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  • P3803 [模板] 多项式乘法 (FFT)

    Rt

    注意len要为2的幂

    #include <bits/stdc++.h>
    using namespace std;
    const double PI = acos(-1.0);
    
    inline int read()
    {
        char c=getchar();int x=0,f=1;
        while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    
    int n, m;
    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[4000005], x2[4000005];
    
    int main()
    {
        scanf("%d%d", &n, &m);
        for(int i = 0; i <= n; i++) {
            int u; u = read();
            x1[i] = Complex(1.0 * u, 0);
        }
        for(int i = 0; i <= m; i++) {
            int u; u = read();
            x2[i] = Complex(1.0 * u, 0);
        }
    
        int len = 1;
        while(len <= n + m) len <<= 1;
    
        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 = 0; i <= n + m; i++) printf("%d ", (int)(x1[i].x + 0.5));
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/lwqq3/p/11335961.html
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