#595382 (C++17) No.215 素数サイコロと合成数サイコロ (3-Hard)

• 問題: No.215 素数サイコロと合成数サイコロ (3-Hard)
• ユーザー: beet
• 提出日時: 2020-12-19 16:28:52
• 言語: C++17 (gcc 13.2.0 + boost 1.83.0)
• 結果: AC
• 実行時間: 1,375 ms / 4,000 ms
• コード長: 7,678 bytes
• コンパイル時間: 3,223 ms
• コンパイル使用メモリ: 227,744 KB
• 実行使用メモリ: 5,148 KB
• 最終ジャッジ日時: 2023-10-21 09:19:40
• 合計ジャッジ時間: 7,831 ms
• ジャッジサーバーID （参考情報）: judge12 / judge15

入力	結果	実行時間; 実行使用メモリ
testcase_00	AC	1,374 ms; 5,148 KB
testcase_01	AC	1,375 ms; 5,148 KB

ソースコード

// verification-helper: PROBLEM https://yukicoder.me/problems/444

#include <bits/stdc++.h>
using namespace std;

#define call_from_test
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;
  }
};

// construct a charasteristic equation from sequence
// return a monic polynomial in O(n^2)
template<typename T>
vector<T> berlekamp_massey(vector<T> &as){
  using Poly = vector<T>;
  int n=as.size();
  Poly bs({-T(1)}),cs({-T(1)});
  T y(1);
  for(int ed=1;ed<=n;ed++){
    int l=cs.size(),m=bs.size();
    T x(0);
    for(int i=0;i<l;i++) x+=cs[i]*as[ed-l+i];
    bs.emplace_back(0);
    m++;
    if(x==T(0)) continue;
    T freq=x/y;
    if(m<=l){
      for(int i=0;i<m;i++)
        cs[l-1-i]-=freq*bs[m-1-i];
      continue;
    }
    auto ts=cs;
    cs.insert(cs.begin(),m-l,T(0));
    for(int i=0;i<m;i++) cs[m-1-i]-=freq*bs[m-1-i];
    bs=ts;
    y=x;
  }
  for(auto &c:cs) c/=cs.back();
  return cs;
}

// Find k-th term of linear recurrence
template<typename T>
struct BostanMori{
  using Poly = vector<T>;
  using Conv = function<Poly(Poly, Poly)>;

  Conv conv;
  BostanMori(Conv conv_):conv(conv_){}

  Poly sub(Poly as,int odd){
    Poly bs((as.size()+!odd)/2);
    for(int i=odd;i<(int)as.size();i+=2) bs[i/2]=as[i];
    return bs;
  }

  // as: initial values
  // cs: monic polynomial
  T build(long long k,Poly as,Poly cs){
    reverse(cs.begin(),cs.end());
    assert(cs[0]==T(1));
    int n=cs.size()-1;
    as.resize(n,0);
    Poly bs=conv(as,cs);
    bs.resize(n);
    while(k){
      Poly ds(cs);
      for(int i=1;i<(int)ds.size();i+=2) ds[i]=-ds[i];
      bs=sub(conv(bs,ds),k&1);
      cs=sub(conv(cs,ds),0);
      k>>=1;
    }
    return bs[0];
  }
};

#undef call_from_test

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(0);

  using M = Mint<int>;
  using Poly = vector<M>;
  ArbitraryMod<M> arb;
  auto conv=[&](auto as,auto bs){return arb.multiply(as,bs);};

  long long n;
  cin>>n;
  n--;

  int p,c;
  cin>>p>>c;

  const int d = 606 * 13;
  auto calc=[&](int l,vector<int> vs){
    int m=vs.size();
    vector<Poly> dp(m,Poly(d));
    for(int i=0;i<m;i++) dp[i][0]=M(1);
    for(int t=0;t<l;t++){
      for(int i=0;i<m;i++){
        for(int j=d-1;j>=0;j--){
          dp[i][j]=0;
          if(i) dp[i][j]+=dp[i-1][j];
          if(j>=vs[i]) dp[i][j]+=dp[i][j-vs[i]];
        }
      }
    }
    return dp.back();
  };

  Poly cf({M(1)});
  cf=conv(cf,calc(p,vector<int>({2,3,5,7,11,13})));
  cf=conv(cf,calc(c,vector<int>({4,6,8,9,10,12})));
  cf.resize(d,M(0));

  Poly dp(d*3,0),as(d*3,0);
  dp[0]=M(1);
  for(int i=0;i<(int)dp.size();i++){
    for(int j=0;j<d&&i+j<(int)dp.size();j++)
      dp[i+j]+=dp[i]*cf[j];

    for(int j=1;j<d&&i+j<(int)dp.size();j++)
      as[i]+=dp[i+j];
  }
  as.resize(d*2);

  BostanMori<M> seq(conv);
  cout<<seq.build(n,as,berlekamp_massey(as))<<endl;
  return 0;
}