FFT Posted on 2020-02-24 | title: 快速傅里叶变换date: 2020/2/24tag: 算法mathjax: true P1919 A×B Problem快速傅里叶变换FFT 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253#include<bits/stdc++.h>using namespace std;const int maxn=(1<<21)+1;const double PI=acos(-1);struct Complex{ double r,i; Complex(){r=0,i=0;} Complex(double real,double imag):r(real),i(imag){}}F[maxn],G[maxn];Complex operator +(Complex a,Complex b){return Complex(a.r+b.r,a.i+b.i);}Complex operator -(Complex a,Complex b){return Complex(a.r-b.r,a.i-b.i);}Complex operator *(Complex a,Complex b){return Complex(a.r*b.r-a.i*b.i,a.r*b.i+a.i*b.r);}int rev[maxn],len,lim=1,totF=0,totG=0,ans[maxn];char n[maxn],m[maxn];void FFT(Complex *a,double opt){ for(int i=0;i<lim;++i) if(rev[i]>i) swap(a[i],a[rev[i]]); for(int dep=1;dep<=log2(lim);++dep){ int m=(1<<dep); Complex wn=Complex(cos(2.0*PI/m),opt*sin(2.0*PI/m)); for(int k=0;k<lim;k+=m){ Complex w=Complex(1,0); for(int j=0;j<(m>>1);++j){ Complex t=w*a[k+j+(m>>1)]; Complex u=a[k+j]; a[k+j]=u+t; a[k+j+(m>>1)]=u-t; w=w*wn; } } } if(opt==-1) for(int i=0;i<lim;++i) a[i].r/=(double)lim;}int main(){ scanf("%s%s",&n,&m); int nlen=strlen(n),mlen=strlen(m); for(int i=nlen-1;i>=0;--i) F[totF++].r=(n[i]^48); for(int i=mlen-1;i>=0;--i) G[totG++].r=(m[i]^48); mlen--;nlen--; while(lim<=nlen+mlen) lim<<=1,++len; for(int i=0;i<lim;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1)); FFT(F,1);FFT(G,1); for(int i=0;i<=lim;++i) F[i]=F[i]*G[i]; FFT(F,-1); int tot=0; for(int i=0;i<=lim;++i){ ans[i]+=(int)(F[i].r+0.5); if(ans[i]>=10) ans[i+1]+=ans[i]/10,ans[i]%=10,lim+=(i==lim); } while((!ans[lim])&&lim>=1) --lim; ++lim; while(lim--) cout<<ans[lim]; return 0;}