FFT(快速傅立叶变换)：HDU 1402 A * B Problem Plus

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
　　
　　唉，模板题，膜的邝斌的模板。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 using namespace std;

 const double PI = acos(-1.0);

 struct complex{
     double r,i;
     complex(double r_=0.0,double i_=0.0)
     {
         r=r_;i=i_;
     }
     complex operator +(const complex &b)
     {
         return complex(r+b.r,i+b.i);
     }
     complex operator -(const complex &b)
     {
         return complex(r-b.r,i-b.i);
     }
     complex operator *(const complex &b)
     {
         return complex(r*b.r-i*b.i,i*b.r+r*b.i);
     }
 };

 void Rader(complex *a,int len)
 {
     int k;
     for(int i=,j=len/;i<len-;i++)
     {
         if(i<j)swap(a[i],a[j]);
         k=len/;
         while(j>=k)
         {
             j-=k;
             k>>=;
         }
         j+=k;
     }
 }

 void FFT(complex *a,int len,int on)
 {
     Rader(a,len);
     for(int h=;h<=len;h<<=)
     {
         complex wn(cos(-on**PI/h),sin(-on**PI/h));
         for(int j=;j<len;j+=h)
         {
             complex w(,);
             for(int k=j;k<j+h/;k++)
             {
                 complex u=a[k];
                 complex v=a[k+h/]*w;
                 a[k]=u+v;
                 a[k+h/]=u-v;
                 w=w*wn;
             }
         }
     }
     if(on==-)
         for(int i=;i<len;i++)
             a[i].r/=len;
 }

 const int maxn = ;
 complex Array1[maxn],Array2[maxn];
 char str1[maxn],str2[maxn];
 int sum[maxn],len,len1,len2;

 int main()
 {
     while(~scanf("%s%s",str1,str2))
     {
         len1=strlen(str1);
         len2=strlen(str2);
         len=;
         while(len<len1*||len<len2*)len<<=;
         for(int i=;i<len1;i++)
             Array1[i]=complex(str1[len1-i-]-'',);
         for(int i=;i<len2;i++)
             Array2[i]=complex(str2[len2-i-]-'',);

         for(int i=len1;i<len;i++)
             Array1[i]=complex(,);
         for(int i=len2;i<len;i++)
             Array2[i]=complex(,);

         FFT(Array1,len,);
         FFT(Array2,len,);
         for(int i=;i<len;i++)
             Array1[i]=Array1[i]*Array2[i];
         FFT(Array1,len,-);
         memset(sum,,sizeof(sum));
         for(int i=;i<len;i++){
             sum[i]+=(int)(Array1[i].r+0.5);
             sum[i+]+=sum[i]/;
             sum[i]%=;
         }
         int p=len;
         while(!sum[p]&&p)p--;
         for(;p!=-;p--)
             printf("%d",sum[p]);
         printf("\n");
     }
     return ;
 }

　　两个月后重打了一遍。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 using namespace std;
 const int maxn=;
 const double PI=acos(-1.0);
 struct complex{
     double r,i;
     complex(double r_=0.0,double i_=0.0){
         r=r_;i=i_;
     }
     complex operator +(complex a){
         return complex(r+a.r,i+a.i);
     }
     complex operator -(complex a){
         return complex(r-a.r,i-a.i);
     }
     complex operator *(complex a){
         return complex(r*a.r-i*a.i,r*a.i+a.r*i);
     }
 }A[maxn],B[maxn];

 void Rader(complex *a,int len){
     int k;
     for(int i=,j=len>>;i<len-;i++){
         if(i<j)swap(a[i],a[j]);
         k=len>>;
         while(j>=k){
             j-=k;
             k>>=;
         }
         j+=k;
     }
 }

 void FFT(complex *a,int len,int on){
     Rader(a,len);
     for(int h=;h<=len;h<<=){
         complex wn(cos(-on*PI*2.0/h),sin(-on*PI*2.0/h));
         for(int j=;j<len;j+=h){
             complex w(,);
             for(int k=j;k<j+(h>>);k++){
                 complex x=a[k];
                 complex y=a[k+(h>>)]*w;
                 a[k]=x+y;
                 a[k+(h>>)]=x-y;
                 w=w*wn;
             }
         }
     }
     if(on==-)
         for(int i=;i<len;i++)
             a[i].r/=len;
     return;
 }

 char s[maxn],t[maxn];
 int ans[maxn];
 int main(){
     while(~scanf("%s%s",s,t)){
     int lens=strlen(s);
     int lent=strlen(t),len=;
     while(len<=lens+lent)len<<=;
     memset(A,,sizeof(A));
     memset(B,,sizeof(B));
     for(int i=;i<lens;i++)
         A[i].r=1.0*(s[lens-i-]-'');
     for(int i=;i<lent;i++)
         B[i].r=1.0*(t[lent-i-]-'');
     FFT(A,len,);FFT(B,len,);
     for(int i=;i<len;i++)
         A[i]=A[i]*B[i];
     FFT(A,len,-);

     memset(ans,,sizeof(ans));
     int in=;
     for(int i=;i<len;i++){
         ans[i]=in+floor(A[i].r+0.5);
         in=ans[i]/;
         ans[i]%=;
     }
     int i;
     for(i=len;i&&!ans[i];i--);
     while(~i)printf("%d",ans[i]),i--;
     printf("\n");
     }
     return ;
 }