FFT

 

看这两篇博客就差不多了

第一篇：链接

第二篇：链接

 

关于虚数：链接

 

以上来自：Candy?

 

这篇里面也有一些链接可以看看： 点我

 

拖了好久...

这里就放个抄来的模板吧=_=

const int maxn = 200010;
const double pi = acos(-1.0);
complex<double> omega[maxn], inverse[maxn];

void init(int n){
    for(int i = 0; i < n; i++){
        omega[i] = complex<double>(cos(2 * pi / n * i), sin(2 * pi / n * i));
        inverse[i] = conj(omega[i]);
    }
}
void transform(complex<double> *a, int n, complex<double> *omega){
    int k = 0;
    while((1 << k) < n) k++;
    for(int i = 0; i < n; i++){
        int t = 0;
        for(int j = 0; j < k; j++) if(i & (1 << j)) t |= (1 << (k - j - 1));
        if(i < t) swap(a[i], a[t]);
    }
    for(int l = 2;  l <= n; l *= 2){
        int m = l / 2;
        for(complex<double> *p = a; p != a + n; p += l){
            for(int i = 0; i < m; i++){
                complex<double> t = omega[n / l * i] * p[m + i];
                p[m + i] = p[i] - t;
                p[i] += t;
            }
        }
    }
}
void dft(complex<double> *a, int n){
    transform(a, n, omega);
}
void idft(complex<double> *a, int n){
    transform(a, n, inverse);
    for(int i = 0; i < n; i++) a[i] /= n;
}

void mul(int *a, int n1, int *b, int n2, int *res){
    int n = 1;
    while(n < n1 + n2) n <<= 1;
    complex<double> c1[maxn], c2[maxn];
    for(int i = 0; i < n1; i++) c1[i].real(a[i]);
    for(int i = 0; i < n2; i++) c2[i].real(b[i]);
    init(n);
    dft(c1, n); dft(c2, n);
    for(int i = 0; i < n; i++) c1[i] *= c2[i];
    idft(c1, n);
    for(int i = 0; i < n1 + n2 - 1; i++) res[i] = (int)(c1[i].real() + 0.5);
}