51Nod-1028-大数乘法 V2

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ACM模版
描述

题解

FFT模版题，不禁赞叹FFT的神奇，但是着实不好理解，算法导论上讲得还好，可以看看。
感觉可以用截位相乘的方法做，但是不知道会不会超时。
代码

#include <iostream>#include <cmath>#include <cstring>using namespace std;const double PI = acos(-1.0);const int MAXN = 4e5 + 10;//  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IDFTvoid 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;        }    }}//  求卷积//  用于大数乘法void conv(Complex a[], Complex b[], int ans[], int len){    fft(a, len, 1);    fft(b, len, 1);    for (int i = 0; i < len; i++)    {        a[i] = a[i] * b[i];    }    fft(a, len, -1);    //  精度复原    for (int i = 0; i < len; i++)    {        ans[i] = a[i].x + 0.5;    }}//  进制恢复//  用语大数乘法void turn(int ans[], int len, int unit){    for (int i = 0; i < len; i++)    {        ans[i + 1] += ans[i] / unit;        ans[i] %= unit;    }}char str_1[MAXN], str_2[MAXN];Complex za[MAXN], zb[MAXN];int ans[MAXN];int len;void init(char str_1[], char str_2[]){    int len_1 = (int)strlen(str_1);    int len_2 = (int)strlen(str_2);    len = 1;    while (len < 2 * len_1 || len < 2 * len_2)    {        len <<= 1;    }    int i = 0;    for (; i < len_1; i++)    {        za[i].x = str_1[len_1 - i - 1] - '0';        za[i].y = 0.0;    }    while (i < len)    {        za[i].x = za[i].y = 0.0;        i++;    }    for (i = 0; i < len_2; i++)    {        zb[i].x = str_2[len_2 - i - 1] - '0';        zb[i].y = 0.0;    }    while (i < len)    {        zb[i].x = zb[i].y = 0.0;        i++;    }    return ;}void solve(){    conv(za, zb, ans, len);    turn(ans, len, 10);    while (ans[len - 1] == 0)    {        len--;    }    for (int i = len - 1; i >= 0; i--)    {        printf("%d", ans[i]);    }    printf("\n");    return ;}int main(){    while (~scanf("%s%s", str_1, str_2))    {        init(str_1, str_2);        solve();    }    return 0;}
参考

《FFT》
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