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  • FTT板子

    FTT板子

    用的是A+B例题

    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    
    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);
        }
    };
    /*
     * 进行 FFT 和 IFFT 前的反置变换
     * 位置 i 和 i 的二进制反转后的位置互换
     *len 必须为 2 的幂
     */
    void change(Complex y[], int len) {
        int i, j, k;
        for (int i = 1, j = len / 2; i < len - 1; i++) {
            if (i < j) std::swap(y[i], y[j]);
            // 交换互为小标反转的元素,i<j 保证交换一次
            // i 做正常的 + 1,j 做反转类型的 + 1,始终保持 i 和 j 是反转的
            k = len / 2;
            while (j >= k) {
                j = j - k;
                k = k / 2;
            }
            if (j < k) j += k;
        }
    }
    /*
     * 做 FFT
     * len 必须是 2^k 形式
     * on == 1 时是 DFT,on == -1 时是 IDFT
     */
    void fft(Complex y[], int len, int on) {
        change(y, len);
        for (int h = 2; h <= len; h <<= 1) {
            Complex wn(cos(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;
                    //模长相乘,辅角相加,模长为1相乘不变,相当于辅角增加了2 * PI / h
                    w = w * wn;
                }
            }
        }
        if (on == -1)
            for (int i = 0; i < len; i++)
                y[i].x /= len;
    }
    
    const int N = 200020;
    Complex x1[N], x2[N];
    char str1[N / 2], str2[N / 2];
    int sum[N];
    
    int main() {
        while (scanf("%s%s", str1, str2) == 2) {
            int len1 = strlen(str1);
            int len2 = strlen(str2);
            int len = 1;
            while (len < len1 * 2 || len < len2 * 2) len <<= 1;
            for (int i = 0; i < len1; i++) x1[i] = Complex(str1[len1 - 1 - i] - '0', 0);
            for (int i = len1; i < len; i++) x1[i] = Complex(0, 0);
            for (int i = 0; i < len2; i++) x2[i] = Complex(str2[len2 - 1 - i] - '0', 0);
            for (int i = len2; i < len; i++) x2[i] = Complex(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 = 0; i < len; i++) sum[i] = int(x1[i].x + 0.5);
            for (int i = 0; i < len; i++) {
                sum[i + 1] += sum[i] / 10;
                sum[i] %= 10;
            }
            len = len1 + len2 - 1;
            while (sum[len] == 0 && len > 0) len--;
            for (int i = len; i >= 0; i--) printf("%c", sum[i] + '0');
            printf("
    ");
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/2aptx4869/p/15032071.html
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