[Explanations] How many $ 1 $ (exBSGS)

[Solution] how many questions \ (1 \) (exBSGS)

Solving the equation:
\ [\ underbrace {. 1 \ dots1} _ {n-} \ equiv K \ MOD m \]
actually
\ [\ dfrac {10 ^ n
-1} {9} \ equiv k \ mod m \] is
\ [10 ^ n \ equiv 9k +
1 \ mod m \] directly exBSGS [summary] crown jewel - Elementary Primary number Theory

//@winlere
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<cmath>

using namespace std;  typedef long long ll;
inline ll qr(){
      register ll ret=0,f=0;
      register char c=getchar();
      while(c<48||c>57)f|=c==45,c=getchar();
      while(c>=48&&c<=57) ret=ret*10+c-48,c=getchar();
      return f?-ret:ret;
}

map < ll , ll > s;
inline ll exBSGS(ll a,ll b,ll m){
      ll AD=1,d;
      ll cnt=0;
      while(d=__gcd(a,m),d!=1) {
        b/=d,m/=d,AD=AD*(a/d),++cnt;
        if(AD==b) return cnt;
      }
      ll sq=sqrt(m)+1,ret=1;
      map < ll ,ll >().swap(s);
      for(ll t=0;t<sq;++t,ret=(__int128)ret*a%m)
        s[(__int128)ret*b%m]=t;
      for(ll t=1,w=AD*ret;t<=sq;++t,w=(__int128)w*ret%m)
        if(s.find(w)!=s.end())
          return cnt+t*sq-s[w];
      return -1;
}

int main(){
      ll k=qr(),m=qr();
      printf("%lld\n",exBSGS(10,9ll*k+1,m));
      return 0;
}

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Origin www.cnblogs.com/winlere/p/11279308.html