Poj3696 The Lukiest Number

传送门

Solution

懒得写啦

Code

#include<iostream>
#include<cstdio>
#include<cmath>
#define Rg register
#define go(i,a,b) for(Rg int i=a;i<=b;i++)
#define yes(i,a,b) for(Rg int i=a;i>=b;i--)
#define ll long long
using namespace std;
int read()
{
    int x=0,y=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=(x<<3)+(x<<1)+c-'0';c=getchar();}
    return x*y;
}
ll T,L,d,n,m,ans,c;
bool flag=0;
ll gcd(ll x,ll y){return y==0?x:gcd(y,x%y);}
ll phi(ll x)
{
    ll sm=x,y=x;
    go(i,2,sqrt(y))
    {
        if(x%i==0)sm=sm/i*(i-1);
        while(x%i==0)x/=i;
    }
    if(x>1)sm=sm/x*(x-1);
    return sm;
}
ll mul(ll x,ll y,ll mod)
{
    ll sm=0;
    while(y)
    {
        if(y&1)sm=(sm+x)%mod;
        x=(x<<1)%mod;y>>=1;
    }
    return sm;
}
ll ksm(ll x,ll y,ll mod)
{
    ll sm=1;
    while(y)
    {
        if(y&1)sm=mul(sm,x,mod);//注意这里直接乘起来会溢出
        x=mul(x,x,mod);y>>=1;
    }
    return sm;
}
int main()
{
    while(1)
    {
        scanf("%lld",&L);if(!L)break;T++;flag=0;ans=0;
        if(8%L==0){printf("Case %lld: 1
",T);continue;}
        d=gcd(L,8);n=9*L/d;m=phi(n);c=sqrt(m);
        go(i,2,c)
        {
            if(m%i==0&&ksm(10,i,n)==1){ans=i;flag=1;break;}
        }
        if(!flag)
            yes(i,c,2)
            {
                if(m%i==0&&ksm(10,m/i,n)==1){ans=m/i;break;}
            }
        printf("Case %lld: %lld
",T,ans);
    }
    return 0;
}