Berlekamp–Massey 算法

学习博客
学习博客
BM算法用于求解常系数线性递推式。
它可以在 $O(n^2)$ 的时间复杂度内解决问题。

先打个比赛，一会再更
学习算法看上边的两个学习链接就行，板子可以参考我下边这个题
比赛的第一个题用这个算法就能过，太tm巧了
A. Amazing Discovery
正解：矩阵快速幂

BM上板子就能过，太牛逼了这个算法，还求个der的公式
code：

#include<bits/stdc++.h>
using namespace std;
#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 SZ(x) ((int)(x).size())
typedef long long ll;
typedef vector<ll> VI;

const ll mod=998244353;

ll n;
namespace linear_seq 
{
    const int N=10010;
    ll res[N],base[N],_c[N],_md[N];
    vector<ll> Md;
    
    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;
	}
    void mul(ll *a,ll *b,int 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 (int 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;
        int k=SZ(a);
        assert(SZ(a)==SZ(b));
        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 (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;
                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-2ll)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0ll);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1ll-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(0ll);
                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)));
    }
};
ll q_pow(int a, int n){
	ll ans = 1;while(n){
		if(n&1)ans=(ans*a)%mod;a=(1ll*a*a)%mod;n>>=1;
	}return ans;
}
ll zh(int n, int m)
{
	ll sum1 = 1, sum2 = 1;
	for(ll i = 1; i <= m; ++i)
		sum1 = (sum1 * (n - i + 1)) % mod,
		sum2 = (sum2 * i) % mod;
	return sum1 * q_pow(sum2,mod-2) % mod;
}
int main() {
        ll a, b, n;
        cin >> a >> b >> n;
        vector<ll>v;
        for(int i=1;i<=20;i++){// 先想办法求出来前20个
        	ll ans = 0;
			for(int j = 0; j <= i; j += 2)
        	{
        		ans = (ans + zh(i, j) * q_pow(a,i-j) % mod * (q_pow(b,j/2) * 2ll) % mod) % mod;
			}
			v.push_back(ans);
        }
       
    //while (t--) {
        //scanf("%lld",&n);
        printf("%lld\n",linear_seq::gao(v,n-1));
    //}
}