BZOJ2179:FFT快速傅立叶(FFT)

Description

给出两个n位10进制整数x和y，你需要计算x*y。

Input

第一行一个正整数n。 第二行描述一个位数为n的正整数x。 第三行描述一个位数为n的正整数y。

Output

输出一行，即x*y的结果。

Sample Input

1
3
4

Sample Output

12
数据范围：
n<=60000

Solution

FFT模板题，做的时候注意处理一下进位和前导零就好

Code

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #define N (240000+100)
 using namespace std;
 double pi=acos(-1.0);
 struct complex
 {
     double x,y;
     complex (double xx=,double yy=)
     {
         x=xx; y=yy;
     }
 }a[N],b[N];
 int n=-,m=-,t,fn,l,r[N];
 char ch;

 complex operator + (complex a,complex b){return complex(a.x+b.x,a.y+b.y);}
 complex operator - (complex a,complex b){return complex(a.x-b.x,a.y-b.y);}
 complex operator * (complex a,complex b){return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
 complex operator / (complex a,double b){return complex(a.x/b,a.y/b);}

 void FFT(int n,complex *a,int opt)
 {
     for (int i=; i<n; ++i)
         if (i<r[i])
             swap(a[i],a[r[i]]);
     for (int k=; k<n; k<<=)
     {
         complex wn=complex(cos(pi/k),opt*sin(pi/k));
         for (int i=; i<n; i+=(k<<))
         {
             complex w=complex(,);
             for (int j=; j<k; ++j,w=w*wn)
             {
                 complex x=a[i+j], y=w*a[i+j+k];
                 a[i+j]=x+y; a[i+j+k]=x-y;
             }
         }
     }
     if (opt==-) for (int i=; i<n; ++i) a[i]=a[i]/n;
 }

 int main()
 {
     scanf("%d",&t); t--;
     for (int i=; i<=t; ++i)
     {
         ch=getchar();
         while (ch<'' || ch>'') ch=getchar();
         if (n==- && ch=='') continue;
         a[++n].x=ch-'';
     }
     for (int i=; i<=t; ++i)
     {
         ch=getchar();
         while (ch<'' || ch>'') ch=getchar();
         if (m==- && ch=='') continue;
         b[++m].x=ch-'';
     }
     if (m==-) m=;
     if (n==-) n=;
     fn=;
     while (fn<=n+m) fn<<=, l++;
     for (int i=;i<fn;++i)
         r[i]=(r[i>>]>>) | ((i&)<<(l-));

     FFT(fn,a,); FFT(fn,b,);
     for (int i=;i<=fn;++i)
         a[i]=a[i]*b[i];
     FFT(fn,a,-);
     for (int i=n+m;i>=;--i)
     {
         a[i-].x+=(int)(a[i].x+0.5)/;
         a[i].x=(int)(a[i].x+0.5)%;
     }
     int p=;
     while ((int)(a[p].x+0.5)== && p<n+m) p++;
     for (int i=p;i<=n+m;++i)
         printf("%d",(int)(a[i].x+0.5));
 }