The Luckiest number POJ - 3696 （欧拉函数）

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本文介绍了一个算法问题，即寻找最小正整数，该整数为给定幸运数L的倍数且仅由数字8组成。文章提供了完整的C++代码实现，涉及欧拉函数等数学概念。

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<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
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*i<a;i++)
        if(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 "<<count++<<": 0"<<endl;
        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 "<<count++<<": "<<ans<<endl;
        }
    }

    return 0;
}