[洛谷4884]多少个1？

题目传送门：https://www.luogu.org/problemnew/show/P4884

题目大意：

求1111...(n个1) mod m=k的最小的n

---

假定有n个1，那么1111...可以写成\(\sum\limits_{i=0}^n10^i=\dfrac{10^{n+1}-1}{9}\)，然后就可以推一波式子

\[\dfrac{10^{n+1}-1}{9}\equiv k(\%m)\\10^{n+1}\equiv9k+1(\%m) \]

然后直接BSGS求解即可

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
#define Fi first
#define Se second
typedef long long ll;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;
template<typename T>inline T min(T x,T y){return x<y?x:y;}
template<typename T>inline T max(T x,T y){return x>y?x:y;}
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
namespace Math{
	using std::map;
	map<ll,int>Mp;
	ll mlt(ll _a,ll _b,ll _p){
		ll _c=(ld)_a*_b/_p;
		ll _Ans=_a*_b-_c*_p;
		if (_Ans<0)	_Ans+=_p;
		return _Ans;
	}
	ll power(ll a,ll b,ll p){
		ll res=1;
		for (;b;b>>=1,a=mlt(a,a,p))	if (b&1)	res=mlt(res,a,p);
		return res;
	}
	ll BSGS(ll y,ll z,ll p){
		int lmt=(ll)sqrt(p)+1; ll tmp=z;
		Mp.insert(map<ll,int>::value_type(tmp,0));
		for (int i=1;i<=lmt;i++){
			tmp=mlt(tmp,y,p);
			map<ll,int>::iterator it=Mp.find(tmp);
			if (it==Mp.end())	it=Mp.insert(map<ll,int>::value_type(tmp,i)).Fi;
			it->Se=i;
		}
		tmp=1; ll K=power(y,lmt,p);
		for (ll i=1;i*lmt<p;i++){
			tmp=mlt(K,tmp,p);
			map<ll,int>::iterator it=Mp.find(tmp);
			if (it!=Mp.end())	return i*lmt-it->Se;
		}
		return -1;
	}
}
int main(){
	ll K=read(0ll),m=read(0ll);
	K=(9*K+1)%m;
	printf("%lld\n",Math::BSGS(10,K,m));
	return 0;
}