拆系数FFT
对于任意模数 (mod)
设(m=sqrt {mod})
把多项式(A(x))和(B(x))的系数都拆成(a imes m+b)的形式，时(a, b)都小于(m)
提出，那么一个多项式就可以拆成两个多项式的加法
一个是(a*m)的，一个是(b)的
直接乘法分配律，(aa)一遍，(ab)一遍，(ba)，(bb)一遍，四遍(FFT)
乘出来不会超过取模范围
然后合并直接
[(a imes m+b)(c imes m+d)=a imes c imes m^2+(a imes c+b imes d)m+b imes d
]
这样子要进行 (7) 次 (DFT)
如果研究一下 (myy) (2016) 年的集训队论文就会发现有 (2) 次 或者 (1.5) 次 (DFT) 的 (FFT) 算法
2次的够了吧
(myy) 巧妙的运用了复数的虚部，优化了算法
具体来说
设
[C(x)=A(x)+iB(x)
]
[D(x)=A(x)-iB(x)
]
假设
(c(w_n^k)) 表示将 (C(x)) (DFT) 后的点值
(d(w_n^k)) 表示将 (D(x)) (DFT) 后的点值
(w) 为 (n) 次单位复数根
设 (conj(x)) 表示 (x) 的共轭复数
那么
[c(w_{2n}^{k})=A(w_{2n}^{k})+iB(w_{2n}^{k})=sum_{j=0}^{2n-1}A_jw_{2n}^{jk}+iB_jw_{2n}^{jk}
]
[=sum_{j=0}^{2n-1}(A_j+iB_j)(cosfrac{pi jk}{n}+isin frac{pi jk}{n})
]
[d(w_{2n}^{k})=A(w_{2n}^{k})-iB(w_{2n}^{k})=sum_{j=0}^{2n-1}(A_j-iB_j)(cosfrac{pi jk}{n}+isin frac{pi jk}{n})
]
设 (x=frac{pi jk}{n})
那么
[d(w_{2n}^{k})=sum_{j=0}^{2n-1}(A_jcosx+B_jsinx)+i(A_jsinx-B_jcosx)
]
[=conj(sum_{j=0}^{2n-1}(A_jcosx+B_jsinx)-i(A_jsinx-B_jcosx))
]
[=conj(sum_{j=0}^{2n-1}(A_jcos(-x)-B_jsin(-x))+i(A_jsin(-x)+B_jcos(-x)))
]
[=conj(sum_{j=0}^{2n-1}(A_j+iB_j)(cos(-x)+isin(-x)))
]
[=conj(sum_{j=0}^{2n-1}(A_j+iB_j)(cos(2pi-x)+isin(2pi-x)))
]
[=conj(c(w_{2n}^{2n-k}))
]
也就是说
只需要一次 (DFT) 求出 (c) 就可以求出 (d)
那么
[DFT(A(k))=frac{c(w^{k})+d(w^{k})}{2}
]
[DFT(B(k))=frac{c(w^{k})-d(w^{k})}{2i}
]
再用一次 (DFT^{-1}) 还原出多项式，多项式乘法只要两次 (DFT)
考虑拆系数 (FFT)
(DFT) 两两合并，从 (4) 次变成 (2) 次
(DFT^{-1}) 合并其中一个，从 (3) 次变成 (2) 次
一共 (4) 次
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
namespace IO {
const int maxn((1 << 21) + 1);
char ibuf[maxn], *iS, *iT, obuf[maxn], *oS = obuf, *oT = obuf + maxn - 1, c, st[65];
int f, tp;
char Getc() {
return (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, maxn, stdin), (iS == iT ? EOF : *iS++)) : *iS++);
}
void Flush() {
fwrite(obuf, 1, oS - obuf, stdout);
oS = obuf;
}
void Putc(char x) {
*oS++ = x;
if (oS == oT) Flush();
}
template <class Int> void In(Int &x) {
for (f = 1, c = Getc(); c < '0' || c > '9'; c = Getc()) f = c == '-' ? -1 : 1;
for (x = 0; c <= '9' && c >= '0'; c = Getc()) x = (x << 3) + (x << 1) + (c ^ 48);
x *= f;
}
template <class Int> void Out(Int x) {
if (!x) Putc('0');
if (x < 0) Putc('-'), x = -x;
while (x) st[++tp] = x % 10 + '0', x /= 10;
while (tp) Putc(st[tp--]);
}
}
using IO :: In;
using IO :: Out;
using IO :: Putc;
using IO :: Flush;
const int maxn(1 << 18);
const double pi(acos(-1));
struct Complex {
double a, b;
inline Complex() {
a = b = 0;
}
inline Complex(double _a, double _b) {
a = _a, b = _b;
}
inline Complex operator +(Complex x) const {
return Complex(a + x.a, b + x.b);
}
inline Complex operator -(Complex x) const {
return Complex(a - x.a, b - x.b);
}
inline Complex operator *(Complex x) const {
return Complex(a * x.a - b * x.b, a * x.b + b * x.a);
}
inline Complex Conj() {
return Complex(a, -b);
}
};
Complex a[maxn], b[maxn], w[maxn], a1[maxn], a2[maxn];
int r[maxn], l, deg, g[maxn], h[maxn], mod;
inline void FFT(Complex *p, int opt) {
register int i, j, k, t;
register Complex wn, x, y;
for (i = 0; i < deg; ++i) if (r[i] < i) swap(p[r[i]], p[i]);
for (i = 1; i < deg; i <<= 1)
for(t = i << 1, j = 0; j < deg; j += t)
for (k = 0; k < i; ++k) {
wn = w[deg / i * k];
if (opt == -1) wn.b *= -1;
x = p[j + k], y = wn * p[i + j + k];
p[j + k] = x + y, p[i + j + k] = x - y;
}
}
inline void Mul(int n, int *p, int *q, int *f) {
register int i, k, v1, v2, v3;
register Complex ca, cb, da1, da2, db1, db2;
for (deg = 1, l = 0; deg < n; deg <<= 1) ++l;
for (i = 0; i < deg; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
for (i = 0; i < deg; ++i) w[i] = Complex(cos(pi * i / deg), sin(pi * i / deg));
for (i = 0; i < n; ++i) a[i] = Complex(p[i] & 32767, p[i] >> 15), b[i] = Complex(q[i] & 32767, q[i] >> 15);
for (FFT(a, 1), FFT(b, 1), i = 0; i < deg; ++i) {
k = (deg - i) & (deg - 1), ca = a[k].Conj(), cb = b[k].Conj();
da1 = (ca + a[i]) * Complex(0.5, 0), da2 = (a[i] - ca) * Complex(0, -0.5);
db1 = (cb + b[i]) * Complex(0.5, 0), db2 = (b[i] - cb) * Complex(0, -0.5);
a1[i] = da1 * db1 + (da1 * db2 + da2 * db1) * Complex(0, 1), a2[i] = da2 * db2;
}
for (FFT(a1, -1), FFT(a2, -1), i = 0; i < deg; ++i) {
v1 = (ll)(a1[i].a / deg + 0.5) % mod, v2 = (ll)(a1[i].b / deg + 0.5) % mod;
v3 = (ll)(a2[i].a / deg + 0.5) % mod, f[i] = (((ll)v3 << 30) + ((ll)v2 << 15) + v1) % mod;
if (f[i] < 0) f[i] += mod;
}
}
int main() {
register int len, i, n, m;
In(n), In(m), In(mod), ++n, ++m;
for (i = 0; i < n; ++i) In(h[i]), h[i] %= mod;
for (i = 0; i < m; ++i) In(g[i]), g[i] %= mod;
for (len = 1, n += m - 1; len < n; len <<= 1);
for (Mul(len, h, g, g), i = 0; i < n; ++i) Out(g[i]), Putc(' ');
return Flush(), 0;
}