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1周前 (07-19) 7次浏览

# FTT板子

``````#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>

const double PI = acos(-1.0);
struct Complex {
double x, y;
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 前的反置变换
* 位置 i 和 i 的二进制反转后的位置互换
*len 必须为 2 的幂
*/
void change(Complex y[], int len) {
int i, j, k;
for (int i = 1, j = len / 2; i < len - 1; i++) {
if (i < j) std::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 做反转类型的 + 1，始终保持 i 和 j 是反转的
k = len / 2;
while (j >= k) {
j = j - k;
k = k / 2;
}
if (j < k) j += k;
}
}
/*
* 做 FFT
* len 必须是 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(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;
//模长相乘，辅角相加，模长为1相乘不变，相当于辅角增加了2 * PI / h
w = w * wn;
}
}
}
if (on == -1)
for (int i = 0; i < len; i++)
y[i].x /= len;
}

const int N = 200020;
Complex x1[N], x2[N];
char str1[N / 2], str2[N / 2];
int sum[N];

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);
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;
}
``````