The Luckiest number POJ

The Luckiest number

 POJ - 3696

题意：给一个数L，问至少几个8（如88，8888，88888）可以整数L。若不能，就输出-1.

参考的两位dalao的

http://blog.csdn.net/Danliwoo/article/details/48865127

http://blog.csdn.net/whai362/article/details/43908013

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
#define ll long long
//const int maxn=44725;  //sqrt(2000000000)=44721.4
//int pri[maxn];
//int cnt;
/*
void init(){
    cnt=0;
    memset(pri,0,sizeof(pri));
    for(ll i=2;i<maxn;i++) if(!pri[i]){
        pri[cnt++]=i;
        for(ll j=i*i;j<maxn;j+=i) pri[j]=1;
    }
}
*/
ll L;
ll gcd(ll a,ll b){
    return b==0?a:gcd(b,a%b);
}

ll phi(ll x){
    ll ans=x;
    int m=sqrt(x+0.5);
    for(int i=2;i<=m;i++) if(x%i==0){
        ans=ans/i*(i-1);
        while(x%i==0) x/=i;
    }
    if(x>1) ans=ans/x*(x-1);
    return ans;
}
ll quickadd(ll a,ll b,ll m){
    ll temp=a%m,res=0;
    while(b){
        if(b&1) res=(res+temp)%m;
        b>>=1;
        temp=(temp+temp)%m;
    }
    return res;
}
ll quickpow(ll a,ll b,ll m){
    ll temp=a%m,res=1;
    while(b){
        if(b&1) res=quickadd(res,temp,m);
        b>>=1;
        temp=quickadd(temp,temp,m);
    }
    return res;
}
int main(){
    int kase=0;
   // init();
    while(scanf("%lld",&L)&&L){
        ll m=L*9/gcd(L,8);
        printf("Case %d: ",++kase);
        if(gcd(m,10)!=1) puts("0");
        else{
            ll ph=phi(m);
            int lim=sqrt(ph);
            ll ans=ph;
            for(ll i=1;i<=lim;i++){
                if(ph%i==0){
                    if(quickpow(10,i,m)==1) ans=min(ans,i);
                    else if(quickpow(10,ph/i,m)==1) ans=min(ans,ph/i);
                }
            }
            printf("%lld
",ans);
        }
    }
    return 0;
}