Number 1? (BSGS)

   I can see the stamp face questions oh

　It took an afternoon thorough understanding of the implementation BSGS, the feeling can be, in fact, did not think so hard before. . .

   BSGS What is it? - can solve high-order congruence equation. concise.

   If want to learn, then recommend the chiefs of the blog , speak of is still very clear.

   This question is for us. . . . .

  We can see by observing  the nature of modulo, we can multiply the original formula with nine sides plus a 

   This is BSGS found the standard formula, as m guarantee is a prime number, so there is no need to expand the use BSGS.

  We find that question head is going to blow the LL

  Bigwigs will write _int128, had to learn a quick ride, la la la

  Quick Take the code:
  

LL mul(LL a,LL b,LL p)
{
    LL L=a*(b>>25LL)%p*(1LL<<25)%p;
    LL R=a*(b&((1LL<<25)-1))%p;
    return (L+R)%p;
}

 

  In this case, this question can be A, and happy.

   

#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL k,P;
LL mul(LL a,LL b,LL p)
{
    LL L=a*(b>>25LL)%p*(1LL<<25)%p;
    LL R=a*(b&((1LL<<25)-1))%p;
    return (L+R)%p;
}
LL ksm(LL x,LL y)
{
    LL res=1;
    for (;y;y>>=1,x=1LL*mul(x,x,P))
    if (y&1)res=1LL*mul(res,x,P);
    return res;
}
map<LL,LL> mp;
int main()
{
    scanf("%lld%lld",&k,&P);
    k=9*k+1;k%=P;
    LL m=ceil(sqrt(P));LL now=k;
    for (LL i=0;i<=m;i++)
    {
        mp[now]=i+1;
        now=mul(now,10,P);
    }
    now=1;LL ans=-1;LL kk=ksm(10,m);
    for (LL i=1;i<=m;i++)
    {
        now=1LL*mul(now,kk,P);
        if (mp[now]) {ans=i*m-mp[now]+1;break;}
    }
    printf("%lld",ans);
    return 0;
}