#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
 
LL n;
 
LL Pow(LL a, LL b, LL mod)
{
    LL res = 1;
    while (b)
    {
        if (b&1)
            res = (res*a)%mod;
        a = (a*a)%mod;
        b >>= 1;
    }
    return res;
}
 
LL Euler(LL x)
{
    LL res = x;
    for (int i = 2;i <= x;i++)
    {
        if (x%i == 0)
        {
            while (x % i == 0)
                x /= i;
            res = res/i*(i - 1);
        }
    }
    if (x > 1)
        res = res/x*(x-1);
    return res;
}
 
LL Solve(int cnt, LL mod)
{
    LL res = 1e18;
    for (LL i = 1;i*i <= cnt;i++)
    {
        if (cnt%i == 0)
        {
//            cout << i << endl;
            if (Pow(10, i, mod) == 1)
                res = min(res, i);
            if (Pow(10, cnt/i, mod) == 1)
                res = min(res, cnt/i);
        }
    }
    if (res == 1e18)
        return 0;
    return res;
}
 
int main()
{
    int cnt = 0;
    while (~scanf("%lld", &n) && n)
    {
        LL eul = Euler((9LL*n)/__gcd(8LL, n*9));
//        cout << eul << endl;
        LL res = Solve(eul, (9LL*n)/__gcd(8LL, n*9));
        printf("Case %d: %lld\n", ++cnt, res);
    }
 
    return 0;
}

#include<iostream> #include<algorithm> #include<cstring>

using namespace std; typedef long long ll;

ll gcd(ll a,ll b) // 非常基础的最大公约数模板 { return b?gcd(b,a%b):a; }

ll get_euler(ll n) // 试除法 求 欧拉函数值模板 { ll s=n; for(ll i=2;i<=n/i;i++) { if(n%i0) { while(n%i0) n/=i; s = s / i * (i - 1); } } if(n>1) s=s/n*(n-1); return s; } ll quick_mul(ll a,ll b,ll c) // 龟速乘模板 { ll s=0;//注意初值为0，因为是累加。 while(b) { if(b&1) s = (s + a)%c; a = (a+a)%c; b/=2; } return s; } ll quick_mi(ll a,ll b,ll c) //快速幂模板 { ll s=1; while(b) { if(b&1) s = quick_mul(s,a,c); a=quick_mul(a,a,c); b/=2; } return s; } int main(void) { ll n; int T = 1; while(cin>>n,n) { ll mini = 1e18;

ll d=gcd(n,8),a=9*n/d,phi=get_euler(a);
    // 分别对应的就是解析中的 d、A、φ(A)

 	if(a%2==0||a%5==0) mini=0; // 如果不是互质，根据定理直接无解。
 	else
 	for(ll i=1;i<=phi/i;i++) // 枚举约数进行判断。
 	    if(phi%i==0) // 因为因数都是一对一对的，找到其中一个就够了，
 	    {
 		    if(quick_mi(10,i,a)==1) mini=min(mini,i); //判断 10^i % a 等不等于1，并且更新mini值。
 		    if(quick_mi(10,phi/i,a)==1) mini=min(mini,phi/i);//同上
	    }
 	printf("Case %d: %lld\n", T ++ , mini);
 	
  } 
return 0;

• [ ] 1.

#include <bits/stdc++.h> using namespace std; typedef long long LL;

LL n;

LL Pow(LL a, LL b, LL mod) { LL res = 1; while (b) { if (b&1) res = (resa)%mod; a = (aa)%mod; b >>= 1; } return res; }

LL Euler(LL x) { LL res = x; for (int i = 2;i <= x;i++) { if (x%i == 0) { while (x % i == 0) x /= i; res = res/i*(i - 1); } } if (x > 1) res = res/x*(x-1); return res; }

LL Solve(int cnt, LL mod) { LL res = 1e18; for (LL i = 1;i*i <= cnt;i++) { if (cnt%i == 0) { // cout << i << endl; if (Pow(10, i, mod) == 1) res = min(res, i); if (Pow(10, cnt/i, mod) == 1) res = min(res, cnt/i); } } if (res == 1e18) return 0; return res; }

int main() { int cnt = 0; while (~scanf("%lld", &n) && n) { LL eul = Euler((9LLn)/__gcd(8LL, n9)); // cout << eul << endl; LL res = Solve(eul, (9LLn)/__gcd(8LL, n9)); printf("Case %d: %lld\n", ++cnt, res); }

return 0;

}