POJ 3696 欧拉函数的应用

The Luckiest number

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10897		Accepted: 2769

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

8
11
16
0

Sample Output

Case 1: 1
Case 2: 2
Case 3: 0

Source

2008 Asia Hefei Regional Contest Online by USTC  

//关键在于欧拉函数a^φ(n)≡1 (mod n)此时不能保证 φ(n)是最小的^x≡1 (mod n)
/*证明；x0为a^x≡1 (mod n)的最小值且x0为x的因子；

若不然 
设φ(n)=ax0+r；r不为0；
a^ax0*a^r≡1(mod n)
==>a^r≡1(mod n) ,与假设x0为最小矛盾，因为r<x0；

*/

#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){
	return !b?a:gcd(b,a%b);
}
ll quimul(ll a,ll b,ll mod)
{
	ll ans=0;
	while(b)
	{
		if(b&1)ans=(ans+a)%mod;
		b>>=1;
		a=(a+a)%mod;
	}
	return ans;
}
ll quipow(ll a, ll n,ll mod)
{
	ll ans=1%mod;
	while(n)
	{
		if(n&1)ans=quimul(ans,a,mod);
		n>>=1;
		a=quimul(a,a,mod);
	}
	return ans;
}
ll euler( ll n)
{
	ll ans=n;
	for(int i=2;i*i<=n;i++)
	{
		if(n%i==0)
		{
			ans=1ll*ans/i*(i-1);
			while(n%i==0)n/=i;
		}
	}
	if(n>1)ans=1ll*ans/n*(n-1);
	return ans;
}
int main()
{
	ll L,tot=1;
	while(~scanf("%lld",&L)&&L)
	{
		ll ans=0;
		ll d=gcd(L,8LL);
		ll mod=1ll*9*L/d,n;
		vector<ll>f;
		ll gd=gcd(10LL,mod);
		
		if(gd!=1){
			printf("Case %lld: %lld\n",tot++,0ll);
			continue;
		}
		n=euler(mod);
		for(ll i=1;i*i<=n;i++)
		{
			if(n%i==0)
			{
				f.push_back(i);
				if(1ll*i*i!=n)f.push_back(n/i);
			}
		}
		sort(f.begin(),f.end());
		for(ll i=0;i<f.size();i++)
		{
			if(quipow(10,f[i],mod)==1)
			{
				ans=f[i];
				break;
			}
		}
		printf("Case %lld: %lld\n",tot++,ans);
	}
	return 0;
}

　　

标签：gcd,ll,number,3696,POJ,ans,test,欧拉,mod
来源： https://www.cnblogs.com/matt-su/p/15969068.html