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  • BM递推

    从别的大佬处看到的模板

    #include<bits/stdc++.h>
    #define fi first
    #define se second
    #define INF 0x3f3f3f3f
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
    #define pqueue priority_queue
    #define NEW(a,b) memset(a,b,sizeof(a))
    #define Pii pair<int,int>
    #define VI vector<int>
    const double pi=4.0*atan(1.0);
    const double e=exp(1.0);
    const int maxn=3e6+8;
    typedef long long LL;
    typedef unsigned long long ULL;
    const LL mod=1e9+7;
    const ULL base=1e7+7;
    using namespace std;
    LL qpow(LL a,LL b){
        LL ans=1;
        a%=mod;
        while(b){
            if(b&1){ans=ans*a%mod;}
            a=a*a%mod;
            b>>=1;
        }
        return ans;
    }
    LL n;
    namespace linear_seq{
        const int N=10010;
        LL res[N],base[N],_c[N],_md[N];
        VI Md;
        void mul(LL *a,LL *b,int k){
            for(int i=0;i<k+k;i++) _c[i]=0;
            for(int i=0;i<k;i++) if(a[i]) for (int j=0;j<k;j++) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
            for(int i=k+k-1;i>=k;i--) if(_c[i])
                for(int j=0;j<(int)(Md).size();j++) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
            for(int i=0;i<k;i++) a[i]=_c[i];
        }
        int solve(LL n,VI a,VI b){
            LL ans=0,pnt=0;
            int k=(int)(a).size();
            assert((int)(a).size()==(int)(b).size());
            for(int i=0;i<k;i++) _md[k-1-i]=-a[i];_md[k]=1;
            Md.clear();
            for (int i=0;i<k;i++) if (_md[i]!=0) Md.push_back(i);
            for (int i=0;i<k;i++) 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 (int j=0;j<(int)(Md).size();j++) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
                }
            }
            for(int i=0;i<k;i++) 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(int n=0;n<(int)(s).size();n++){
                LL d=0;
                for(int i=0;i<L+1;i++) 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*qpow(b,mod-2)%mod;
                    while ((int)(C).size()<(int)(B).size()+m) C.push_back(0);
                    for (int i=0;i<(int)(B).size();i++) 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*qpow(b,mod-2)%mod;
                    while ((int)(C).size()<(int)(B).size()+m) C.push_back(0);
                    for (int i=0;i<(int)(B).size();i++) C[i+m]=(C[i+m]+c*B[i])%mod;
                    ++m;
                }
            }
            return C;
        }
        int gao(VI a,LL n){
            VI c=BM(a);
            c.erase(c.begin());
            for(int i=0;i<(int)(c).size();i++) c[i]=(mod-c[i])%mod;
            return solve(n,c,VI(a.begin(),a.begin()+(int)(c).size()));
        }
    };
    int main(){
        int t;
        for(scanf("%d",&t);t;t--){
            scanf("%lld",&n);
            printf("%d
    ",linear_seq::gao(VI{1,3,5,7},n-1));
        }
    }

     https://ac.nowcoder.com/acm/contest/889/A

    取余可以不是质数的高级BM/LR

    #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()
    {
        cin.tie(0);
        cout.tie(0);
        ll n, m;
        cin >> n >> m;
        std::vector<long long> f = {0, 1};
        for (int i = 2; i < 1000; i++)
            f.push_back((f[i - 1] + f[i - 2]) % mod);
        for (auto& t : f) t = gmod(t, m);
        for (int i = 1; i < 1000; i++)
            f[i] = (f[i - 1] + f[i]) % mod;
        //LinearRecurrence qw(f,mod,false);
        printf("%lld
    ",linear_seq::gao(f,n,mod,false));
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Profish/p/9738143.html
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