2019牛客多校第九场AThe power of Fibonacci（广义BM）

2019牛客多校第九场AThe power of Fibonacci（广义BM）

题目大意

求斐波那契数列m次方的前n项和

解题思路

显然，斐波那契的m次方前缀和依然是线性递推，因此考虑用exBM求解，套板子就行

AC代码

#include
using namespace std;
typedef long long LL;
const int mod=1e9;
struct LinearRecurrence {
    using vec = std::vector<LL>;
    static void extand(vec &a, size_t d, LL value = 0) {
        if (d <= a.size()) return;
        a.resize(d, value);
    }
    static void exgcd(LL a, LL b, LL &g, LL &x, LL &y) {
        if (!b) x = 1, y = 0, g = a;
        else {
            exgcd(b, a % b, g, y, x);
            y -= x * (a / b);
        }
    }
    static LL crt(const vec &c, const vec &m) {
        int n = c.size();
        LL M = 1, ans = 0;
        for (int i = 0; i < n; ++i) M *= m[i];
        for (int i = 0; i < n; ++i) {
            LL x, y, g, tm = M / m[i];
            exgcd(tm, m[i], g, x, y);
            ans = (ans + tm * x * c[i] % M) % M;
        }
        return (ans + M) % M;
    }
    static vec ReedsSloane(const vec &s, LL mod) {
        auto inverse = [](LL a, LL m) {
            LL d, x, y;
            exgcd(a, m, d, x, y);
            return d == 1 ? (x % m + m) % m : -1;
        };
        auto L = [](const vec &a, const vec &b) {
            int da = (a.size() > 1 || (a.size() == 1 && a[0])) ? a.size() - 1 : -1000;
            int db = (b.size() > 1 || (b.size() == 1 && b[0])) ? b.size() - 1 : -1000;
            return std::max(da, db + 1);
        };
        auto prime_power = [&](const vec &s, LL mod, LL p, LL e) {
            // linear feedback shift register mod p^e, p is prime
            std::vector<vec> a(e), b(e), an(e), bn(e), ao(e), bo(e);
            vec t(e), u(e), r(e), to(e, 1), uo(e), pw(e + 1);
            pw[0] = 1;
            for (int i = pw[0] = 1; i <= e; ++i) pw[i] = pw[i - 1] * p;
            for (LL i = 0; i < e; ++i) {
                a[i] = {pw[i]}, an[i] = {pw[i]};
                b[i] = {0}, bn[i] = {s[0] * pw[i] % mod};
                t[i] = s[0] * pw[i] % mod;
                if (t[i] == 0)
                    t[i] = 1, u[i] = e;
                else
                    for (u[i] = 0; t[i] % p == 0; t[i] /= p, ++u[i]);
            }
            for (size_t k = 1; k < s.size(); ++k) {
                for (int g = 0; g < e; ++g)
                    if (L(an[g], bn[g]) > L(a[g], b[g])) {
                        ao[g] = a[e - 1 - u[g]];
                        bo[g] = b[e - 1 - u[g]];
                        to[g] = t[e - 1 - u[g]];
                        uo[g] = u[e - 1 - u[g]];
                        r[g] = k - 1;
                    }
                a = an, b = bn;
                for (int o = 0; o < e; ++o) {
                    LL d = 0;
                    for (size_t i = 0; i < a[o].size() && i <= k; ++i) {
                        d = (d + a[o][i] * s[k - i]) % mod;
                    }
                    if (d == 0) {
                        t[o] = 1, u[o] = e;
                    } else {
                        for (u[o] = 0, t[o] = d; t[o] % p == 0; t[o] /= p, ++u[o]);
                        int g = e - 1 - u[o];
                        if (L(a[g], b[g]) == 0) {
                            extand(bn[o], k + 1);
                            bn[o][k] = (bn[o][k] + d) % mod;
                        } else {
                            LL coef =t[o] * inverse(to[g], mod) % mod * pw[u[o] - uo[g]] % mod;
                            int m = k - r[g];
                            extand(an[o], ao[g].size() + m);
                            extand(bn[o], bo[g].size() + m);
                            for (size_t i = 0; i < ao[g].size(); ++i) {
                                an[o][i + m] -= coef * ao[g][i] % mod;
                                if (an[o][i + m] < 0) an[o][i + m] += mod;
                            }
                            while (an[o].size() && an[o].back() == 0) an[o].pop_back();
                            for (size_t i = 0; i < bo[g].size(); ++i) {
                                bn[o][i + m] -= coef * bo[g][i] % mod;
                                if (bn[o][i + m] < 0) bn[o][i + m] -= mod;
                            }
                            while (bn[o].size() && bn[o].back() == 0) bn[o].pop_back();
                        }
                    }
                }
            }
            return std::make_pair(an[0], bn[0]);
        };
        std::vector<std::tuple<LL, LL, int>> fac;
        for (LL i = 2; i * i <= mod; ++i)
            if (mod % i == 0) {
                LL cnt = 0, pw = 1;
                while (mod % i == 0) mod /= i, ++cnt, pw *= i;
                fac.emplace_back(pw, i, cnt);
            }
        if (mod > 1) fac.emplace_back(mod, mod, 1);
        std::vector<vec> as;
        size_t n = 0;
        for (auto &&x: fac) {
            LL mod, p, e;
            vec a, b;
            std::tie(mod, p, e) = x;
            auto ss = s;
            for (auto &&x: ss) x %= mod;
            std::tie(a, b) = prime_power(ss, mod, p, e);
            as.emplace_back(a);
            n = std::max(n, a.size());
        }
        vec a(n), c(as.size()), m(as.size());
        for (size_t i = 0; i < n; ++i) {
            for (size_t j = 0; j < as.size(); ++j) {
                m[j] = std::get<0>(fac[j]);
                c[j] = i < as[j].size() ? as[j][i] : 0;
            }
            a[i] = crt(c, m);
        }
        return a;
    }
    LinearRecurrence(const vec &s, const vec &c, LL mod) :
            init(s), trans(c), mod(mod), m(s.size()) {}
  
    LinearRecurrence(const vec &s, LL mod, bool is_prime = true) : mod(mod) {
        vec A;
        A = ReedsSloane(s, mod);
        if (A.empty()) A = {0};
        m = A.size() - 1;
        trans.resize(m);
        for (int i = 0; i < m; ++i) trans[i] = (mod - A[i + 1]) % mod;
          
        std::reverse(trans.begin(), trans.end());
        init = {s.begin(), s.begin() + m};
    }
   
    LL calc(LL n) {
        if (mod == 1) return 0;
        if (n < m) return init[n];
        vec v(m), u(m << 1);
        int msk = !!n;
        for (LL m = n; m > 1; m >>= 1) msk <<= 1;
        v[0] = 1 % mod;
        for (int x = 0; msk; msk >>= 1, x <<= 1) {
            std::fill_n(u.begin(), m * 2, 0);
            x |= !!(n & msk);
            if (x < m) u[x] = 1 % mod;
            else {// can be optimized by fft/ntt
                for (int i = 0; i < m; ++i)
                    for (int j = 0, t = i + (x & 1); j < m; ++j, ++t)
                        u[t] = (u[t] + v[i] * v[j]) % mod;
                for (int i = m * 2 - 1; i >= m; --i)
                    for (int j = 0, t = i - m; j < m; ++j, ++t)
                        u[t] = (u[t] + trans[j] * u[i]) % mod;
            }
            v = {u.begin(), u.begin() + m};
        }
        LL ret = 0;
        for (int i = 0; i < m; ++i) ret = (ret + v[i] * init[i]) % mod;
        return ret;
    }
    vec init, trans;
    LL mod;
    int m;
};
int quick_pow(int a,int b)
{
    int ans=1;
    while(b)
    {
        if(b&1) ans=1LL*a*ans%mod;
        a=1LL*a*a%mod;
        b>>=1;
    }
    return ans;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    vector<LL> f;
    f.push_back(0),f.push_back(1);
    for(int i=2;i<=2000;i++)
    {
        f.push_back((f[i-1]+f[i-2])%mod);
    }
    for(int i=1;i<=2000;i++) f[i]=quick_pow(f[i],m);
    for(int i=1;i<=2000;i++) f[i]=(f[i]+f[i-1])%mod;
    LinearRecurrence qw(f,mod,false);
    printf("%lld\n",qw.calc(n));
}