hdu----(1402)A * B Problem Plus（FFT模板）

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12665    Accepted Submission(s): 2248

Problem Description

Calculate A * B.

 

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

 

Output

For each case, output A * B in one line.

 

Sample Input

1 2 1000 2

 

Sample Output

2 2000

 

Author

DOOM III

快速傅里叶转换算法.....

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
#define N 50500*2
const 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.);
         memset(ans,0,sizeof(int)*(n+1));
        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;
}