bzoj 2179 FFT快速傅立叶 FFT

题面

题目传送门

解法

题如其名……

不妨将多项式的\(x^i\)变成\(10^i\)，然后就是一个比较简单的FFT了

md读进来的是一个字符串，并且要倒序

最后注意进位问题

时间复杂度：\(O(n\ log\ n)\)

代码

#include <bits/stdc++.h>
#define N 1 << 18
using namespace std;
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
const double pi = acos(-1);
struct Complex {
	double x, y;
	Complex (double tx = 0, double ty = 0) {x = tx, y = ty;}
} a[N], b[N];
char s1[N], s2[N];
int ans[N], rev[N];
Complex operator + (Complex a, Complex b) {return (Complex) {a.x + b.x, a.y + b.y};}
Complex operator - (Complex a, Complex b) {return (Complex) {a.x - b.x, a.y - b.y};}
Complex operator * (Complex a, Complex b) {return (Complex) {a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x};}
void getrev(int l) {
	for (int i = 0; i < (1 << l); i++)
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << l - 1);
}
void FFT(Complex *a, int n, int key) {
	for (int i = 0; i < n; i++)
		if (i < rev[i]) swap(a[i], a[rev[i]]);
	for (int i = 1; i < n; i <<= 1) {
		Complex wn(cos(pi / i), key * sin(pi / i));
		for (int r = i << 1, j = 0; j < n; j += r) {
			Complex w(1, 0);
			for (int k = 0; k < i; k++, w = w * wn) {
				Complex x = a[j + k], y = w * a[i + j + k];
				a[j + k] = x + y, a[i + j + k] = x - y;
			}
		}
	}
	if (key == -1)
		for (int i = 0; i < n; i++) a[i].x /= n;
}
int main() {
	int n; read(n);
	scanf(" %s %s", s1, s2);
	for (int i = 0; i < n; i++)
		a[n - i - 1].x = s1[i] - '0', b[n - i - 1].x = s2[i] - '0';
	int len = 1, l = 0;
	while (len <= 2 * n - 2) len <<= 1, l++;
	getrev(l); FFT(a, len, 1), FFT(b, len, 1);
	for (int i = 0; i <= len; i++) a[i] = a[i] * b[i];
	FFT(a, len, -1);
	for (int i = 0; i <= len; i++) ans[i] = (int)(a[i].x + 0.5);
	for (int i = 0; i <= len; i++)
		ans[i + 1] += ans[i] / 10, ans[i] %= 10;
	while (len > 0 && ans[len] == 0) len--;
	for (int i = len; i >= 0; i--) cout << ans[i]; cout << "\n";
	return 0;
}