【数学】多项式（模板）

【数学】多项式（模板）
不打开O2的话效率不够⾼，但是这种程度上的优化确实应该留给编译器去做了，学个计算机假如连编译器能做的优化都要⼿动去做并且牺牲程序的简洁性，那还是早点改⾏去⼯地搬砖吧。

const int MAXLOGN = 20;
const int MAXN = 1 << MAXLOGN;
const int MOD = 998244353;
const int G = 3;
inline int qadd(const int &x, const int &y) {
int r = x + y;
return r >= MOD ? r - MOD : r;
}
inline int qsub(const int &x, const int &y) {
int r = x - y;
return r < 0 ? r + MOD : r;
}
inline int qmul(const int &x, const int &y) {
ll r = 1LL * x * y;
if(r >= MOD)
r %= MOD;
return r;
}
int qpow(int x, int n) {
ll res = 1;
while(n) {
if(n & 1)
res = qmul(res, x);
x = qmul(x, x);
n >>= 1;
}
return res;
}
using poly_t = int[MAXN];
using poly = int *const;
/* Show h[0...n-1], be careful that the length of h is less than n. */
void polyshow(poly &h, int n) {
for(int i = 0; i != n; ++i)
printf("%d%c", h[i], " \n"[i == n - 1]);
}
void FNTT(poly &h, int n, int op) {
static int gl[MAXLOGN], rev[MAXN], revl;
if(gl[0] == 0) {
for(int l = 0; l < MAXLOGN; ++l)
gl[l] = qpow(G, (MOD - 1) / (1 << (l + 1)));
}
if((1 << revl) != n) {
for(revl = 0; (1 << revl) != n; ++revl);
for(int i = 0; i < n; i++)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (revl - 1));
}
for(int i = 0; i < n; ++i) {
if(i < rev[i])
swap(h[i], h[rev[i]]);
}
for(int l = 0, len; (len = 1 << l) < n; ++l) {
for(int i = 0, Gl = gl[l]; i < n; i += (len << 1)) {
for(int j = i, w = 1; j < i + len; ++j, w = qmul(w, Gl)) {
int u = h[j], t = qmul(h[j + len], w);
h[j] = qadd(u, t), h[j + len] = qsub(u, t);
}
}
}
if(op == -1) {
reverse(h + 1, h + n);
for(int i = 0, inv = qpow(n, MOD - 2); i < n; ++i)
h[i] = qmul(h[i], inv);
}
}
/* Enlarge n to the smallest power of 2. */
void pretreat(poly &h, int &n) {
int tn = 1;
while(tn < n)
tn <<= 1;
fill(h + n, h + tn, 0);
n = tn;
}
/* Add h1 and h2, and store the result in f. */
int polyadd(poly &h1, int n1, poly &h2, int n2, poly &f) {
int n = max(n1, n2);
fill(h1 + n1, h1 + n, 0);
fill(h2 + n2, h2 + n, 0);
for(int i = 0; i != n; ++i)
f[i] = qadd(h1[i], h2[i]);
return n;
}
/* Substract h2 from h1, and store the result in f. */
int polysub(poly &h1, int n1, poly &h2, int n2, poly &f) {
int n = max(n1, n2);
fill(h1 + n1, h1 + n, 0);
fill(h2 + n2, h2 + n, 0);
for(int i = 0; i != n; ++i)
f[i] = qsub(h1[i], h2[i]);
return n;
}
/* Multiply h1 and h2, and store the result in f. */
int polymul(poly &h1, int n1, poly &h2, int n2, poly &f) {
int n = n1 + n2 - 1, tn = 1;
while(tn < n)
tn <<= 1;
fill(h1 + n1, h1 + tn, 0), FNTT(h1, tn, 1);
fill(h2 + n2, h2 + tn, 0), FNTT(h2, tn, 1);
for(int i = 0; i != tn; ++i)
f[i] = qmul(h1[i], h2[i]);
FNTT(f, tn, -1);
return n;
}
/* The following methods are solved in the sense of modulo x^n */ /* Get the inverse of h, and store the result in f. */
void polyinv(poly &h, int n, poly &f) {
pretreat(h, n);
static poly_t tmp;
fill(f, f + n + n, 0), f[0] = qpow(h[0], MOD - 2);
for(int t = 2; t <= n; t <<= 1) {
copy(h, h + t, tmp), fill(tmp + t, tmp + t + t, 0);
FNTT(f, t + t, 1), FNTT(tmp, t + t, 1);
for(int i = 0; i != t + t; ++i)
f[i] = qmul(f[i], qsub(2, qmul(f[i], tmp[i])));
FNTT(f, t + t, -1);
fill(f + t, f + t + t, 0);
}
}
/* Get the derivative of h, and store the result in f. */
void polyder(poly & h, int n, poly & f) {
for(int i = 1; i != n; ++i)
f[i - 1] = qmul(h[i], i);
f[n - 1] = 0;
}
/* Get the integral of h, and store the result in f. */
void polyint(poly &h, int n, poly &f, int C = 0) {
static int inv[MAXN];
if(inv[0] == 0) {
inv[1] = 1;
for(int i = 2; i < MAXN; ++i)
inv[i] = qmul(inv[MOD % i], (MOD - MOD / i));
inv[0] = -1;
}
for(int i = n - 1; i != 0; --i)
f[i] = qmul(h[i - 1], inv[i]); /* or inv[i] = qpow(i, MOD - 2) */ f[0] = C; /* constant C */
}
/* Get the logarithm of h, and store the result in f. */
void polylog(poly &h, int n, poly & f) {
pretreat(h, n);
static poly_t tmp;
polyder(h, n, tmp), polyinv(h, n, f);
fill(tmp + n, tmp + n + n, 0);
FNTT(tmp, n + n, 1), FNTT(f, n + n, 1);
for(int i = 0; i != n + n; ++i)
tmp[i] = qmul(tmp[i], f[i]);
FNTT(tmp, n + n, -1);
polyint(tmp, n, f);
}
/* Get the exponent of h, and store the result in f. */ void polyexp(poly & h, int n, poly & f) {
pretreat(h, n);
static poly_t tmp;
fill(f, f + n + n, 0), f[0] = 1;
for(int t = 2; t <= n; t <<= 1) {
polylog(f, t, tmp);
tmp[0] = qsub(qadd(h[0], 1), tmp[0]);
for(int i = 1; i != t; ++i)
tmp[i] = qsub(h[i], tmp[i]);
fill(tmp + t, tmp + t + t, 0);
FNTT(f, t + t, 1), FNTT(tmp, t + t, 1);
for(int i = 0; i != t + t; ++i)
f[i] = qmul(f[i], tmp[i]);
FNTT(f, t + t, -1);
fill(f + t, f + t + t, 0);
}
}
void polypow1(poly &h, int n, int k1, int k2, poly &f) {
static poly_t tmp;
polylog(h, n, tmp);
for(int i = 0; i != n; ++i)
tmp[i] = qmul(tmp[i], k1);
polyexp(tmp, n, f);
}
void polypow2(poly &h, int n, int k1, int k2, poly &f) {
for(int i = 0, invh = qpow(h[0], MOD - 2); i != n; ++i) f[i] = qmul(h[i], invh);
polypow1(h, n, k1, k2, f);
for(int i = 0, hk = qpow(h[0], k2); i != n; ++i)
f[i] = qmul(f[i], hk);
}
/* Get the power k of h, and store the result in f.
k1 = k mod MOD, k2 = k mod (MOD - 1), k = true k */ void polypow(poly &h, int n, int k1, int k2, int k, poly &f) { int t = 0;
for(; t < n && h[t] == 0; ++t);
if(1LL * k * t >= n)
fill(f, f + n, 0);
else {
static poly_t tmp;
for(int i = 0; i + t != n; ++i)
tmp[i] = h[i + t];
fill(tmp + n - t, tmp + n, 0);
polypow2(tmp, n, k1, k2, f);
for(int i = n - 1; i >= k * t; --i)
f[i] = f[i - k * t];
fill(f, f + k * t, 0);
}
}。