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  • BZOJ 2179: FFT快速傅立叶

    二次联通门 : BZOJ 2179: FFT快速傅立叶

    权限题放题面

    #include <cstdio>
    #include <iostream>
    #include <cmath>
    
    #define Max 1000005
    #define rg register
    typedef double flo;
    struct vec 
    {
        flo x, y; vec (flo a = 0, flo b = 0) : x (a), y (b) { }
    
        vec operator + (const vec &rhs)
        { return vec (x + rhs.x, y + rhs.y); }
    
        vec operator - (const vec &rhs)
        { return vec (x - rhs.x, y - rhs.y); }
    
        vec &operator /= (const flo &k)
        { return x /= k, y /= k, *this; }
    
        vec operator * (const vec &rhs)
        { return vec (x * rhs.x - y * rhs.y, x * rhs.y + y * rhs.x); }
    
    } a[Max], b[Max];
    int rd[Max];
    char s[Max];
    const flo PI = acos (-1.0);
    void DFT (vec *a, int N, int f = 1)
    {
        rg int i, j, k; vec wn, w, x, y;
        for (i = 0; i < N; ++ i) if (rd[i] > i) std :: swap (a[i], a[rd[i]]);
    
        for (k = 1; k < N; k <<= 1)
        {
            wn = vec (cos (PI / k), f * sin (PI / k));
            for (j = 0; j < N; j += k << 1)
                for (i = 0, w = vec (1, 0); i < k; ++ i, w = w * wn)
                    x = a[j + i], y = w * a[i + j + k], a[i + j] = x + y, a[i + j + k] = x - y;
        }    
        if (f == -1) for (i = 0; i < N; ++ i) a[i] /= N;
    }
    void FFT (vec *a, vec *b, int N)
    {
        int p = N << 1, L = 0; rg int i;
        for (N = 1; N <= p; N <<= 1) ++ L;
        for (i = 0; i < N; ++ i)
            rd[i] = (rd[i >> 1] >> 1) | ((i & 1) << (L - 1));
        DFT (a, N), DFT (b, N);
        for (i = 0; i <= N; ++ i) a[i] = a[i] * b[i];
        DFT (a, N, -1);
    }
    int c[Max];
    int main (int argc, char *argv[])
    {
        int N; scanf ("%d", &N); -- N; rg int i;
        scanf ("%s", s);
        for (i = 0; i <= N; ++ i) a[i].x = s[N - i] - '0';
        scanf ("%s", s);
        for (i = 0; i <= N; ++ i) b[i].x = s[N - i] - '0';
        FFT (a, b, N);
        int p = N * 2;
        for (i = 0; i <= p; ++ i) c[i] = (int) (a[i].x + 0.5);
        for (i = 0; i <= p; ++ i)
            if (c[i] >= 10)
            {
                c[i + 1] += c[i] / 10, c[i] %= 10;
                if (i == p) ++ p;
            }
        for (; c[p] == 0; -- p);
        for (i = p; i >= 0; -- i) printf ("%d", c[i]);
        putchar ('
    ');
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ZlycerQan/p/8044845.html
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