poj 3696 The Luckiest number（欧拉函数+快速幂取模）

8
11
16
0

Case 1: 1
Case 2: 2
Case 3: 0

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll long long
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char ch;
    int a = 0;
    while((ch = getchar()) == ' ' | ch == '\n');
    a += ch - '0';
    while((ch = getchar()) != ' ' && ch != '\n')
    {
        a *= 10;
        a += ch - '0';
    }
    return a;
}

void Print(int a)    //输出外挂
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}

ll L;
ll mod;
ll gcd(ll a,ll b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}
ll mult(ll a,ll b)
{
    ll n=0;
    while(b)
    {
        if(b&1)
            n=(n+a)%mod;
        a=a*2%mod;
        b>>=1;
    }
    return n;
}
ll quick_mod(ll x)
{
    ll n=1,m=10;
    while(x)
    {
        if(x&1)
            n=mult(n,m);
        m=mult(m,m);
        x>>=1;
    }
    return n;
}

int main()
{
    //fread;
    int cs=1;
    while(scanf("%lld",&L)&&L)
    {
        printf("Case %d: ",cs++);
        mod=9*L/gcd(8,L);
        if(gcd(10,mod)!=1)
        {
            printf("0\n");
            continue;
        }
        ll rea=mod,n=mod;
        for(ll i=2;i*i<=n;i++)
            if(n%i==0)
        {
            rea=rea-rea/i;
            while(n%i==0)
                n/=i;
        }
        if(n>1)
            rea=rea-rea/n;//rea此时为m欧拉函数的值
        n=rea;
        ll p[100][2]; int cnt=0;
        for(ll i=2;i*i<=n;i++)
            if(n%i==0)
        {
            p[cnt][0]=i;
            p[cnt][1]=0;
            while(n%i==0)
            {
                n/=i;
                p[cnt][1]++;
            }
            cnt++;
        }
        if(n>1)
        {
            p[cnt][0]=n;
            p[cnt][1]=1;
            cnt++;
        }//p[i][0]存储n中素因子的值  p[i][1]存储该素因子的个数
        for(int i=0;i<cnt;i++)
        {
            for(int j=1;j<=p[i][1];j++)
                if(quick_mod(rea/p[i][0])==1)
                    rea/=p[i][0];
        }
        printf("%lld\n",rea);

    }
    return 0;
}