[洛谷P4723]【模板】线性递推
题目大意：求一个满足$k$阶齐次线性递推数列$a_i$的第$n$项。
即：$a_n=\sum\limits_{i=1}^{k}f_i \times a_{n-i}$
题解：线性齐次递推，先见洛谷题解，下回再补
卡点：数组大小计算错误，求逆中途计算时忘记加$mod$等
C++ Code：（这份全部是板子，可以用来测试，但是常数巨大）
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#define maxk 32010
#define maxn 131072
const int mod = 998244353;
#define mul(x, y) static_cast<long long> (x) * (y) % mod
namespace Math {
inline int pw(int base, int p) {
static int res;
for (res = 1; p; p >>= 1, base = mul(base, base)) if (p & 1) res = mul(res, base);
return res;
}
inline int inv(int x) { return pw(x, mod - 2); }
}
inline void reduce(int &x) { x += x >> 31 & mod; }
namespace Poly {
#define N maxn
int lim, s, rev[N], Wn[N];
inline void init(const int n) {
lim = 1, s = -1; while (lim < n) lim <<= 1, ++s;
for (register int i = 1; i < lim; ++i) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
const int t = Math::pw(3, (mod - 1) / lim);
*Wn = 1; for (register int *i = Wn + 1; i != Wn + lim; ++i) *i = mul(*(i - 1), t);
}
inline void FFT(int *A, const int op = 1) {
for (register int i = 1; i < lim; ++i) if (i < rev[i]) std::swap(A[i], A[rev[i]]);
for (register int mid = 1; mid < lim; mid <<= 1) {
const int t = lim / mid >> 1;
for (register int i = 0; i < lim; i += mid << 1)
for (register int j = 0; j < mid; ++j) {
const int X = A[i + j], Y = mul(A[i + j + mid], Wn[t * j]);
reduce(A[i + j] += Y - mod), reduce(A[i + j + mid] = X - Y);
}
}
if (!op) {
const int ilim = Math::inv(lim);
for (register int *i = A; i != A + lim; ++i) *i = mul(*i, ilim);
std::reverse(A + 1, A + lim);
}
}
void INV(int *A, int *B, int n) {
if (n == 1) { *B = Math::inv(*A); return ; }
static int C[N], D[N];
const int len = n + 1 >> 1;
INV(A, B, len), init(len * 3);
std::memcpy(C, A, n << 2), std::memset(C + n, 0, lim - n << 2);
std::memcpy(D, B, len << 2), std::memset(D + len, 0, lim - len << 2);
FFT(C), FFT(D);
for (int i = 0; i < lim; ++i) D[i] = (2 - mul(D[i], C[i]) + mod) * D[i] % mod;
FFT(D, 0);
std::memcpy(B + len, D + len, n - len << 2);
}
void DIV(int *A, int *B, int *Q, int n, int m) {
static int C[N], D[N], E[N];
const int len = n - m + 1;
std::reverse_copy(A, A + n, C), std::reverse_copy(B, B + m, D);
INV(D, E, len), init(len << 1);
std::memset(C + len, 0, lim - len << 2), std::memset(E + len, 0, lim - len << 2);
FFT(C), FFT(E);
for (int i = 0; i < lim; ++i) Q[i] = mul(C[i], E[i]);
FFT(Q, 0), std::reverse(Q, Q + len);
}
void DIV_MOD(int *A, int *B, int *Q, int *R, int n, int m) {
static int C[N], D[N], E[N];
const int len = n - m + 1;
DIV(A, B, Q, n, m), init(n << 1);
std::memcpy(C, A, n << 2), std::memset(C + n, 0, lim - n << 2);
std::memcpy(D, B, m << 2), std::memset(D + m, 0, lim - m << 2);
std::memcpy(E, Q, len << 2), std::memset(E + len, 0, lim - len << 2);
FFT(C), FFT(D), FFT(E);
for (int i = 0; i < lim; ++i) reduce(R[i] = C[i] - mul(D[i], E[i]));
FFT(R, 0);
}
void MOD(int *A, int *B, int m) {
static int Q[N], R[N];
DIV_MOD(A, B, Q, R, (m << 1) - 1, m + 1);
std::memcpy(A, R, m << 2);
}
void POW(int *base, int p, int *Mod, int m) {
static int res[N], T[N];
res[0] = 1;
while (p) {
if (p & 1) {
init(m << 1), std::memset(res + m, 0, lim - m << 2);
std::memcpy(T, base, m << 2), std::memset(T + m, 0, lim - m << 2);
FFT(T), FFT(res);
for (int i = 0; i < lim; ++i) res[i] = mul(res[i], T[i]);
FFT(res, 0); MOD(res, Mod, m);
}
p >>= 1;
if (p) {
init(m << 1), std::memset(base + m, 0, lim - m << 2);
FFT(base);
for (int i = 0; i < lim; ++i) base[i] = mul(base[i], base[i]);
FFT(base, 0), MOD(base, Mod, m);
}
}
std::memcpy(base, res, m << 2);
}
int solve(int *f, int *a, int n, int k) { //a为递推式0~k-1项，f为转移数组1~k项
static int A[maxn], G[maxn];
for (int i = 1; i <= k; ++i) reduce(G[k - i] = -f[i]);
G[k] = A[1] = 1;
Poly::POW(A, n, G, k);
int ans = 0;
for (int i = 0; i < k; ++i) reduce(ans += mul(A[i], a[i]) - mod);
return ans;
}
#undef N
}
int n, k;
int f[maxk], a[maxk];
int main()
std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
std::cin >> n >> k;
for (int i = 1; i <= k; ++i) std::cin >> f[i];
for (int i = 0; i < k; ++i) std::cin >> a[i], reduce(a[i]);
std::cout << Poly::solve(f, a, n, k) << '\n';
return 0;
}
发现取模的那一个多项式是一定的，可以预处理出它的逆元以及点值表达式等，减小常数。
C++ Code：（这一份常数还算正常）
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#define maxk 32010
#define maxn 65536
const int mod = 998244353;
#define mul(x, y) static_cast<long long> (x) * (y) % mod
namespace Math {
inline int pw(int base, int p) {
static int res;
for (res = 1; p; p >>= 1, base = mul(base, base)) if (p & 1) res = mul(res, base);
return res;
}
inline int inv(int x) { return pw(x, mod - 2); }
}
inline void reduce(int &x) { x += x >> 31 & mod; }
namespace Poly {
#define N maxn
int lim, s, rev[N], Wn[N];
inline void init(const int n) {
lim = 1, s = -1; while (lim < n) lim <<= 1, ++s;
for (register int i = 1; i < lim; ++i) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
const int t = Math::pw(3, (mod - 1) / lim);
*Wn = 1; for (register int *i = Wn + 1; i != Wn + lim; ++i) *i = mul(*(i - 1), t);
}
inline void FFT(int *A, const int op = 1) {
for (register int i = 1; i < lim; ++i) if (i < rev[i]) std::swap(A[i], A[rev[i]]);
for (register int mid = 1; mid < lim; mid <<= 1) {
const int t = lim / mid >> 1;
for (register int i = 0; i < lim; i += mid << 1)
for (register int j = 0; j < mid; ++j) {
const int X = A[i + j], Y = mul(A[i + j + mid], Wn[t * j]);
reduce(A[i + j] += Y - mod), reduce(A[i + j + mid] = X - Y);
}
}
if (!op) {
const int ilim = Math::inv(lim);
for (register int *i = A; i != A + lim; ++i) *i = mul(*i, ilim);
std::reverse(A + 1, A + lim);
}
}
void INV(int *A, int *B, int n) {
if (n == 1) { *B = Math::inv(*A); return ; }
static int C[N], D[N];
const int len = n + 1 >> 1;
INV(A, B, len), init(len * 3);
std::memcpy(C, A, n << 2), std::memset(C + n, 0, lim - n << 2);
std::memcpy(D, B, len << 2), std::memset(D + len, 0, lim - len << 2);
FFT(C), FFT(D);
for (int i = 0; i < lim; ++i) D[i] = (2 - mul(D[i], C[i]) + mod) * D[i] % mod;
FFT(D, 0);
std::memcpy(B + len, D + len, n - len << 2);
}
int G[N], INVG[N];
void DIV(int *A, int *Q, int n, int m) {
static int C[N];
const int len = n - m + 1;
std::reverse_copy(A, A + n, C), std::memset(C + len, 0, lim - len << 2);
FFT(C);
for (int i = 0; i < lim; ++i) Q[i] = mul(C[i], INVG[i]);
FFT(Q, 0), std::reverse(Q, Q + len);
}
void DIV_MOD(int *A, int *R, int n, int m) {
static int Q[N];
const int len = n - m + 1;
DIV(A, Q, n, m), std::memset(Q + len, 0, lim - len << 2);
FFT(Q);
for (int i = 0; i < lim; ++i) R[i] = mul(G[i], Q[i]);
FFT(R, 0);
for (int i = 0; i < m; ++i) reduce(R[i] = A[i] - R[i]);
}
void POW(int *A, int p, int m) {
if (!p) return ;
POW(A, p >> 1, m);
static int T[N];
std::memcpy(T, A, m << 2), std::memset(T + m, 0, lim - m << 2);
FFT(T);
for (int i = 0; i < lim; ++i) T[i] = mul(T[i], T[i]);
FFT(T, 0);
if (p & 1) {
for (int i = 2 * m - 1; ~i; --i) T[i] = T[i - 1];
T[0] = 0;
}
DIV_MOD(T, A, 2 * m, m + 1);
}
int solve(int *f, int *a, int n, int k) { //a为递推式0~k-1项，f为转移数组1~k项
static int A[maxn], B[maxn];
for (int i = 1; i <= k; ++i) reduce(G[k - i] = -f[i]);
G[k] = A[0] = 1;
std::reverse_copy(G, G + k + 1, B), B[k] = 0;
INV(B, INVG, k), init(k << 1);
FFT(G), FFT(INVG);
Poly::POW(A, n, k);
int ans = 0;
for (int i = 0; i < k; ++i) reduce(ans += mul(A[i], a[i]) - mod);
return ans;
}
#undef N
}
int n, k;
int f[maxk], a[maxk];
int main() {
std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
std::cin >> n >> k;
for (int i = 1; i <= k; ++i) std::cin >> f[i];
for (int i = 0; i < k; ++i) std::cin >> a[i], reduce(a[i]);
std::cout << Poly::solve(f, a, n, k) << '\n';
return 0;
}
