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【模板】快速傅立叶变换 (FFT)
2025-08-15
统计加载中...

日后可能发展成 acwing 的数学专题打卡记录,和计算几何一样,所以 slug 先给了 math-acwing,但是目前还没多余的时间学数学,最可能用到什么学什么。

FFT#

AcWing 3122. 多项式乘法#

#include <iostream>
#include <cmath>
using namespace std;
const int N = 300000;
const double PI = acos(-1);
struct Complex {
double x, y;
Complex operator +(const Complex &t) const {
return {x + t.x, y + t.y};
}
Complex operator -(const Complex &t) const {
return {x - t.x, y - t.y};
}
Complex operator *(const Complex &t) const {
return {x * t.x - y * t.y, x * t.y + y * t.x};
}
} a[N], b[N];
int rev[N], tot, bit;
void fft(Complex a[], int inv) {
for (int i = 0; i < tot; ++i) {
if (i < rev[i]) swap(a[i], a[rev[i]]);
}
for (int mid = 1; mid < tot; mid <<= 1) {
Complex w1 = {cos(PI / mid), inv * sin(PI / mid)};
for (int i = 0; i < tot; i += mid * 2) {
Complex wk = {1, 0};
for (int j = 0; j < mid; ++j, wk = wk * w1) {
Complex x = a[i + j], y = wk * a[i + j + mid];
a[i + j] = x + y, a[i + j + mid] = x - y;
}
}
}
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i <= n; ++i) scanf("%lf", &a[i].x);
for (int i = 0; i <= m; ++i) scanf("%lf", &b[i].x);
while ((1 << bit) < n + m + 1) bit++;
tot = 1 << bit;
for (int i = 0; i < tot; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
fft(a, 1), fft(b, 1);
for (int i = 0; i < tot; ++i) a[i] = a[i] * b[i];
fft(a, -1);
for (int i = 0; i < n + m + 1; ++i) {
printf("%d ", int(a[i].x / tot + 0.5));
}
printf("\n");
return 0;
}

AcWing 3123. 高精度乘法II#

复数乘法运算符写挂,回扣计算几何的向量叉积写挂。

#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 300000;
const double PI = acos(-1);
struct Complex {
double x, y;
Complex operator +(const Complex &t) const {
return {x + t.x, y + t.y};
}
Complex operator -(const Complex &t) const {
return {x - t.x, y - t.y};
}
Complex operator *(const Complex &t) const {
return {x * t.x - y * t.y, x * t.y + y * t.x};
}
} a[N], b[N];
char s[N], t[N];
int rev[N], bit, tot;
int res[N];
void fft(Complex a[], int inv) {
for (int i = 0; i < tot; ++i) {
if (i < rev[i]) swap(a[i], a[rev[i]]);
}
for (int mid = 1; mid < tot; mid <<= 1) {
Complex w1 = {cos(PI / mid), inv * sin(PI / mid)};
for (int i = 0; i < tot; i += mid * 2) {
Complex wk = {1, 0};
for (int j = 0; j < mid; ++j, wk = wk * w1) {
Complex x = a[i + j], y = wk * a[i + j + mid];
a[i + j] = x + y, a[i + j + mid] = x - y;
}
}
}
}
int main() {
scanf("%s%s", s, t);
int n = strlen(s) - 1, m = strlen(t) - 1;
for (int i = 0; i <= n; ++i) a[i] = {double(s[n - i] - 48), 0};
for (int i = 0; i <= m; ++i) b[i] = {double(t[m - i] - 48), 0};
while ((1 << bit) < n + m + 1) bit++;
tot = 1 << bit;
for (int i = 0; i < tot; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
fft(a, 1), fft(b, 1);
for (int i = 0; i < tot; ++i) a[i] = a[i] * b[i];
fft(a, -1);
int len = 0;
for (int i = 0; i <= n + m; ++i) {
res[i] = int(a[i].x / tot + 0.5);
}
for (len = 0; len <= n + m || res[len]; ++len) {
res[len + 1] += res[len] / 10, res[len] %= 10;
}
while (len > 1 && res[len - 1] == 0) len--;
for (int i = len - 1; i >= 0; --i) putchar(res[i] + 48);
putchar('\n');
return 0;
}
【模板】快速傅立叶变换 (FFT)
https://starlab.top/posts/math-acwing/
作者
Star
发布于
2025-08-15
许可协议
CC BY-NC-SA 4.0

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