即日起在codingBlog上分享您的技术经验即可获得积分,积分可兑换现金哦。

The Luckiest number POJ – 3696 (欧拉函数)

编程语言 Coldfresh 14℃ 0评论

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


又是欧拉函数 。。。

#include
#include
#include
#include
using namespace std;
typedef long long ll;
ll gcd(ll x,ll y)
{
    return x%y?gcd(y,x%y):y;
}
ll euler(ll n)
{
    ll res=n,a=n;
    for(int i=2;i*iif(a%i==0)
    {
        res=res/i*(i-1);
        while(a%i==0)
        a/=i;
    }
    if(a>1)
        res=res/a*(a-1);
    return res;
}
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 poww(long long a,long long b,long long mod) {
    long long ret=1;
    while(b) {
        if(b&1)
            ret=multi(ret,a,mod);  //直接相乘的话可能会溢出
        a=multi(a,a,mod);
        b=b>>1;
    }
    return ret;
}
int main()
{
    ll n;
    int count=1;
    for(;;)
    {
        scanf("%lld",&n);
        if(n==0)break;
        ll d=gcd(8,n);
        n=n*9/d;
        if(gcd(10,n)!=1)
            cout<<"Case "<": 0"<else
        {
            ll phi=euler(n);
            ll up=sqrt((double)phi);
            ll ans=phi;
            bool sign=false;
            for(int i=1;i<=up;i++)
                if(phi%i==0&&poww(10,(ll)i,n)==1)
                {
                    ans=i;
                    sign=true;
                    break;
                }
            if(!sign)
            for(int i=up;i>=2;i--)
                if(phi%i==0&&poww(10,phi/i,n)==1)
                {
                    ans=phi/i;
                    break;
                }
            cout<<"Case "<": "<return 0;
}

转载请注明:CodingBlog » The Luckiest number POJ – 3696 (欧拉函数)

喜欢 (0)or分享 (0)
发表我的评论
取消评论

*

表情