FFT与NTT

讲解:http://www.cnblogs.com/poorpool/p/8760748.html

递归版FFT

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
using namespace std;
const int MAXN = 4000005;
const double PI = acos(-1);
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return rv * fh;
}
struct Complex{
	double x, y;
	Complex (double xx = 0.0, double yy = 0.0) {
		x = xx; y = yy;
	}
	Complex operator + (const Complex &u) const{
		return Complex(x + u.x, y + u.y);
	}
	Complex operator - (const Complex &u) const{
		return Complex(x - u.x, y - u.y);
	}
	Complex operator * (const Complex &u) const{
		return Complex(x * u.x - y * u.y, x * u.y + y * u.x);
	}
}a[MAXN], b[MAXN], buf[MAXN];
int n, m;
void fft(Complex a[], int lim, int opt) {
	if(lim == 1) return;
	int tmp = lim / 2;
	for(int i = 0; i < tmp; i++) {
		buf[i] = a[i * 2];
		buf[tmp + i] = a[i * 2 + 1];
	}
	for(int i = 0; i < lim; i++) {
		a[i] = buf[i];
	}
	fft(a, tmp, opt);
	fft(a + tmp, tmp, opt);
	Complex wn = Complex(cos(PI * 2.0 / lim), opt * sin(PI * 2.0 / lim)), w = Complex(1.0, 0.0);
	for(int i = 0; i < tmp; i++) {
		buf[i] = a[i] + w * a[i + tmp];
		buf[i + tmp] = a[i] - w * a[i + tmp];
		w = w * wn;
	}
	for(int i = 0; i < lim; i++) {
		a[i] = buf[i];
	}
}
int main() {
	n = init(); m = init();
	for(int i = 0; i <= n; i++) a[i].x = init();
	for(int i = 0; i <= m; i++) b[i].x = init();
	int lim = 1;
	while(lim <= n + m) lim <<= 1;
	fft(a, lim, 1);
	fft(b, lim, 1);
	for(int i = 0; i <= lim; i++) {
		a[i] = a[i] * b[i];
	}
	fft(a, lim, -1);
	for(int i = 0; i <= n + m; i++) {
		printf("%d ", (int)(a[i].x / lim + 0.5));
	}
	printf("\n");
	return 0;
}

迭代版FFT

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int MAXN = 4000005;
const double PI = acos(-1);
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return rv * fh;
}
struct Complex{
	double x, y;
	Complex (double xx = 0.0, double yy = 0.0) {
		x = xx; y = yy;
	}
	Complex operator + (const Complex &u) const {
		return Complex(x + u.x, y + u.y);
	}
	Complex operator - (const Complex &u) const{
		return Complex(x - u.x, y - u.y);
	}
	Complex operator * (const Complex &u) const{
		return Complex(x * u.x - y * u.y, x * u.y + y * u.x);
	}
}a[MAXN], b[MAXN], buf[MAXN];
int n, m, rev[MAXN], lim, limcnt;
void fft(Complex a[], int opt) {
	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) {
		Complex wn = Complex(cos(PI / mid), opt * sin(PI / mid));
		for(int R = mid << 1, j = 0; j < lim; j += R) {
			Complex w = Complex(1.0, 0.0);
			for(int k = 0; k < mid; k++) {
				Complex x = a[j + k], y = w * a[j + mid + k];
				a[j + k] = x + y;
				a[j + mid + k] = x - y;
				w = w * wn;
			}
		}
	}
}
int main() {
	n = init(); m = init();
	for(int i = 0; i <= n; i++) a[i].x = init();
	for(int i = 0; i <= m; i++) b[i].x = init();
	lim = 1;
	while(lim <= n + m) {lim <<= 1; limcnt++;}
	for(int i = 0; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (limcnt - 1));
	fft(a, 1);
	fft(b, 1);
	for(int i = 0; i <= lim; i++) a[i] = a[i] * b[i];
	fft(a, -1);
	for(int i = 0; i <= n + m; i++) {
		printf("%d ", (int)(a[i].x / lim + 0.5));
	}
	return 0;
}

NTT

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define ll long long
using namespace std;
const int MAXN = 4000005, MOD = 998244353, gg = 3, gi = 332748118;
ll init() {
	ll rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return fh * rv;
}
ll lim = 1, limcnt, rev[MAXN], n, m, a[MAXN], b[MAXN];
ll ksm(ll a, ll k) {
	ll ans = 1;
	while(k) {
		if(k & 1ll) {
			(ans *= a) %= MOD;
		}
		(a *= a) %= MOD;
		k >>= 1;
	}
	return ans;
}
void ntt(ll a[], int opt) {
	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) {
		ll wn = ksm(opt == 1 ? gg : gi, (MOD - 1) / (mid << 1));
		for(int R = mid << 1, j = 0; j < lim; j += R) {
			ll w = 1;
			for(int k = 0; k < mid; k++) {
				ll x = a[j + k], y = w * a[j + mid + k] % MOD;
				a[j + k] = (x + y) % MOD;
				a[j + mid + k] = (x - y + MOD) % MOD;
				(w *= wn) %= MOD;
			}
		}
	}
	if(opt == -1) {
		ll inv = ksm(lim, MOD - 2);
		for(int i = 0; i <= lim; i++) {
			(a[i] *= inv) %= MOD;
		}
	}
}
int main() {
	n = init(); m = init();
	for(int i = 0; i <= n; i++) {
		a[i] = init();
	}
	for(int i = 0; i <= m; i++) b[i] = init();
	while(lim <= (n + m)) lim <<= 1, limcnt++;
	for(int i = 0; i <= lim; i++) 
		rev[i] = (rev[i>>1]>>1) | ((i&1)<<(limcnt-1));
	ntt(a, 1);
	ntt(b, 1);
	for(int i = 0; i <= lim; i++) (a[i] = a[i] * b[i]) %= MOD;
	ntt(a, -1);
	for(int i = 0; i <= n + m; i++) {
		printf("%lld ", a[i]);
	}
	printf("\n");
	return 0;
}
posted @ 2018-05-23 16:50  Mr_Wolfram  阅读(201)  评论(1编辑  收藏  举报