FFT（板子）

Posted 2021-09-03 zjj0624

FFT
FFT其实就是求两个多项式的乘积的一个算法，里面具体的证明和思路我还不太懂，只是会用模板。
多项式的存储

struct Complex
{
    double x, y;
    Complex operator+ (const Complex& t) const //复数的运算
    {
        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], bit, tot; //辅助变量

FFT函数

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)
    {
        auto w1 = Complex({cos(PI / mid), inv * sin(PI / mid)});
        for (int i = 0; i < tot; i += mid * 2)
        {
            auto wk = Complex({1, 0});
            for (int j = 0; j < mid; j ++, wk = wk * w1)
            {
                auto x = a[i + j], y = wk * a[i + j + mid];
                a[i + j] = x + y, a[i + j + mid] = x - y;
            }
        }
    }
}

对两个多项式进行乘法运算

int main()
{
    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); //把多项式转换成系数的形式
    for (int i = 0; i < tot; i ++ ) a[i] = a[i] * b[i];//对两个多项式的系数进行乘积
    fft(a, -1); //把系数形式转换成多项式形式，结果就存储在a数组里面
    for (int i = 0; i <= n + m; i ++ )
        printf("%d ", (int)(a[i].x / tot + 0.5));

    return 0;
}