快速傅里叶转换 模版

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
using namespace std;
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);
    }
};
class myfft
{
public:
    Complex x1[100],x2[100];
    Complex x[100];
    int sum[100];
    void fft(Complex p[],int len,int on)
    {
        Complex pcpy[100];
        for(int width=2;width<=len;width*=2)
        {
            memcpy(pcpy,p,sizeof(pcpy));
            Complex wn(cos(on*2*PI/width),sin(on*2*PI/width));
            int jiange=len/width;
            for(int i=0;i<jiange;i++)
            {
                Complex w(1,0);
                int ap=i;
                int bp=ap+jiange;
                int cp=i;
                int dp=i+width/2*jiange;
                for(int j=0;j<width/2;j++)
                {
                    p[cp]=pcpy[ap]+w*pcpy[bp];
                    p[dp]=pcpy[ap]-w*pcpy[bp];
                    w=w*wn;
                    ap+=2*jiange;
                    bp+=2*jiange;
                    cp+=jiange;
                    dp+=jiange;
                }
            }
        }
    }

    void solve(int a,int b)
    {
        int len1=0;
        int len2=0;

        for(int i=0; i<100; i++)
            x1[i].x=x1[i].y=x2[i].x=x2[i].y=0;
        while(a)
        {
            x1[len1++].x=a%10;
            a/=10;
        }
        while(b)
        {
            x2[len2++].x=b%10;
            b/=10;
        }
        int len=1;
        while(len<len1*2||len<len2*2)len*=2;
        fft(x1,len,1);
        fft(x2,len,1);
        for(int i=0; i<len; i++)
            x[i]=x1[i]*x2[i];
        fft(x,len,-1);
        for(int i=0; i<len; i++)
            x[i].x=x[i].x/len;
        for(int i=0; i<len; i++)
            sum[i]=(int)(x[i].x+0.5);
        for(int i=0; i<len; i++)
        {
            sum[i+1]+=sum[i]/10;
            sum[i]=sum[i]%10;
        }
        int i;
        for(i=len-1; i>=0; i--)
            if(sum[i])
                break;
        for(; i>=0; i--)
            printf("%d",sum[i]);
        printf("\n");
    }
};
myfft m;

int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    m.solve(a,b);

    return 0;
}