B-M （特征方程教程）模板_b-m方程-优快云博客

本文链接：https://blog.youkuaiyun.com/sdut_jk17_zhangming/article/details/88080025

本文介绍了一种高效的模数运算方法，并详细解释了如何使用此方法解决线性序列问题。通过定义和实现特定的算法，如模幂运算、多项式乘法和线性序列求解，文章展示了如何在大规模数据集上快速准确地找到线性序列的第n项。此外，还提供了一个完整的C++代码示例，用于演示算法的实际应用。

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出处 https://jhcloud.top/blog/?p=1271

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,a,n)for(int i=a;i<n;i++)
#define per(i,a,n)for(int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((ll)(x).size())
#define all(x) (x).begin(),(x).end()
typedef vector<ll> VI;
typedef pair<ll,ll>PII;
const ll mod = 1e9+7;

ll powmod(ll A,ll b)
{
    ll res = 1;
    A%=mod;
    for(; b; b>>=1)
    {
        if(b&1)
            res  =res*A%mod;
        A = A*A%mod;
    }
    return res;
}

namespace liner_seq
{
    const int N = 100010;
    ll res[N],base[N],_c[N],_md[N];
    vector<ll>Md;
    void mul(ll *a,ll *b,ll k)
    {
        rep(i,0,k+k)_c[i] = 0;
        rep(i,0,k)if(a[i])
            rep(j,0,k)_c[i+j] = (_c[i+j]+a[i]*b[j])%mod;
        for(ll i=k+k-1; i>=k; i--)
        {
            if(_c[i])
                rep(j,0,SZ(Md))_c[i-k+Md[j]] = (_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        }
        rep(i,0,k)a[i]=_c[i];
    }

    ll solve(ll n,VI a,VI b)
    {
        ll ans =0,pnt =0;
        ll k = SZ(a);
        rep(i,0,k) _md[k-1-i] = -a[i];
        _md[k] = 1;
        Md.clear();
        rep(i,0,k)if(_md[i]!=0)
            Md.push_back(i);
        rep(i,0,k)res[i ] =base[i] =0;
        res[0] = 1;
        while((1ll<<pnt)<=n)
            pnt++;
        for(ll p=pnt; p>=0; p--)
        {
            mul(res,res,k);
            if((n>>p)&1)
            {
                for(ll i=k-1; i>=0; i--)
                    res[i+1] = res[i];
                res[0 ] =0;
                rep(j,0,SZ(Md))res[Md[j]] = (res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,0,k)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);
        ll L  =0,m=1,b = 1;
        rep(n,0,SZ(s))
        {
            ll d =0;
            rep(i,0,L+1)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);
                rep(i,0,SZ(B))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);
                rep(i,0,SZ(B))C[i+m] = (C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    ll gao(VI a,ll n)
    {
        VI  c = BM(a);
        c.erase(c.begin());
        rep(i,0,SZ(c))c[i] = (mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
}

VI a;

int main()
{
    for(int i=0; i<18; i++)
    {
        a.push_back(i);
    }
    int T;
    cin>>T;
    while(T--)
    {
        long long n;
        cin>>n;
        printf("%lld\n",liner_seq::gao(a,n)%mod);
    }
    return 0;
}