[AHOI2001]多项式乘法 题解
题目地址:洛谷:【P2553】[AHOI2001]多项式乘法 – 洛谷 题目描 …
May all the beauty be blessed.
题目地址:洛谷:【P2553】[AHOI2001]多项式乘法 – 洛谷 题目描 …
题目地址:洛谷:【P4245】【模板】MTT – 洛谷
多项式乘法。
输入样例#1:
5 8 28 19 32 0 182 99 95 77 54 15 3 98 66 21 20 38
输出样例#1:
7 18 25 19 5 13 12 2 9 22 5 27 6 26
数据范围1e5。
由于NTT对模数有要求,我们需要使用任意模数卷积算法,也就是俗称的MTT。具体做法是,把一个数拆成32768x+y的形式,对拆出来的四组数分别卷积后拼起来。这样可以避免计算过程中的越界问题,一般的数据能够满足精度。
单位根最好预处理出来,这样才不会有精度问题。
// Code by KSkun, 2018/3
#include <cstdio>
#include <cmath>
#include <algorithm>
typedef long long LL;
inline char fgc() {
static char buf[100000], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF
: *p1++;
}
inline LL readint() {
register LL res = 0, neg = 1;
char c = fgc();
while(c < '0' || c > '9') {
if(c == '-') neg = -1;
c = fgc();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = fgc();
}
return res * neg;
}
const int MAXN = 1 << 20;
const double PI = acos(-1);
int MO;
struct Complex {
double real, imag;
Complex(double real = 0, double imag = 0) : real(real), imag(imag) {}
inline Complex conj() {
return Complex(real, -imag);
}
inline Complex operator+(Complex rhs) const {
return Complex(real + rhs.real, imag + rhs.imag);
}
inline Complex operator-(Complex rhs) const {
return Complex(real - rhs.real, imag - rhs.imag);
}
inline Complex operator*(Complex rhs) const {
return Complex(real * rhs.real - imag * rhs.imag, imag * rhs.real + real * rhs.imag);
}
inline Complex operator*=(Complex rhs) {
return (*this) = (*this) * rhs;
}
friend inline Complex operator*(double x, Complex cp) {
return Complex(x * cp.real, x * cp.imag);
}
inline Complex operator/(double x) const {
return Complex(real / x, imag / x);
}
inline Complex operator/=(double x) {
return (*this) = (*this) / x;
}
friend inline Complex operator/(double x, Complex cp) {
return x * cp.conj() / (cp.real * cp.real - cp.imag * cp.imag);
}
inline Complex operator/(Complex rhs) const {
return (*this) * rhs.conj() / (rhs.real * rhs.real - rhs.imag * rhs.imag);
}
inline Complex operator/=(Complex rhs) {
return (*this) = (*this) / rhs;
}
inline double length() {
return sqrt(real * real + imag * imag);
}
};
int n, m, len, rev[MAXN];
LL x[MAXN], y[MAXN], z[MAXN];
Complex a[MAXN], b[MAXN], c[MAXN], d[MAXN], e[MAXN], f[MAXN], g[MAXN], h[MAXN], wn[MAXN];
inline void fft(Complex *arr, int f) {
for(int i = 0; i < n; i++) {
if(i < rev[i]) std::swap(arr[i], arr[rev[i]]);
}
for(int i = 1; i < n; i <<= 1) {
for(int j = 0; j < n; j += i << 1) {
for(int k = 0; k < i; k++) {
Complex w = Complex(wn[n / i * k].real, f * wn[n / i * k].imag),
x = arr[j + k], y = w * arr[j + k + i];
arr[j + k] = x + y;
arr[j + k + i] = x - y;
}
}
}
if(f == -1) {
for(int i = 0; i < n; i++) {
arr[i] /= n;
}
}
}
int main() {
n = readint(); m = readint(); MO = readint();
for(int i = 0; i <= n; i++) {
x[i] = readint() % MO;
}
for(int i = 0; i <= m; i++) {
y[i] = readint() % MO;
}
m += n;
for(n = 1; n <= m; n <<= 1) len++;
for(int i = 0; i < n; i++) {
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
}
for(int i = 1; i < n; i <<= 1) {
for(int k = 0; k < i; k++) {
wn[n / i * k] = Complex {cos(PI * k / i), sin(PI * k / i)};
}
}
for(int i = 0; i < n; i++) {
a[i].real = x[i] >> 15; b[i].real = x[i] & 0x7fff;
c[i].real = y[i] >> 15; d[i].real = y[i] & 0x7fff;
}
fft(a, 1); fft(b, 1); fft(c, 1); fft(d, 1);
for(int i = 0; i < n; i++) {
e[i] = a[i] * c[i]; f[i] = b[i] * c[i];
g[i] = a[i] * d[i]; h[i] = b[i] * d[i];
}
fft(e, -1); fft(f, -1); fft(g, -1); fft(h, -1);
for(int i = 0; i < n; i++) {
z[i] = (((LL(round(e[i].real)) % MO) << 30) % MO
+ ((LL(round(f[i].real)) % MO) << 15) % MO
+ ((LL(round(g[i].real)) % MO) << 15) % MO
+ LL(round(h[i].real)) % MO) % MO;
}
for(int i = 0; i <= m; i++) {
printf("%lld ", z[i]);
}
return 0;
}
题目地址:洛谷:【P1919】【模板】A*B Problem升级版(FFT快速傅里叶) & …
这篇文章写得很烂,而且大部分来自于算导,如有感兴趣者可以直接阅读算导或参考资料QAQ!本文 …
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