多项式乘积 FFT

在做题打比赛的时候，FFT（快速傅里叶变换）这个名词超级常见，这个算法实际上是在结合复数，多项式的点值表示等知识，加速两个多项式的乘法运算的过程

而两个高精大数的乘法也可以看成两个多项式的乘积，所以也可以用FFT来优化

#include <stdio.h> 
#include <string.h> 
#include <iostream> 
#include <algorithm> 
#include <math.h> 
using namespace std;

const double PI = acos(-1.0);
//复数结构体 
struct Complex {
	double x, y; //实部和虚部 x+yi 
	Complex(double _x = 0.0, double _y = 0.0) {
		x = _x;
		y = _y;
	}
	Complex operator -(const Complex &b) const {
		return Complex(x - b.x, y - b.y);
	}
	Complex operator +(const Complex &b) const {
		return Complex(x + b.x, y + b.y);
	}
	Complex operator *(const Complex &b) const {
		return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
	}
};
/* 
 * 进行FFT和IFFT前的反转变换。 
 * 位置i和 （i二进制反转后位置）互换
 * len必须去2的幂 
 */
void change(Complex y[], int len) {
	int i, j, k;
	for (i = 1, j = len / 2; i < len - 1; i++) {
		if (i < j)
			swap(y[i], y[j]);
		//交换互为小标反转的元素，i<j保证交换一次 
		//i做正常的+1，j左反转类型的+1,始终保持i和j是反转的 
		k = len / 2;
		while (j >= k) {
			j -= k;
			k /= 2;
		}
		if (j < k)
			j += k;
	}
}
/* 
 * 做FFT 
 * len必须为2^k形式， 
 * on==1时是DFT，on==-1时是IDFT 
 */
void fft(Complex y[], int len, int on) {
	change(y, len);
	for (int h = 2; h <= len; h <<= 1) {
		Complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
		for (int j = 0; j < len; j += h) {
			Complex w(1, 0);
			for (int k = j; k < j + h / 2; k++) {
				Complex u = y[k];
				Complex t = w * y[k + h / 2];
				y[k] = u + t;
				y[k + h / 2] = u - t;
				w = w * wn;
			}
		}
	}
	if (on == -1)
		for (int i = 0; i < len; i++)
			y[i].x /= len;
}
const int MAXN = 200010;
Complex x1[MAXN], x2[MAXN];
char str1[MAXN / 2], str2[MAXN / 2];
int sum[MAXN];
int main() {
	while (scanf("%s%s", str1, str2) == 2) {
		int len1 = strlen(str1);
		int len2 = strlen(str2);
		int len = 1;
		while (len < len1 * 2 || len < len2 * 2)
			len <<= 1;
		for (int i = 0; i < len1; i++)
			x1[i] = Complex(str1[len1 - 1 - i] - '0', 0);
		for (int i = len1; i < len; i++)
			x1[i] = Complex(0, 0);
		for (int i = 0; i < len2; i++)
			x2[i] = Complex(str2[len2 - 1 - i] - '0', 0);
		for (int i = len2; i < len; i++)
			x2[i] = Complex(0, 0);
		//求DFT 
		fft(x1, len, 1);
		fft(x2, len, 1);
		for (int i = 0; i < len; i++)
			x1[i] = x1[i] * x2[i];
		fft(x1, len, -1);
		for (int i = 0; i < len; i++)
			sum[i] = (int) (x1[i].x + 0.5);
		for (int i = 0; i < len; i++) {
			sum[i + 1] += sum[i] / 10;
			sum[i] %= 10;
		}
		len = len1 + len2 - 1;
		while (sum[len] <= 0 && len > 0)
			len--;
		for (int i = len; i >= 0; i--)
			printf("%c", sum[i] + '0');
		printf("\n");
	}
	return 0;
}

参考博客：

Blog 1

Blog 2

Blog 3