BZOJ2194: 快速傅立叶之二 (FFT)

题意:题如其名 求c_k = sigma (a_i * b_i - k)

题解:一般FFT 要满足c_k = (sigma a_i * b_k - i)的形式

   于是这个题的技巧就是把b数组翻转一下 b_i = b_n - 1 - i (下标从0)

   题目就变形为c_k = sigma (a_i * b_n - 1 - i + k) = c_n - 1 + k 于是就上板子了

   kuangbin的板子还是很厉害的

 

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;
const double PI = acos(-1.0);

struct Complex
{
    double x, y;
    Complex(double _x = 0.0, double _y = 0.0)
    {
        x = _x;
        y = _y;
    }
    Complex operator + (const Complex &b) const
    {
        return Complex(x + b.x, y + b.y);
    }
    Complex operator - (const Complex &b) const
    {
        return Complex(x - b.x, y - b.y);
    }
    Complex operator * (const Complex &b) const
    {
        return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
    }
};

void change(Complex y[], int len)
{
    int i, j, k;
    for(i = 1, j = len / 2; i < len - 1; i++)
    {
        if(i < j) swap(y[i], y[j]);
        k = len / 2;
        while(j >= k)
        {
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}

void fft(Complex y[], int len, int on)
{
    change(y, len);
    for(int h = 2; h <= len; h <<= 1)
    {
        Complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
        for(int j = 0; j < len; j += h)
        {
            Complex w(1, 0);
            for(int k = j; k < j + h / 2; k++)
            {
                Complex u = y[k];
                Complex t = w * y[k + h / 2];
                y[k] = u + t;
                y[k + h / 2] = u - t;
                w = w * wn;
            }
        }
    }

    if(on == -1)
        for(int i = 0; i < len; i++)
            y[i].x /= len;
}

Complex x1[400005], x2[400005];

int main()
{
    int n; scanf("%d", &n);
    int nn = n;
    for(int i = 0; i < n; i++)
    {
        double u, v;
        scanf("%lf%lf", &u, &v);
        x1[i] = Complex(u, 0);
        x2[n - i - 1] = Complex(v, 0);
    }
    n = 2 * n - 1;
    int len = 1;
    while(len < n) len <<= 1;
    for(int i = n; i < len; i++) x1[i] = x2[i] = Complex(0.0, 0.0);

    fft(x1, len, 1);
    fft(x2, len, 1);
    for(int i = 0; i < len; i++) x1[i] = x1[i] * x2[i];
    fft(x1, len, -1);

    for(int i = nn - 1; i < n; i++) printf("%d\n", (int)(x1[i].x + 0.5));
    return 0;
}
View Code

 

   

posted @ 2018-07-16 15:59  lwqq3  阅读(146)  评论(0编辑  收藏  举报