POJ - 3696 同余

给定\(L\),求最小的\(x\)满足\(L|8(10^x-1)\)

/*H E A D*/
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll euler(ll n){
    ll ans=n;
    for(ll i = 2; i*i <= n; i++){
        if(n%i==0){
            ans=ans/i*(i-1);
            while(n%i==0) n/=i;
        }
    }
    if(n>1) ans=ans/n*(n-1);
    return ans;
}
ll fmp(ll a,ll b,ll m){
    ll ans=0;
    while(b){
        if(b&1) ans=(ans+a)%m;
        a=(a+a)%m;
        b>>=1;
    }
    return ans;
}
ll fpw(ll a,ll n,ll m){
    ll ans=1;
    while(n){
        if(n&1) ans=fmp(ans,a,m);
        a=fmp(a,a,m);
        n>>=1;
    }
    return ans;
}
int main(){
    ll L,kase=0;
    while(cin>>L){
        if(L==0) break;
        L=9ll*L/gcd(L,8);
        printf("Case %lld: ",++kase);
        if(gcd(10,L)!=1){
            println(0);
            continue;
        }
        ll p=euler(L);
        ll ans=p;
        for(ll i=1; i*i<=p; i++){
            if(p%i!=0)continue;
            if(fpw(10,i,L)==1){
                ans=min(ans,i);
            }
            if(i*i!=p&&fpw(10,p/i,L)==1){
                ans=min(ans,p/i);
            }
        }
        println(ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/caturra/p/8448591.html