fft模板+讲解

讲解：https://blog.csdn.net/qq_37136305/article/details/81184873 有点小问题，一个是代码不太对，另一个是倍增的图不太对
讲解+代码：https://www.cnblogs.com/zwfymqz/p/8244902.html

#include "bits/stdc++.h"

using namespace std;
const int MAXN = 1e7 + 10;
const double Pi = acos(-1.0);

struct Complex
{
    double x, y;

    Complex(double xx = 0, double yy = 0)
    { x = xx, y = yy; }
} a[MAXN], b[MAXN];

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); }

int l, r[MAXN], N, M;
int limit = 1;

void fft(Complex *A, int type)
{
    for (int i = 0; i < limit; i++)
        if (i < r[i])
        { swap(A[i], A[r[i]]); } //求出要迭代的序列
    for (int mid = 1; mid < limit; mid <<= 1)
    { 
        Complex Wn(cos(Pi / mid), type * sin(Pi / mid)); //单位根
        for (int R = mid << 1, j = 0; j < limit; j += R)
        { //R是区间的长度，j表示前已经到哪个位置了
            Complex w(1, 0); //幂
            for (int k = 0; k < mid; k++, w = w * Wn)//不断向上合并
            { 
                Complex x = A[j + k], y = w * A[j + mid + k];
                A[j + k] = x + y;
                A[j + mid + k] = x - y;
            }
        }
    }
}

int main()
{
    cin >> N >> M;
    for (int i = 0; i <= N; i++) cin >> a[i].x;
    for (int i = 0; i <= M; i++) cin >> b[i].x;
    while (limit <= N + M)//补足位数 必须是2^x次方长度
    { limit <<= 1, l++; }
    for (int i = 0; i < limit; i++)//翻转二进制下标 为倍增迭代做准备
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    //两次fft 系数表示-》点值表示
    fft(a, 1);
    fft(b, 1);
    for (int i = 0; i <= limit; i++) a[i] = a[i] * b[i];
    fft(a, -1);//逆变换 点值表示-》系数表示
    for (int i = 0; i <= N + M; i++)
        printf("%d ", (int) (a[i].x / limit + 0.5));
    return 0;
}

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