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FFT模板

时间:2018-08-20 13:43:42      阅读:201      评论:0      收藏:0      [点我收藏+]

标签:oid   dft   one   bsp   多项式   bool   display   log   alt   

FFT 实际是 DFT 的一种快速实现方法

可以将多项式的乘法从 O(n^2) 优化到 O(nlogn)

暂时没有看到很好的科普文章、原理自行百度吧

 

技术分享图片
#define L(x) (1 << (x))
const double PI = acos(-1.0);
const int maxn = (1<<17) + (int)1e3;
double ax[maxn], ay[maxn], bx[maxn], by[maxn];
int revv(int x, int bits)
{
    int ret = 0;
    for (int i = 0; i < bits; i++){
        ret <<= 1;
        ret |= x & 1;
        x >>= 1;
    }
    return ret;
}

void fft(double * a, double * b, int n, bool rev)
{
    int bits = 0;
    while (1 << bits < n) ++bits;
    for (int i = 0; i < n; i++){
        int j = revv(i, bits);
        if (i < j) swap(a[i], a[j]), swap(b[i], b[j]);
    }

    for (int len = 2; len <= n; len <<= 1){
        int half = len >> 1;
        double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);
        if (rev) wmy = -wmy;
        for (int i = 0; i < n; i += len){
            double wx = 1, wy = 0;
            for (int j = 0; j < half; j++){
                double cx = a[i + j], cy = b[i + j];
                double dx = a[i + j + half], dy = b[i + j + half];
                double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;
                a[i + j] = cx + ex, b[i + j] = cy + ey;
                a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;
                double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;
                wx = wnx, wy = wny;
            }
        }
    }
    if (rev){
        for (int i = 0; i < n; i++)
            a[i] /= n, b[i] /= n;
    }
}

int Convolution(int a[],int na,int b[],int nb,int ans[]) //两个数组求卷积,有时ans数组要开成long long
{
    int len = max(na, nb), ln;
    for(ln=0; L(ln)<len; ++ln);
    len=L(++ln);
    for (int i = 0; i < len ; ++i){
        if (i >= na) ax[i] = 0, ay[i] =0;
        else ax[i] = a[i], ay[i] = 0;
    }
    fft(ax, ay, len, 0);
    for (int i = 0; i < len; ++i){
        if (i >= nb) bx[i] = 0, by[i] = 0;
        else bx[i] = b[i], by[i] = 0;
    }
    fft(bx, by, len, 0);
    for (int i = 0; i < len; ++i){
        double cx = ax[i] * bx[i] - ay[i] * by[i];
        double cy = ax[i] * by[i] + ay[i] * bx[i];
        ax[i] = cx, ay[i] = cy;
    }
    fft(ax, ay, len, 1);
    for (int i = 0; i < len; ++i)
        ans[i] = (int)(ax[i] + 0.5);
    return len;
}


int Convolution_self(long long a[], int na, int ans[]) //自己跟自己求卷积,有时候ans数组要开成long long
{
    int len = na, ln;
    for(ln = 0; L(ln) < na; ++ln);
    len=L(++ln);
    for(int i = 0; i < len; ++i){
        if (i >= na) ax[i] = 0, ay[i] = 0;
        else ax[i] = a[i], ay[i] = 0;
    }
    fft(ax, ay, len, 0);
    for(int i=0; i<len; ++i){
        double cx = ax[i] * ax[i] - ay[i] * ay[i];
        double cy = 2 * ax[i] * ay[i];
        ax[i] = cx, ay[i] = cy;
    }
    fft(ax, ay, len, 1);

    for(int i=0; i<len; ++i)
        ans[i] = ax[i] + 0.5;
    return len;
}
View Code

 

FFT模板

标签:oid   dft   one   bsp   多项式   bool   display   log   alt   

原文地址:https://www.cnblogs.com/Rubbishes/p/9505036.html

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