洛谷P4723 【模板】线性递推(多项式取模 线性代数)

题意

题目链接

Sol

Orz shadowice

注意，下面的代码自带O(随时TLE)倍大常数。。

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
#define ull unsigned long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 4e5 + 10, INF = 1e9 + 10, INV2 = 499122177;
const double eps = 1e-9, pi = acos(-1);
const int G = 3, mod = 998244353;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
namespace Poly {
    int rev[MAXN], GPow[MAXN], A[MAXN], B[MAXN], C[MAXN], lim;
    template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
    template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
    template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
    template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
    int fp(int a, int p, int P = mod) {
        int base = 1;
        for(; p; p >>= 1, a = 1ll * a * a % P) if(p & 1) base = 1ll * base *  a % P;
        return base;
    }
    int GetLen(int x) {
        int lim = 1;
        while(lim < x) lim <<= 1;
        return lim;
    }
    int GetLen2(int x) {
    	int lim = 1;
    	while(lim <= x) lim <<= 1;
    	return lim;
    }
    int GetOrigin(int x) {//¼ÆËãÔ­¸ù
        static int q[MAXN]; int tot = 0, tp = x - 1;
        for(int i = 2; i * i <= tp; i++) if(!(tp % i)) {q[++tot] = i;while(!(tp % i)) tp /= i;}
        if(tp > 1) q[++tot] = tp;
        for(int i = 2, j; i <= x - 1; i++) {
            for(j = 1; j <= tot; j++) if(fp(i, (x - 1) / q[j], x) == 1) break;
            if(j == tot + 1) return i;
        }
    }
    void Init(int Lim) {
        for(int i = 1; i <= Lim; i++) GPow[i] = fp(G, (mod - 1) / i);
    }
    void NTT(int *A, int lim, int opt) {
        int len = 0; for(int N = 1; N < lim; N <<= 1) ++len;
        for(int i = 1; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
        for(int i = 0; i <= lim; i++) if(i < rev[i]) swap(A[i], A[rev[i]]);
        for(int mid = 1; mid < lim; mid <<= 1) {
            int Wn = GPow[mid << 1];
            for(int i = 0; i < lim; i += (mid << 1)) {
                for(int j = 0, w = 1; j < mid; j++, w = mul(w, Wn)) {
                    int x = A[i + j], y = mul(w, A[i + j + mid]);
                    A[i + j] = add(x, y), A[i + j + mid] = add(x, -y);
                }
            }
        }
        if(opt == -1) {
        	reverse(A + 1, A + lim);
            int Inv = fp(lim, mod - 2);
            for(int i = 0; i <= lim; i++) mul2(A[i], Inv);
        }
    }
    void Mul(int *a, int *b, int N, int M) {
        memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
        int lim = 1, len = 0;
        while(lim <= N + M) len++, lim <<= 1;
        for(int i = 0; i <= N; i++) A[i] = a[i];
        for(int i = 0; i <= M; i++) B[i] = b[i];
        NTT(A, lim, 1); NTT(B, lim, 1);
        for(int i = 0; i <= lim; i++) B[i] = mul(B[i], A[i]);
        NTT(B, lim, -1);
        for(int i = 0; i <= N + M; i++) b[i] = B[i];
        memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
    }
    void Inv(int *a, int *b, int len) {//B1 = 2B - A1 * B^2
        if(len == 1) {b[0] = fp(a[0], mod - 2); return ;}
        Inv(a, b, len >> 1);
        for(int i = 0; i < len; i++) A[i] = a[i], B[i] = b[i];
        NTT(A, len << 1, 1); NTT(B, len << 1, 1);
        for(int i = 0; i < (len << 1); i++) mul2(A[i], mul(B[i], B[i]));
        NTT(A, len << 1, -1);
        for(int i = 0; i < len; i++) add2(b[i], add(b[i], -A[i]));
        for(int i = 0; i < (len << 1); i++) A[i] = B[i] = 0;
    }
    void Dao(int *a, int *b, int len) {
    	for(int i = 1; i < len; i++) b[i - 1] = mul(i, a[i]); b[len - 1] = 0;
    }
    void Ji(int *a, int *b, int len) {
        for(int i = 1; i < len; i++) b[i] = mul(a[i - 1], fp(i, mod - 2)); b[0] = 0;
    }
    void Ln(int *a, int *b, int len) {//G(A) = \frac{A}{A'} qiudao zhihou jifen
    	static int A[MAXN], B[MAXN];
    	Dao(a, A, len);
        Inv(a, B, len);
    	NTT(A, len << 1, 1); NTT(B, len << 1, 1);
    	for(int i = 0; i < (len << 1); i++) B[i] = mul(A[i], B[i]);
    	NTT(B, len << 1, -1);
    	Ji(B, b, len << 1);
    	memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
    }
    void Exp(int *a, int *b, int len) {//F(x) = F_0 (1 - lnF_0 + A) but code ..why....
    	if(len == 1) return (void) (b[0] = 1);
        Exp(a, b, len >> 1); Ln(b, C, len);
        C[0] = add(a[0] + 1, -C[0]);
        for(int i = 1; i < len; i++) C[i] = add(a[i], -C[i]);
        NTT(C, len << 1, 1); NTT(b, len << 1, 1);
        for(int i = 0; i < (len << 1); i++) mul2(b[i], C[i]);
        NTT(b, len << 1, -1);
        for(int i = len; i < (len << 1); i++) C[i] = b[i] = 0;
    }
    void Sqrt(int *a, int *b, int len) {
    	static int B[MAXN];
        Ln(a, B, len);
        for(int i = 0; i < len; i++) B[i] = mul(B[i], INV2);
        Exp(B, b, len);
    }
    void Div(int *F, int *G, int *Q, int *R, int N, int M) {//F(n) = G(m) * Q(n - m + 1) + R(m)
        static int Ginv[MAXN], TF[MAXN], TG[MAXN]; memset(Ginv, 0, sizeof(Ginv));
        memcpy(TF, F, (N + 1) << 2); memcpy(TG, G, (M + 1) << 2);

        reverse(F, F + N + 1); reverse(G, G + M + 1);
        Inv(G, Ginv, GetLen2(N - M));//why not M
        Mul(F, Ginv, N - M, N - M);
        for(int i = 0; i <= N - M; i++) Q[i] = Ginv[i];
        reverse(Q, Q + N - M + 1);
        reverse(F, F + N + 1); reverse(G, G + M + 1);
        Mul(Q, G, N - M, M);
        for(int i = 0; i < M; i++) R[i] = add(F[i], -G[i]);
    	memcpy(F, TF, (N + 1) << 2); memcpy(G, TG, (M + 1) << 2);
	}
    void PowNum(int *a, int *b, int P, int N, int len) {
    	static int tx[MAXN], ty[MAXN]; memset(tx, 0, sizeof(tx)); memset(ty, 0, sizeof(ty));
		Ln(a, tx, len);
		for(int i = 0; i < N; i++) ty[i] = mul(P, tx[i]);
		Exp(ty, b, len);
	}
	void MOD(int *A,  int *B, int n, int m) {
		static int Q[MAXN], R[MAXN];
		Div(A, B, Q, R, n, m);
		memcpy(A, R, m << 2);
	}
	void PowPoly(int *base, LL p, int *Mod, int m) {
		static int res[MAXN], T[MAXN]; res[0] = 1;
		while(p) {
			if(p & 1) {
				int lim = GetLen(m << 1);
				memset(res + m, 0, lim - m << 2);;
				memcpy(T, base, m << 2); memset(T + m, 0, lim - m << 2);
				NTT(T, lim, 1); NTT(res, lim, 1);
				for(int i = 0; i < lim; i++) res[i] = mul(res[i], T[i]);
				NTT(res, lim, -1);
				MOD(res, Mod, lim, m);
			}
			p >>= 1;
			if(p) {
				int lim = GetLen(m << 1);
				memset(base + m, 0, lim - m << 2);
				NTT(base, lim, 1);
				for(int i = 0; i < lim; i++) base[i] = mul(base[i], base[i]);
				NTT(base, lim, -1);
				MOD(base, Mod, (m << 1) , m);
			}

		}	

		memcpy(base, res, m << 2);
	}
	int solve(int *f, int *a, LL n, int k) {
		static int AA[MAXN], G[MAXN];
		for(int i = 1; i <= k; i++) G[k - i] = (-f[i] + mod) % mod;
		G[k] = AA[1] = 1;
		PowPoly(AA, n, G, k);
		int ans = 0;
		for(int i = 0; i < k; i++) add2(ans, mul(AA[i], a[i]) - mod);
		return ans;
	}
	int LRec(LL N, int K) {//a_n = \sum_{i=1}^k f_i * a_{n-i}
		static int f[MAXN], a[MAXN];
		Init(8 * K);
		for(int i = 1; i <= K; i++) f[i] = read();
		for(int i = 0; i < K; i++)  a[i] = (read() + mod) % mod;
		return solve(f, a, N, K);
	}
};
using namespace Poly;
signed main() {
	LL N = read();int K = read();
    cout << LRec(N, K);
	return 0;
}