牛客多校第九场 A The power of Fibonacci 杜教bm解线性递推

题意：
计算斐波那契数列前n项和的m次方模1e9

题解：

$F[i] – F[i-1] – F[i-2] = 0$ 

$F[i]^2 – 2 F[i-1]^2 – 2 F[i-2]^2 + F[i-3] = 0$

$F[i]^3 – 3 F[i-1]^3 – 6 F[i-2]^3 + 3 F[i-3] + F[i-4] = 0$

可以看出，斐波那契数列的高次幂依然是可以线性递推出来的，可以推广到任意幂次的情况，具体证明参见Fibonomial Coefficient

硬套杜教bm即可。

#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <unordered_map>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef vector<long long> VI;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const ll mod=1e9;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define SZ(x) ((long long)(x).size())
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
ll powmod(ll a,ll b)
{
    ll res=1ll;
    while(b)
    {
        if(b&1) res=res*a%mod;
        a=a*a%mod,b>>=1;
    }
    return res;
}
namespace linear_seq {
    const int N=10010;
    using int64 = long long;
    using vec = std::vector<int64>;
    ll res[N],base[N],_c[N],_md[N];
    vector<int> Md;
    void mul(ll *a,ll *b,int k) {
        FOR(i,0,k+k-1) _c[i]=0;
        FOR(i,0,k-1) if (a[i]) FOR(j,0,k-1) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (int i=k+k-1;i>=k;i--) if (_c[i])
            FOR(j,0,SZ(Md)-1) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        FOR(i,0,k-1) a[i]=_c[i];
    }
    int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d
",SZ(b));
        ll ans=0,pnt=0;
        int k=SZ(a);
        assert(SZ(a)==SZ(b));
        FOR(i,0,k-1) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        FOR(i,0,k-1) if (_md[i]!=0) Md.push_back(i);
        FOR(i,0,k-1) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (int p=pnt;p>=0;p--) {
            mul(res,res,k);
            if ((n>>p)&1) {
                for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                FOR(j,0,SZ(Md)-1) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        FOR(i,0,k-1) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    VI BM(VI s) {
        VI C(1,1),B(1,1);
        int L=0,m=1,b=1;
        FOR(n,0,SZ(s)-1) {
            ll d=0;
            FOR(i,0,L) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n) {
                VI T=C;
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                FOR(i,0,SZ(B)-1) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            } else {
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                FOR(i,0,SZ(B)-1) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    static void extand(vec &a, size_t d, int64 value = 0) {
        if (d <= a.size()) return;
        a.resize(d, value);
    }
    static void exgcd(int64 a, int64 b, int64 &g, int64 &x, int64 &y) {
        if (!b) x = 1, y = 0, g = a;
        else {
            exgcd(b, a % b, g, y, x);
            y -= x * (a / b);
        }
    }
    static int64 crt(const vec &c, const vec &m) {
        int n = c.size();
        int64 M = 1, ans = 0;
        for (int i = 0; i < n; ++i) M *= m[i];
        for (int i = 0; i < n; ++i) {
            int64 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, int64 mod) {
        auto inverse = [](int64 a, int64 m) {
            int64 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, int64 mod, int64 p, int64 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 (int64 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) {
                    int64 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 {
                            int64 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<int64, int64, int>> fac;
        for (int64 i = 2; i * i <= mod; ++i)
            if (mod % i == 0) {
                int64 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) {
            int64 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;
    }
    ll gao(VI a,ll n,ll mod,bool prime=true) {
        VI c;
        if(prime) c=BM(a);
        else c=ReedsSloane(a,mod);
        c.erase(c.begin());
        FOR(i,0,SZ(c)-1) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};
ll gmod(ll a,ll b)
{
    ll res=1;
    while(b)
    {
        if(b&1) res=res*a%mod;
        a=a*a%mod,b>>=1;
    }
    return res;
}
int main()
{
   vector<ll> v;
    v.push_back(0);
    v.push_back(1);
    ll n, m;
    cin >> n >> m;
    for (int i = 2; i < 1000; ++i) v.push_back((v[i - 1] + v[i - 2]) % mod);
    for (auto& t : v) t = powmod(t, m);
    for (int i = 1; i < 1000; ++i) v[i] = (v[i - 1] + v[i]) % mod;
    printf("%lld
", linear_seq::gao(v,n,mod,false));
}

话说杜教BM真是神器啊。

补充：

模数为1e9+9时的计算方法（zoj3774）

考虑斐波那契数列的通项公式：$frac{1}{sqrt5} left [ left ( frac{1+sqrt5}{2} ight ) ^n+left (frac{1-sqrt5}{2} ight) ^n ight ]$

记

$a=left ( frac{1+sqrt5}{2} ight ) ^n$
$b=left ( frac{1-sqrt5}{2} ight ) ^n$

然后可以将斐波那契数列每一项二项式展开，共有m+1项，

$(Feb_i)^m=left ( a^i + b^i ight )^m=sum _{r=0}^m(-1)^rC_m^r(a^{m-r}b^r)^i$

发现，对于数列从1-n的每一项，它们二次展开后的m+1项中，系数相同的项是一个等比数列，比如斐波那契数列第1-n项二次展开后的第r项系数均为$(-1)^rC_m^r$

$(-1)^rC_m^r(a^{m-r}b^r)$　　$(-1)^rC_m^r(a^{m-r}b^r)^2$　　$(-1)^rC_m^r(a^{m-r}b^r)^3$　　$(-1)^rC_m^r(a^{m-r}b^r)^4$　　balabala......　　$(-1)^rC_m^r(a^{m-r}b^r)^n$

可以用等比数列计算，记$t_r=a^rb^{m-r}$

$ans_r=(-1)^rC_m^rfrac{t_r(t_r^n-1)}{t_r-1}$

$ans=sum _{r=0}^m ans_r$

枚举r从0到m并计算，求和即可。

注意，根号5在模1e9+9意义下可用383008016代替（二次剩余），根号1/5可用276601605代替（逆元）

然而在模1e9意义下无法用整数表示这两个数，故此方法失效。

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;
const int P=1000000009;
const int INV2=500000005;
const int SQRT5=383008016;
const int INVSQRT5=276601605;
const int A=691504013;
const int B=308495997;

const int N=100005;

ll n,K;
ll fac[N],inv[N];
ll pa[N],pb[N];

inline void Pre(int n){
    fac[0]=1; 
    for (int i=1;i<=n;i++) fac[i]=fac[i-1]*i%P;
    inv[1]=1; 
    for (int i=2;i<=n;i++) inv[i]=(P-P/i)*inv[P%i]%P;
    inv[0]=1; 
    for (int i=1;i<=n;i++) inv[i]=inv[i]*inv[i-1]%P;
    pa[0]=1; 
    for (int i=1;i<=n;i++) pa[i]=pa[i-1]*A%P;
    pb[0]=1; 
    for (int i=1;i<=n;i++) pb[i]=pb[i-1]*B%P;
}

inline ll C(int n,int m){
    return fac[n]*inv[m]%P*inv[n-m]%P;
}

inline ll Pow(ll a,ll b){
    ll ret=1;
    for (;b;b>>=1,a=a*a%P)
        if (b&1)
            ret=ret*a%P;
    return ret;
}

inline ll Inv(ll x){
    return Pow(x,P-2);
}

inline void Solve(){
    ll Ans=0;
    for (int j=0;j<=K;j++){
        ll t=pa[K-j]*pb[j]%P,tem;
        tem=t==1?n%P:t*(Pow(t,n)-1+P)%P*Inv(t-1)%P;
        if (~j&1)
            Ans+=C(K,j)*tem%P,Ans%=P;
        else
            Ans+=P-C(K,j)*tem%P,Ans%=P;
    }
    Ans=Ans*Pow(INVSQRT5,K)%P;
    printf("%lld
",Ans);
}

int main(){
    int T;
    #ifdef LOCAL
    freopen("t.in","r",stdin);
    freopen("t.out","w",stdout);
    #endif
    Pre(100000);
    scanf("%d",&T);
    while (T--){
        scanf("%lld%lld",&n,&K);
        Solve();
    }
}