[POJ3696] The Luckiest number
\(\text{Description}\)
- \(\text{Given a number }L(L\leqslant 2,000,000,000),\text{find a number }x,\text{ so that }L|8\times(10^x-1)/9.\)
- \(\text{Output the length of }x.\text{If x doesn't exist, output 0.}\)
\(\text{Method}\)
\[L|8\times(10^x-1)/9
\]
\(\text{Let }M=\dfrac{9L}{\gcd(L,8)},\)
\[M|10^x-1
\]
\[10^x\equiv 1\pmod{M}
\]
\(\text{Use Euler's Theorem}\quad \gcd(a,m)=1\Rightarrow a^{\varphi(m)}\equiv 1\pmod{m},\)
\(\text{If }\gcd(10,M)\text{ equals }1:\)
\[x|\varphi(M)
\]
\(\text{Then count the divisors of }\varphi(M),\text{ and find the smallest }x.\)
\(\text{Else}:\)
\[x \text{ doesn't exist.}
\]
\(\text{Code}\)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define int long long
using namespace std;
int qmul(int a,int b,int mod)
{
if(a==0||b==0||mod==1ll)return 0;
if(b==1ll)return a%mod;
int ans=qmul(a,b/2ll,mod);
ans+=ans,ans%=mod;
if(b%2ll)ans+=a,ans%=mod;
return ans;
}
int qpow(int a,int b,int mod)
{
if(a==0||mod==1ll)return 0;
if(b==0)return 1ll;
int ans=qpow(a,b/2ll,mod);
ans=qmul(ans,ans,mod),ans%=mod;
if(b%2ll)ans=qmul(ans,a,mod),ans%=mod;
return ans;
}
int gcd(int a,int b)
{
if(b==0)return a;
else return gcd(b,a%b);
}
int geteuler(int n)
{
if(n==1ll)return 0;
int limit=sqrt(n),ans=n;
for(int i=2ll;i<=limit;i++)
if(n%i==0)
{
ans-=ans/i;
while(n%i==0)n/=i;
}
if(n>1ll)ans-=ans/n;
return ans;
}
int calc(int l)
{
int x=l/gcd(l,8ll)*9ll;
int flag=gcd(10ll,x);
if(flag!=1ll)return 0;
int phi=geteuler(x);
int limit=sqrt(phi),smallest;
for(int i=1ll;i<=limit;i++)
if(phi%i==0)
{
if(qpow(10,i,x)==1)return i;
int another=phi/i;
if(qpow(10,another,x)==1)smallest=another;
}
return smallest;
}
int n,cnt;
signed main()
{
while(~scanf("%lld",&n))
{
if(n==0)break;
cnt++;
printf("Case %lld: %lld\n",cnt,calc(n));
}
return 0;
}