

For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the length of Bob's luckiest number. If Bob can't construct his luckiest number, print a zero.

Sample Input

Sample Output

Case 1: 1
Case 2: 2
Case 3: 0//(10^x-1)=L*p*9/8;
//所以m=(9*L)/(gcd(L,8))
//所以10^x-1=m*p1
//所以转化成求同余方程 (10^x=1)modm
#include<vector> 
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
vector<LL>mp;
bool book[50000];
LL p[20000];
LL gcd(LL a,LL b){return b==0?a:gcd(b,a%b);}
void prim()
{
	memset(book,false,sizeof(book));
	book[0]=book[1]=1;
	int k=0;
	for(int i=2;i<50000;i++)
	{
		if(!book[i])p[k++]=i;
		for(int j=0;j<k&&i*p[j]<50000;j++)
		{
			book[i*p[j]]=1;
			if(!(i%p[j]))break;
		}
	}
}
LL mul(LL a,LL b,LL MOD)
{
	LL ans=0;
	while(b)
	{
		if(b&1)ans+=a,ans%=MOD;
		b>>=1;
		a+=a,a%=MOD;
	}
	return (ans%MOD+MOD)%MOD;
}
LL getphi(LL n)
{
	LL rea=n;
	for(int i=0;p[i]*p[i]<=n;i++)if(n%p[i]==0)
	{
		rea-=rea/p[i];
		while(n%p[i]==0)n/=p[i];
	}
	if(n>1)rea=rea-rea/n;
	return rea;
}
LL quickpow(LL n,LL m,LL MOD)
{
	LL b=1;
	while(m)
	{
		if(m&1)b=mul(b,n,MOD);
		m>>=1;
		n=mul(n,n,MOD);
	}
	return (b%MOD+MOD)%MOD;
}
void getfac(LL m)
{
	mp.clear();
	for(int i=0;p[i]*p[i]<=m;i++)if(m%p[i]==0)mp.push_back(p[i]),m/=p[i];
	if(m>1)mp.push_back(m);  
}
int main()
{
	LL L;
	prim();
	while(scanf("%lld",&L)==1)
	{
		LL m=(9*L)/(gcd(L,8));
		LL phi=getphi(m);
		LL ans=phi;
		bool flag=true;
		while(flag)
		{
			getfac(phi);
			flag=false;
			for(int i=0;i<mp.size();i++)
			{
				if(quickpow(10,phi/mp[i],m)==1){flag=true;if(phi/mp[i]<ans)ans=(phi/mp[i]);} 
			}
			phi=ans;
		}
		if(gcd(10,m)!=1)ans=0;
		printf("%lld\n",ans);
	}
}