链接
算法概述
这个算法是利用了转移矩阵 的特殊性，写出转移矩阵的特征多项式
然后再利用“将 作为自变量带入其特征多项式，得到的函数值为 ”这个特点，将 化为一个 次多项式
进而把 转化成
再利用 这 个向量的第一维恰好就是原始给出的 个向量
从而得出答案的
由于这里的大佬们实在讲的太好了，我觉得我不一定比他们写得好，所以就不再详细描述算法了
流程
令
用多项式带余除法+快速幂，求出 除以 的余数 ， 的 次方项系数即为
计算答案
时间复杂度
十分有效的常数优化
因为每次整除的时候，除的都是同一个多项式，没必要每次都求逆，直接预处理即可
实践证明，时间可以缩短一半
代码
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
ll c, f(1);
for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
for(;isdigit(c);c=getchar())x=x*10+c-0x30;
return f*x;
}
#define mod 998244353ll
struct EasyMath
{
ll prime[maxn], phi[maxn], mu[maxn];
bool mark[maxn];
ll fastpow(ll a, ll b, ll c)
{
ll t(a%c), ans(1ll);
for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
return ans;
}
void exgcd(ll a, ll b, ll &x, ll &y)
{
if(!b){x=1,y=0;return;}
ll xx, yy;
exgcd(b,a%b,xx,yy);
x=yy, y=xx-a/b*yy;
}
ll inv(ll x, ll p) //p是素数
{return fastpow(x%p,p-2,p);}
ll inv2(ll a, ll p)
{
ll x, y;
exgcd(a,p,x,y);
return (x+p)%p;
}
void shai(ll N)
{
ll i, j;
for(i=2;i<=N;i++)mark[i]=false;
*prime=0;
phi[1]=mu[1]=1;
for(i=2;i<=N;i++)
{
if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
for(j=1;j<=*prime and i*prime[j]<=N;j++)
{
mark[i*prime[j]]=true;
if(i%prime[j]==0)
{
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
mu[i*prime[j]]=-mu[i];
phi[i*prime[j]]=phi[i]*(prime[j]-1);
}
}
}
ll CRT(vector<ll> a, vector<ll> m) //要求模数两两互质
{
ll M=1, ans=0, n=a.size(), i;
for(i=0;i<n;i++)M*=m[i];
for(i=0;i<n;i++)(ans+=a[i]*(M/m[i])%M*inv2(M/m[i],m[i]))%=M;
return ans;
}
}em;
struct NTT
{
ll n;
vector<ll> R;
void init(ll bound) //bound是积多项式的最高次幂
{
ll L(0);
for(n=1;n<=bound;n<<=1,L++);
R.resize(n);
for(ll i=0;i<n;i++)R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
}
void ntt(vector<ll>& a, int opt)
{
ll i, j, k, wn, w, x, y, inv(em.fastpow(n,mod-2,mod));
for(i=0;i<n;i++)if(i>R[i])swap(a[i],a[R[i]]);
for(i=1;i<n;i<<=1)
{
if(opt==1)wn=em.fastpow(3,(mod-1)/(i<<1),mod);
else wn=em.fastpow(3,(mod-1-(mod-1)/(i<<1)),mod);
for(j=0;j<n;j+=i<<1)
for(w=1,k=0;k<i;k++,w=w*wn%mod)
{
x=a[k+j], y=a[k+j+i]*w%mod;
a[k+j]=(x+y)%mod, a[k+j+i]=(x-y)%mod;
}
}
if(opt==-1)for(i=0;i<n;i++)(a[i]*=inv)%=mod;
}
};
struct Poly
{
vector<ll> v;
ll n; //n是最高次项的次数
Poly(ll N){v.resize(N+1);n=N;}
Poly(const Poly& p){v=p.v; n=p.n;}
void resize(ll N){n=N; v.resize(N+1);}
ll& operator[](ll id){return v[id];}
void show()
{
printf("n=%lld\n",n);
ll i; rep(i,0,n-1)printf("%lldx^%lld + ",(v[i]+mod)%mod,i);
printf("%lldx^%lld\n",(v[n]+mod)%mod,n);
}
};
Poly operator+(Poly A, Poly B)
{
ll i;
Poly C(max(A.n,B.n));
A.resize(C.n), B.resize(C.n);
rep(i,0,C.n)C[i]=A[i]+B[i];
return C;
}
Poly operator*(Poly A, Poly B)
{
NTT ntt;
ll i, n=A.n+B.n;
ntt.init(A.n+B.n);
A.resize(ntt.n-1), B.resize(ntt.n-1);
ntt.ntt(A.v,1), ntt.ntt(B.v,1);
Poly C(ntt.n-1);
rep(i,0,C.n)C[i]=(A[i]*B[i])%mod;
ntt.ntt(C.v,-1);
C.resize(n);
return C;
}
Poly operator~(Poly F)
{
Poly H(0), f(0);
ll i, tar, j;
H[0]=em.inv(F[0],mod);
f.resize(0), f[0]=F[0];
for(tar=1;tar<=F.n;tar*=2);
F.resize(tar-1);
for(i=1;i<=tar;i*=2)
{
f.resize(i-1);
rep(j,i>>1,i-1)f[j]=F[j];
Poly t=H*H*f; t.resize(i-1);
H.resize(i-1);
rep(j,0,i-1)H[j]=(2*H[j]-t[j])%mod;
}
H.resize(F.n);
return H;
}
Poly gg(0);
pair<Poly,Poly> operator/(Poly F, Poly G)
{
ll n=F.n, m=G.n, i;
if(m>n)return make_pair(Poly(0),F);
Poly f=F, g=gg;
reverse(f.v.begin(),f.v.end());
f.resize(n-m), g.resize(n-m);
Poly Q=f*g;
Q.resize(n-m);
reverse(Q.v.begin(),Q.v.end());
Poly t(Q*G), R(m-1);
rep(i,0,m-1)R[i]=F[i]-t[i];
return make_pair(Q,R);
}
ll calc(vector<ll> f, vector<ll> a, ll n)
{
ll k=f.size(), i, j;
Poly t(1), r(k-1), F(k);
rep(i,0,k-1)F[i]=-f[k-1-i];
F[k]=1;
gg=F;
reverse(gg.v.begin(),gg.v.end());
gg=~gg;
t[1]=1;
r[0]=1;
for(ll b=n;b;b>>=1,t=(t*t/F).second)
if(b&1)r=(r*t/F).second;
ll ans=0;
rep(i,0,k-1)(ans+=r[i]*a[i])%=mod;
return ans;
}
int main()
{
ll n, k, i;
n=read(), k=read();
vector<ll> f(k), a(k);
rep(i,0,k-1)f[i]=read()%mod;
rep(i,0,k-1)a[i]=read()%mod;
printf("%lld",(calc(f,a,n)+mod)%mod);
return 0;
}