HDU1402 FFT高精度乘法模板题

Posted 2021-01-17 json-five

#include<bits/stdc++.h>
using namespace std;
//HDU 1402 求高精度乘法
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 二进制反转后位置）互换
* len 必须为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]);
//交换互为小标反转的元素，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;
    }
}
/*
* 做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 = 50005*2;
Complex x1[MAXN],x2[MAXN];
char str1[MAXN/2],str2[MAXN/2];
int sum[MAXN];
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);
//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‘);
        printf("
");
    }
    return 0;
}