bzoj2179: FFT快速傅立叶

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 using namespace std;
 const double PI=acos(-);
 const int maxn=;
 int n,len,m,rev[maxn],ans[maxn];
 struct node{
     double real,imag;
     node operator +(const node &x){return (node){real+x.real,imag+x.imag};}
     node operator -(const node &x){return (node){real-x.real,imag-x.imag};}
     node operator *(const node &x){return (node){real*x.real-imag*x.imag,real*x.imag+imag*x.real};}
 }a[maxn],b[maxn],c[maxn],w,wn,t1,t2;
 void read(int &x){
     x=; int f=; char ch;
     for (ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') f=-;
     for (;isdigit(ch);ch=getchar()) x=x*+ch-''; x*=f;
 }
 void Read(node *a){
     char ch;
     for (int i=m-;i>=;i--){
         for (ch=getchar();!isdigit(ch);ch=getchar());
         a[i].real=(double)(ch-'');
     }
 }
 int Rev(int x){
     int temp=;
     for (int i=;i<len;i++) temp<<=,temp+=(x&),x>>=;
     return temp;
 }
 void FFT(node *a,int op){
     for (int i=;i<n;i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
     for (int s=;s<=n;s<<=){
         wn=(node){cos(op**PI/s),sin(op**PI/s)};
         for (int i=;i<n;i+=s){
             w=(node){,};
             for (int j=i;j<i+s/;j++,w=w*wn){
                 t1=a[j],t2=w*a[j+s/];
                 a[j]=t1+t2,a[j+s/]=t1-t2;
             }
         }
     }
 }
 int main(){
     read(m); n=,len=;
     while (n<(m<<)) n<<=,len++;
     Read(a),Read(b);
     for (int i=;i<n;i++) rev[i]=Rev(i);
     FFT(a,),FFT(b,);
     for (int i=;i<n;i++) c[i]=a[i]*b[i];
     FFT(c,-);
     for (int i=;i<n;i++) ans[i]=(int)round(c[i].real/n);
     for (int i=;i<n;i++) ans[i+]+=ans[i]/,ans[i]=ans[i]%;
     int j; for (j=n-;j>=;j--) if (ans[j]) break;
     if (j==-) puts("");
     else{for (;j>=;j--) printf("%d",ans[j]);puts("");}
     return ;
 }

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2179

题目大意：给定两个大数a，b，求a*b，位数n<=60000;

做法：FFT入门题，FFT的做法可以自己去看算法导论，这题是裸的卷积，直接上DFT或者NTT，我贴的是DFT的。