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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'.

InputThe 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.OutputFor 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


参考网上的题解：
思路;

注意到凡是那种11111..... 22222..... 33333.....之类的序列都可用这个式子来表示：k*(10^x-1)/9
进而简化：8 * (10^x-1)/9=L * k (k是一个整数)
8*(10^x-1)=9L*k
d=gcd(9L,8)=gcd(8,L)
8*(10^x-1)/d=9L/d*k
令p=8/d q=9L/d p*(10^x-1)=q*k
因为p,q互质，所以q|(10^x-1)，即10^x-1=0(mod q)，也就是10^x=1(mod 9*L/d)
由欧拉定理可知，当q与10互质的时候，10^(φ(q))=1 (mod q)，即必定存在一个解x。
而题目中要求的是最小的解，设为min，那么有a^min=1%q，因为要满足a^φ(q)=1%q，那么a^φ(q)肯定能变换成(a^min)^i。
所以接下来只要枚举φ(q)的因子，找出符合条件的最小者即可。

无解的时候就是q与10不互质的时候，因为若q与10有公因子d：
1.若d=2，q=2*k，那么10^x=2^x*5^x=1%2k
   即2^x*5^x=1+2k*m，左边为偶数，右边为奇数，显然矛盾。
2.若d=5，q=5*k，那么10^x=2^x*5^x=1%5k
   即2^x*5^x=1+5k*m，左边是5的倍数，右边不是5的倍数，显然矛盾。

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
ll L;
long long multi(long long a,long long b,long long mod) {
    long long ret=0;
    while(b) {
        if(b&1)
            ret=(ret+a)%mod;
        a=(a<<1)%mod;
        b=b>>1;
    }
    return ret;
}
long long quickPow(long long a,long long b,long long mod) {
    long long ret=1;
    while(b) {
        if(b&1)
            ret=multi(ret,a,mod); 
        a=multi(a,a,mod);
        b=b>>1;
    }
    return ret;
}

long long eular(long long n) {
    long long ret=1,i;
    for(i=2; i*i<=n; i++) {
        if(n%i==0) {
            n=n/i;
            ret*=i-1;
            while(n%i==0) {
                n=n/i;
                ret*=i;
            }
        }
    }
    if(n>1)
        ret*=n-1;
    return ret;
}

int main() {
    int t=0;
    while(scanf("%lld",&L)!=EOF) {
        if(L==0)
            break;
        long long p=9*L/__gcd(L,(ll)8);
        long long d=__gcd((ll)10,p);
        if(d==1) {
            long long phi=eular(p);
            long long ans=phi;
            long long m=sqrt((double)phi);
            bool flag=false;
            for(int i=1; i<=m; i++) {
                if(phi%i==0 && quickPow(10,i,p)==1) {
                    ans=i;
                    flag=true;
                    break;
                }
            }
            if(!flag) {
                for(int i=m; i>=2; i--) {
                    if(phi%i==0 && quickPow(10,phi/i,p)==1) {
                        ans=phi/i;
                        break;
                    }
                }
            }
            printf("Case %d: %lld
",++t,ans);
        } else {
            printf("Case %d: 0
",++t);
        }
    }
    return 0;
}