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=.,double _im=.):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=;i<len;++i)
     {
         int t=i,p=;
         for(int j=;j<loglen;++j)
         {
             p<<=;
             p=p|(t&);
             t>>=;
         }
         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=,m=;s<=loglen;++s,m<<=)
     {
         Vir wn=Vir(cos(*PI*on/m),sin(*PI*on/m));
         for(int i=;i<len;i+=m)
         {
             Vir w=Vir(1.0,);
             for(int j=;j<m/;++j)
             {
                 Vir u=a[i+j];
                 Vir v=w*a[i+j+m/];
                 a[i+j]=u+v;
                 a[i+j+m/]=u-v;
                 w=w*wn;
             }
         }
     }
     if(on==-)
     {
         for(int i=;i<len;++i){
           a[i].re/=len;
           a[i].im/=len;
         }
     }
 }
 char a[N*],b[N*];
 Vir pa[N*],pb[N*];
 int ans[N*];
 int main ()
 {
     while(scanf("%s%s",a,b)!=EOF)
     {
         int lena=strlen(a);
         int lenb=strlen(b);
         int n=,loglen=;
         while(n<lena+lenb) n<<=,loglen++;
         for(int i=,j=lena-;i<n;++i,--j)
             pa[i]=Vir(j>=?a[j]-'':.,.);
         for(int i=,j=lenb-;i<n;++i,--j)
             pb[i]=Vir(j>=?b[j]-'':.,.);
          memset(ans,,sizeof(int)*(n+));
         FFT(pa,loglen,n,);
         FFT(pb,loglen,n,);
         for(int i=;i<n;++i)
             pa[i]=pa[i]*pb[i];
         FFT(pa,loglen,n,-);

         for(int i=;i<n;++i)
             ans[i]=pa[i].re+0.5;
         for(int i=;i<n;++i)
         {
              ans[i+]+=ans[i]/;
              ans[i]%=;
         }
         int pos=lena+lenb-;
         for(;pos>&&ans[pos]<=;--pos) ;
         for(;pos>=;--pos)
             printf("%d",ans[pos]);
         puts("");
     }
     return ;
 }