[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;
}
posted @ 2019-03-31 08:47 Wolfycz 阅读(...) 评论(...) 编辑 收藏