#446418 (C++17) No.1011 Infinite Stairs

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問題	No.1011 Infinite Stairs
コンテスト
ユーザー	beet
提出日時	2020-03-20 21:31:43
言語	C++17 (gcc 13.3.0 + boost 1.87.0)
結果	AC
実行時間	610 ms / 2,000 ms
コード長	9,586 bytes
コンパイル時間	4,051 ms
コンパイル使用メモリ	218,128 KB
最終ジャッジ日時	2025-01-09 08:15:27
ジャッジサーバーID （参考情報）	judge5 / judge3
このコードへのチャレンジ
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ファイルパターン	結果
sample	AC * 3
other	AC * 24
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ソースコード

raw source code
#include <bits/stdc++.h>
using namespace std;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
using Int = long long;
const char newl = '\n';


template<typename T,T MOD = 1000000007>
struct Mint{
  static constexpr T mod = MOD;
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }

  static Mint add_identity(){return Mint(0);}
  static Mint mul_identity(){return Mint(1);}

  Mint inv(){return pow(MOD-2);}

  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}

  Mint operator+(Mint a) const{return Mint(v)+=a;}
  Mint operator-(Mint a) const{return Mint(v)-=a;}
  Mint operator*(Mint a) const{return Mint(v)*=a;}
  Mint operator/(Mint a) const{return Mint(v)/=a;}

  Mint operator-() const{return v?Mint(MOD-v):Mint(v);}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}

  static Mint comb(long long n,int k){
    Mint num(1),dom(1);
    for(int i=0;i<k;i++){
      num*=Mint(n-i);
      dom*=Mint(i+1);
    }
    return num/dom;
  }
};
template<typename T,T MOD> constexpr T Mint<T, MOD>::mod;
template<typename T,T MOD>
ostream& operator<<(ostream &os,Mint<T, MOD> m){os<<m.v;return os;}


namespace FFT{
  using dbl = double;

  struct num{
    dbl x,y;
    num(){x=y=0;}
    num(dbl x,dbl y):x(x),y(y){}
  };

  inline num operator+(num a,num b){
    return num(a.x+b.x,a.y+b.y);
  }
  inline num operator-(num a,num b){
    return num(a.x-b.x,a.y-b.y);
  }
  inline num operator*(num a,num b){
    return num(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
  }
  inline num conj(num a){
    return num(a.x,-a.y);
  }

  int base=1;
  vector<num> rts={{0,0},{1,0}};
  vector<int> rev={0,1};

  const dbl PI=asinl(1)*2;

  void ensure_base(int nbase){
    if(nbase<=base) return;

    rev.resize(1<<nbase);
    for(int i=0;i<(1<<nbase);i++)
      rev[i]=(rev[i>>1]>>1)+((i&1)<<(nbase-1));

    rts.resize(1<<nbase);
    while(base<nbase){
      dbl angle=2*PI/(1<<(base+1));
      for(int i=1<<(base-1);i<(1<<base);i++){
        rts[i<<1]=rts[i];
        dbl angle_i=angle*(2*i+1-(1<<base));
        rts[(i<<1)+1]=num(cos(angle_i),sin(angle_i));
      }
      base++;
    }
  }

  void fft(vector<num> &as){
    int n=as.size();
    assert((n&(n-1))==0);

    int zeros=__builtin_ctz(n);
    ensure_base(zeros);
    int shift=base-zeros;
    for(int i=0;i<n;i++)
      if(i<(rev[i]>>shift))
        swap(as[i],as[rev[i]>>shift]);

    for(int k=1;k<n;k<<=1){
      for(int i=0;i<n;i+=2*k){
        for(int j=0;j<k;j++){
          num z=as[i+j+k]*rts[j+k];
          as[i+j+k]=as[i+j]-z;
          as[i+j]=as[i+j]+z;
        }
      }
    }
  }

  template<typename T>
  vector<long long> multiply(vector<T> &as,vector<T> &bs){
    int need=as.size()+bs.size()-1;
    int nbase=0;
    while((1<<nbase)<need) nbase++;
    ensure_base(nbase);

    int sz=1<<nbase;
    vector<num> fa(sz);
    for(int i=0;i<sz;i++){
      T x=(i<(int)as.size()?as[i]:0);
      T y=(i<(int)bs.size()?bs[i]:0);
      fa[i]=num(x,y);
    }
    fft(fa);

    num r(0,-0.25/sz);
    for(int i=0;i<=(sz>>1);i++){
      int j=(sz-i)&(sz-1);
      num z=(fa[j]*fa[j]-conj(fa[i]*fa[i]))*r;
      if(i!=j)
        fa[j]=(fa[i]*fa[i]-conj(fa[j]*fa[j]))*r;
      fa[i]=z;
    }
    fft(fa);

    vector<long long> res(need);
    for(int i=0;i<need;i++)
      res[i]=round(fa[i].x);

    return res;
  }

};


template<typename T>
struct ArbitraryMod{
  using dbl=FFT::dbl;
  using num=FFT::num;

  vector<T> multiply(vector<T> as,vector<T> bs){
    int need=as.size()+bs.size()-1;
    int sz=1;
    while(sz<need) sz<<=1;
    vector<num> fa(sz),fb(sz);
    for(int i=0;i<(int)as.size();i++)
      fa[i]=num(as[i].v&((1<<15)-1),as[i].v>>15);
    for(int i=0;i<(int)bs.size();i++)
      fb[i]=num(bs[i].v&((1<<15)-1),bs[i].v>>15);

    fft(fa);fft(fb);

    dbl ratio=0.25/sz;
    num r2(0,-1),r3(ratio,0),r4(0,-ratio),r5(0,1);
    for(int i=0;i<=(sz>>1);i++){
      int j=(sz-i)&(sz-1);
      num a1=(fa[i]+conj(fa[j]));
      num a2=(fa[i]-conj(fa[j]))*r2;
      num b1=(fb[i]+conj(fb[j]))*r3;
      num b2=(fb[i]-conj(fb[j]))*r4;
      if(i!=j){
        num c1=(fa[j]+conj(fa[i]));
        num c2=(fa[j]-conj(fa[i]))*r2;
        num d1=(fb[j]+conj(fb[i]))*r3;
        num d2=(fb[j]-conj(fb[i]))*r4;
        fa[i]=c1*d1+c2*d2*r5;
        fb[i]=c1*d2+c2*d1;
      }
      fa[j]=a1*b1+a2*b2*r5;
      fb[j]=a1*b2+a2*b1;
    }
    fft(fa);fft(fb);

    vector<T> cs(need);
    using ll = long long;
    for(int i=0;i<need;i++){
      ll aa=T(llround(fa[i].x)).v;
      ll bb=T(llround(fb[i].x)).v;
      ll cc=T(llround(fa[i].y)).v;
      cs[i]=T(aa+(bb<<15)+(cc<<30));
    }
    return cs;
  }
};


template<typename M_>
class Enumeration{
  using M = M_;
protected:
  static vector<M> fact,finv,invs;
public:
  static void init(int n){
    n=min<decltype(M::mod)>(n,M::mod-1);

    int m=fact.size();
    if(n<m) return;

    fact.resize(n+1,1);
    finv.resize(n+1,1);
    invs.resize(n+1,1);

    if(m==0) m=1;
    for(int i=m;i<=n;i++) fact[i]=fact[i-1]*M(i);
    finv[n]=M(1)/fact[n];
    for(int i=n;i>=m;i--) finv[i-1]=finv[i]*M(i);
    for(int i=m;i<=n;i++) invs[i]=finv[i]*fact[i-1];
  }

  static M Fact(int n){
    init(n);
    return fact[n];
  }
  static M Finv(int n){
    init(n);
    return finv[n];
  }
  static M Invs(int n){
    init(n);
    return invs[n];
  }

  static M C(int n,int k){
    if(n<k||k<0) return M(0);
    init(n);
    return fact[n]*finv[n-k]*finv[k];
  }

  static M P(int n,int k){
    if(n<k||k<0) return M(0);
    init(n);
    return fact[n]*finv[n-k];
  }

  // put n identical balls into k distinct boxes
  static M H(int n,int k){
    if(n<0||k<0) return M(0);
    if(!n&&!k) return M(1);
    init(n+k);
    return C(n+k-1,n);
  }
};
template<typename M>
vector<M> Enumeration<M>::fact=vector<M>();
template<typename M>
vector<M> Enumeration<M>::finv=vector<M>();
template<typename M>
vector<M> Enumeration<M>::invs=vector<M>();


template<typename M_>
struct FormalPowerSeries : Enumeration<M_> {
  using M = M_;
  using super = Enumeration<M>;
  using super::fact;
  using super::finv;
  using super::invs;

  using Poly = vector<M>;
  using Conv = function<Poly(Poly, Poly)>;
  Conv conv;
  FormalPowerSeries(Conv conv):conv(conv){}

  Poly pre(const Poly &as,int deg){
    return Poly(as.begin(),as.begin()+min((int)as.size(),deg));
  }

  Poly add(Poly as,Poly bs){
    int sz=max(as.size(),bs.size());
    Poly cs(sz,M(0));
    for(int i=0;i<(int)as.size();i++) cs[i]+=as[i];
    for(int i=0;i<(int)bs.size();i++) cs[i]+=bs[i];
    return cs;
  }

  Poly sub(Poly as,Poly bs){
    int sz=max(as.size(),bs.size());
    Poly cs(sz,M(0));
    for(int i=0;i<(int)as.size();i++) cs[i]+=as[i];
    for(int i=0;i<(int)bs.size();i++) cs[i]-=bs[i];
    return cs;
  }

  Poly mul(Poly as,Poly bs){
    return conv(as,bs);
  }

  Poly mul(Poly as,M k){
    for(auto &a:as) a*=k;
    return as;
  }

  // F(0) must not be 0
  Poly inv(Poly as,int deg){
    assert(as[0]!=M(0));
    Poly rs({M(1)/as[0]});
    for(int i=1;i<deg;i<<=1)
      rs=pre(sub(add(rs,rs),mul(mul(rs,rs),pre(as,i<<1))),i<<1);
    return rs;
  }

  // not zero
  Poly div(Poly as,Poly bs){
    while(as.back()==M(0)) as.pop_back();
    while(bs.back()==M(0)) bs.pop_back();
    if(bs.size()>as.size()) return Poly();
    reverse(as.begin(),as.end());
    reverse(bs.begin(),bs.end());
    int need=as.size()-bs.size()+1;
    Poly ds=pre(mul(as,inv(bs,need)),need);
    reverse(ds.begin(),ds.end());
    return ds;
  }

  Poly mod(Poly as,Poly bs){
    if(as==Poly(as.size(),0)) return Poly({0});
    as=sub(as,mul(div(as,bs),bs));
    if(as==Poly(as.size(),0)) return Poly({0});
    while(as.back()==M(0)) as.pop_back();
    return as;
  }

  // F(0) must be 1
  Poly sqrt(Poly as,int deg){
    assert(as[0]==M(1));
    M inv2=M(1)/M(2);
    Poly ss({M(1)});
    for(int i=1;i<deg;i<<=1){
      ss=pre(add(ss,mul(pre(as,i<<1),inv(ss,i<<1))),i<<1);
      for(M &x:ss) x*=inv2;
    }
    return ss;
  }

  Poly diff(Poly as){
    int n=as.size();
    Poly rs(n-1);
    for(int i=1;i<n;i++) rs[i-1]=as[i]*M(i);
    return rs;
  }

  Poly integral(Poly as){
    super::init(as.size()+1);
    int n=as.size();
    Poly rs(n+1);
    rs[0]=M(0);
    for(int i=0;i<n;i++) rs[i+1]=as[i]*invs[i+1];
    return rs;
  }

  // F(0) must be 1
  Poly log(Poly as,int deg){
    return pre(integral(mul(diff(as),inv(as,deg))),deg);
  }

  // F(0) must be 0
  Poly exp(Poly as,int deg){
    Poly f({M(1)});
    as[0]+=M(1);
    for(int i=1;i<deg;i<<=1)
      f=pre(mul(f,sub(pre(as,i<<1),log(f,i<<1))),i<<1);
    return f;
  }
};

//INSERT ABOVE HERE
signed main(){
  cin.tie(0);
  ios::sync_with_stdio(0);
  using M = Mint<int>;
  ArbitraryMod<M> arb;
  auto conv=[&](auto as,auto bs){return arb.multiply(as,bs);};
  FormalPowerSeries<M> FPS(conv);
  using Poly = decltype(FPS)::Poly;

  int n,d,k;
  cin>>n>>d>>k;
  const int deg = 1<<17;
  Poly ps(deg,0);
  //ps[0]=M(1);
  for(int i=0;i<d;i++) ps[i]=M(1);
  //ps=FPS.inv(ps,deg);
  ps=FPS.log(ps,deg);
  ps=FPS.mul(ps,M(n));
  ps=FPS.exp(ps,deg);

  cout<<ps[k-n]<<newl;
  return 0;
}
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結果

ソースコード
