結果

問題 No.720 行列のできるフィボナッチ数列道場 (2)
ユーザー fumofumofuni
提出日時 2022-10-23 02:49:36
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,055 ms / 2,000 ms
コード長 9,388 bytes
コンパイル時間 3,363 ms
コンパイル使用メモリ 234,352 KB
最終ジャッジ日時 2025-02-08 11:39:44
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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 * 20
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ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[9]={1,0,-1,0,1,1,-1,-1,0};
const ll dx[9]={0,1,0,-1,1,-1,1,-1,0};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
const int mod = MOD;
const int max_n = 200005;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
  bool operator==(const mint &p) const { return x == p.x; }
  bool operator!=(const mint &p) const { return x != p.x; }
  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
using vm=vector<mint>;
using vvm=vector<vm>;
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
}comb(max_n);
namespace NTT {
    // int32型のmodが取れるFFT。auto c=NTT::mul(a,b,mod)で受け取り。TIME指定。
    // ChineseRemと組み合わせてlong longにもできる
    std::vector<int> tmp;
    size_t sz = 1;
 
    inline int powMod(int n, int p, int m) {
        int res = 1;
        while (p) {
            if (p & 1) res = (ll)res * n % m;
            n = (ll)n * n % m;
            p >>= 1;
        }
        return (int)res;
    }
    inline int invMod(int n, int m) {
        return powMod(n, m - 2, m);
    }
 
    template <int Mod, int PrimitiveRoot>
    struct NTTPart {
        static std::vector<int> ntt(std::vector<int> a, bool inv = false) {
            size_t mask = sz - 1;
            size_t p = 0;
            for (size_t i = sz >> 1; i >= 1; i >>= 1) {
                auto& cur = (p & 1) ? tmp : a;
                auto& nex = (p & 1) ? a : tmp;
                int e = powMod(PrimitiveRoot, (Mod - 1) / sz * i, Mod);
                if (inv) e = invMod(e, Mod);
                int w = 1;
                for (size_t j = 0; j < sz; j += i) {
                    for (size_t k = 0; k < i; ++k) {
                        nex[j + k] = (cur[((j << 1) & mask) + k] + (ll)w * cur[(((j << 1) + i) & mask) + k]) % Mod;
                    }
                    w = (ll)w * e % Mod;
                }
                ++p;
            }
            if (p & 1) std::swap(a, tmp);
            if (inv) {
                int invSz = invMod(sz, Mod);
                for (size_t i = 0; i < sz; ++i) a[i] = (ll)a[i] * invSz % Mod;
            }
            return a;
        }
        static std::vector<int> mul(std::vector<int> a, std::vector<int> b) {
            a = ntt(a);
            b = ntt(b);
            for (size_t i = 0; i < sz; ++i) a[i] = (ll)a[i] * b[i] % Mod;
            a = ntt(a, true);
            return a;
        }
    };
 
    constexpr int M[] = {1224736769, 469762049, 167772161};
    constexpr int PR[] = {3, 3, 3};
    constexpr int NTT_CONVOLUTION_TIME = 3;
    /*
        X := max(a)*max(b)*max(|a|, |b|) のとき,
        NTT_CONVOLUTION_TIME <- 1: X <         1224736769 = 1.2*10^ 9 ~ 2^30
        NTT_CONVOLUTION_TIME <- 2: X < 575334854091079681 = 5.8*10^17 ~ 2^59
        NTT_CONVOLUTION_TIME <- 3: X < 2^86 (32bit * 32bit * 10^7くらいまで)
    */
 
    inline void garner(std::vector<int> *c, int mod) {
        if (NTT_CONVOLUTION_TIME == 1) {
            for(auto& x : c[0]) x %= mod;
        }
        else if (NTT_CONVOLUTION_TIME == 2) {
            const int r01 = invMod(M[0], M[1]);
            for (size_t i = 0; i < sz; ++i) {
                c[1][i] = (c[1][i] - c[0][i]) * (ll)r01 % M[1];
                if (c[1][i] < 0) c[1][i] += M[1];
                c[0][i] = (c[0][i] + (ll)c[1][i] * M[0]) % mod;
            }
        }
        else if (NTT_CONVOLUTION_TIME == 3) {
            const int R01 = invMod(M[0], M[1]);
            const int R02 = invMod(M[0], M[2]);
            const int R12 = invMod(M[1], M[2]);
            const int M01 = (ll)M[0] * M[1] % mod;
            for (size_t i = 0; i < sz; ++i) {
                c[1][i] = (c[1][i] - c[0][i]) * (ll)R01 % M[1];
                if (c[1][i] < 0) c[1][i] += M[1];
                c[2][i] = ((c[2][i] - c[0][i]) * (ll)R02 % M[2] - c[1][i]) * R12 % M[2];
                if (c[2][i] < 0) c[2][i] += M[2];
                c[0][i] = (c[0][i] + (ll)c[1][i] * M[0] + (ll)c[2][i] * M01) % mod;
            }
        }
    }
    std::vector<int> mul(std::vector<int> a, std::vector<int> b, int mod) {
        for (auto& x : a) x %= mod;
        for (auto& x : b) x %= mod;
 
        size_t m = a.size() + b.size() - 1;
        sz = 1;
        while (m > sz) sz <<= 1;
        tmp.resize(sz);
        a.resize(sz, 0);
        b.resize(sz, 0);
 
        std::vector<int> c[NTT_CONVOLUTION_TIME];
        if (NTT_CONVOLUTION_TIME >= 1) c[0] = NTTPart<M[0], PR[0]>::mul(a, b);
        if (NTT_CONVOLUTION_TIME >= 2) c[1] = NTTPart<M[1], PR[1]>::mul(a, b);
        if (NTT_CONVOLUTION_TIME >= 3) c[2] = NTTPart<M[2], PR[2]>::mul(a, b);
        for (auto& v : c) v.resize(m);
        garner(c, mod);
        return c[0];
    }
}; // !!! CHECK NTT_CONVOLUTION_TIME !!!
vector<mint> BerlekampMassey(const vector<mint> &s) {
  const int N = (int)s.size();
  vector<mint> b, c;
  b.reserve(N + 1);
  c.reserve(N + 1);
  b.push_back(mint(1));
  c.push_back(mint(1));
  mint y = mint(1);
  for (int ed = 1; ed <= N; ed++) {
    int l = int(c.size()), m = int(b.size());
    mint x = 0;
    for (int i = 0; i < l; i++) x += c[i] * s[ed - l + i];
    b.emplace_back(mint(0));
    m++;
    if (x == mint(0)) continue;
    mint freq = x / y;
    if (l < m) {
      auto tmp = c;
      c.insert(begin(c), m - l, mint(0));
      for (int i = 0; i < m; i++) c[m - 1 - i] -= freq * b[m - 1 - i];
      b = tmp;
      y = x;
    } else {
      for (int i = 0; i < m; i++) c[l - 1 - i] -= freq * b[m - 1 - i];
    }
  }
  reverse(begin(c), end(c));
  return c;
}
template <typename mint>
vector<mint> kitamasa(vector<mint> Q,vector<mint> a) {
  assert(!Q.empty() && Q[0] != 0);
  assert((int)a.size() >= int(Q.size()) - 1);
  vector<mint> P(Q.size()*2-2);
  for(ll i=0;i<Q.size()-1;i++){
    for(ll j=0;j<Q.size();j++){
     P[i+j]+=a[i]*Q[j];
    }
  } 
  P.resize(Q.size() - 1);
  return P;
}
template<class T>
struct bostan_mori {
  vector<T> p, q;
  bostan_mori(vector<T> &_p, vector<T> &_q) : p(_p), q(_q) {}
  void rever(vector<T> &f) const {
    int d = f.size();
    rep(i, d) if (i&1) f[i] = -f[i];
  }
  void even(vector<T> &f) const {
    int d = (f.size() + 1) >> 1;
    rep(i, d) f[i] = f[i<<1];
    f.resize(d);
  }
  void odd(vector<T> &f) const {
    int d = f.size() >> 1;
    rep(i, d) f[i] = f[i<<1|1];
    f.resize(d);
  }
  vector<T> convolution(vector<T> a,vector<T> b) const{
    ll n=a.size(),m=b.size();
    ll z=0,x=0;
    rep(i,n)if(a[i].x)z++;
    rep(j,m)if(b[j].x)x++;
    if(z*x<10000){
        vm c(n+m-1);
        vl na;rep(i,n)if(a[i].x)na.emplace_back(i);
        rep(j,m){
            if(b[j].x==0)continue;
            for(auto p:na){
                c[j+p]+=a[p]*b[j];
            }
        }
        return c;
    }
    vector<int> na(n),nb(m);
    rep(i,n)na[i]=a[i].x;
    rep(i,m)nb[i]=b[i].x;
    auto f=NTT::mul(na,nb,MOD);
    vector<T> c(n+m-1);
    rep(i,n+m-1)c[i]=f[i];
    return c;
  }
  T operator[] (ll n) const {
    vector<T> _p(p), _q(q), _q_rev(q);
    rever(_q_rev);
    for (; n; n >>= 1) {
      _p = convolution(move(_p), _q_rev);
      if (n&1) odd(_p);
      else     even(_p);
      _q = convolution(move(_q), move(_q_rev));
      even(_q);
      _q_rev = _q; rever(_q_rev);
    }
    return _p[0] / _q[0];
  }
};
ll m;
//https://nyaannyaan.github.io/library/fps/kitamasa.hpp
//https://atcoder.jp/contests/tdpc/submissions/34362182
//線形漸化式のprefixからn項目を復元できる。
bostan_mori<mint> interpolation(vm a){
  vm p={0,1};
  vm q(m+3);
  q[0]+=1;q[1]-=1;q[2]-=1;
  q[m]-=1;q[m+1]+=1;q[m+2]+=1;
  return bostan_mori<mint>(p,q);
}
int main(){
    ll n;cin >> n>> m;
    ll base=8000;
    vm v(base);v[1]=1;
    for(ll i=2;i<base;i++)v[i]=v[i-1]+v[i-2];
    vm dp(base);
    rep(i,base){
        dp[i]+=v[i];
        if(i+m<base)dp[i+m]+=dp[i];
    }
    //rep(i,100)cout << dp[i] <<" ";cout << endl;
    //dp.resize(100);
    auto f=interpolation(dp);
    //rep(i,100)cout << f[i] <<" ";cout << endl;
    cout << f[n*m] << endl;
}
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