HDU－1402－A*B Problem Plus

ACM模版

题目链接

A*B Problem Plus

题解

高精度乘法，使用FFT可以快速解决。

代码

#include <iostream>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

// FFT

/* * HDU 1402 求高精度乘法 * A * B Problem Plus */

const double PI = acos(-1.0);

// 复数结构体
struct Complex
{
    double x, y;    // 实部和虚部 x + yi
    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前的反转变换
// 位置i和（i二进制反转后的位置）互换
// len必须去2的幂
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]);
        }
        // 交换护卫小标反转的元素，i < j保证交换一次
        // i做正常的+1，j左反转类型的+1，始终保持i和j是反转的
        k = len / 2;
        while (j >= k)
        {
            j -= k;
            k /= 2;
        }
        if (j < k)
        {
            j += k;
        }
    }
    return ;
}

// FFT
// len必须为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(-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;
        }
    }
}

const int MAXN = 200010;
Complex x1[MAXN], x2[MAXN];
char str1[MAXN / 2], str2[MAXN];
int sum[MAXN];

int main(int argc, const char * argv[])
{
    while (cin >> str1 >> str2)
    {
        int len1 = (int)strlen(str1);
        int len2 = (int)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);
        }
        // 求DFT
        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');
        }
        putchar('\n');
    }

    return 0;
}

参考

《FFT》