【题解】多少个$1$(exBSGS)

【题解】多少个(1)(exBSGS)

解方程：

[underbrace {1dots1}_{n}equiv k mod m ]

其实就是

[dfrac {10^n-1} {9}equiv k mod m ]

就是

[10^nequiv 9k+1 mod m ]

直接exBSGS【总结】皇冠上的明珠——初等数论初步

//@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
",exBSGS(10,9ll*k+1,m));
      return 0;
}