hdu1402 大数相乘 快速傅里叶变换FFT

FFT入门题，FFT模板

#include <iostream>#include <stdio.h>#include <cmath>#include <algorithm>#include <cstring>#include <vector>using namespace std;#define N 50500*2const double PI=acos(-1.0);struct Vir{    double re,im;    Vir(double _re=0.,double _im=0.):re(_re),im(_im){}    Vir operator*(Vir r) { return Vir(re*r.re-im*r.im,re*r.im+im*r.re);}    Vir operator+(Vir r) { return Vir(re+r.re,im+r.im);}    Vir operator-(Vir r) { return Vir(re-r.re,im-r.im);}};void bit_rev(Vir *a,int loglen,int len){    for(int i=0;i<len;++i)    {        int t=i,p=0;        for(int j=0;j<loglen;++j)        {            p<<=1;            p=p|(t&1);            t>>=1;        }        if(p<i)        {            Vir temp=a[p];            a[p]=a[i];            a[i]=temp;        }    }}void FFT(Vir *a,int loglen,int len,int on){    bit_rev(a,loglen,len);    for(int s=1,m=2;s<=loglen;++s,m<<=1)    {        Vir wn=Vir(cos(2*PI*on/m),sin(2*PI*on/m));        for(int i=0;i<len;i+=m)        {            Vir w=Vir(1.0,0);            for(int j=0;j<m/2;++j)            {                Vir u=a[i+j];                Vir v=w*a[i+j+m/2];                a[i+j]=u+v;                a[i+j+m/2]=u-v;                w=w*wn;            }        }    }    if(on==-1)    {        for(int i=0;i<len;++i) a[i].re/=len,a[i].im/=len;    }}char a[N*2],b[N*2];Vir pa[N*2],pb[N*2];int ans[N*2];int main (){    while(scanf("%s%s",a,b)!=EOF)    {        int lena=strlen(a);        int lenb=strlen(b);        int n=1,loglen=0;        while(n<lena+lenb) n<<=1,loglen++;        for(int i=0,j=lena-1;i<n;++i,--j)            pa[i]=Vir(j>=0?a[j]-'0':0.,0.);        for(int i=0,j=lenb-1;i<n;++i,--j)            pb[i]=Vir(j>=0?b[j]-'0':0.,0.);        for(int i=0;i<=n;++i) ans[i]=0;        FFT(pa,loglen,n,1);        FFT(pb,loglen,n,1);        for(int i=0;i<n;++i)            pa[i]=pa[i]*pb[i];        FFT(pa,loglen,n,-1);        for(int i=0;i<n;++i) ans[i]=pa[i].re+0.5;        for(int i=0;i<n;++i) ans[i+1]+=ans[i]/10,ans[i]%=10;        int pos=lena+lenb-1;        for(;pos>0&&ans[pos]<=0;--pos) ;        for(;pos>=0;--pos) printf("%d",ans[pos]);        puts("");    }    return 0;}