UVALIVE6886 Golf Bot (FFT)

题意:打高尔夫 给你n个距离表示你一次可以把球打远的距离

　　 然后对于m个询问 问能否在两杆内把球打进洞

题解:平方一下就好 注意一下x0的系数为1表示打一杆

　　 才发现数组应该开MAXN * 4 之前写的题数据有点不严谨了

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <math.h>
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-5;

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[800005];

int main()
{
    int n, m;
    while(~scanf("%d", &n))
    {
        int zd = 0;
        for(int i = 0; i <= 200002; i++) x1[i] = Complex(0, 0);
        for(int i = 1; i <= n; i++)
        {
            int x; scanf("%d", &x);
            zd = max(zd, x);
            x1[x] = Complex(1, 0);
        }
        x1[0] = Complex(1, 0);

        int len = 1;
        while(len < zd + 1) len <<= 1;
        len <<= 1;
        for(int i = zd + 1; i < len; i++) x1[i] = Complex(0, 0);

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

        scanf("%d", &m);
        int ans = 0;
        for(int i = 1; i <= m; i++)
        {
            int xx; scanf("%d", &xx);
            if(fabs(x1[xx].x) > eps) ans++;
        }
        printf("%d
", ans);
    }
    return 0;
}