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【模板】快速傅立叶变换 (FFT)
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/ 部分信息可能已经过时







