1. 程式人生 > >洛谷 3803 【模板】多項式乘法(FFT)

洛谷 3803 【模板】多項式乘法(FFT)

題目:https://www.luogu.org/problemnew/show/P3803

https://www.cnblogs.com/zwfymqz/p/8244902.html

http://www.cnblogs.com/RabbitHu/p/FFT.html

http://picks.logdown.com/posts/177631-fast-fourier-transform

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define
db double using namespace std; const int N=1e6+5;const db pi=acos(-1.0); int n,m,len,r[N<<2];//<<2! for (n+m)<<1 struct cpl{ db x,y; }I,a[N<<2],b[N<<2]; cpl operator+ (cpl a,cpl b){return (cpl){a.x+b.x,a.y+b.y};} cpl operator- (cpl a,cpl b){return (cpl){a.x-b.x,a.y-b.y};} cpl
operator* (cpl a,cpl b){return (cpl){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};} int rdn() { int ret=0;bool fx=1;char ch=getchar(); while(ch>'9'||ch<'0'){if(ch=='-')fx=0;ch=getchar();} while(ch>='0'&&ch<='9') ret=(ret<<3)+(ret<<1)+ch-'0',ch=getchar(); return fx?ret:-ret; }
void fft(cpl *a,bool fx) { for(int i=0;i<len;i++) if(i<r[i])swap(a[i],a[r[i]]); for(int R=2;R<=len;R<<=1)//<= { int m=R>>1; cpl Wn=(cpl){ cos(pi/m),(fx?-1:1)*sin(pi/m) }; for(int i=0;i<len;i+=R) { cpl w=I; for(int j=0;j<m;j++,w=w*Wn) { cpl tmp=w*a[i+m+j]; a[i+m+j]=a[i+j]-tmp; a[i+j]=a[i+j]+tmp; } } } } int main() { n=rdn(); m=rdn(); I.x=1; I.y=0; for(int i=0;i<=n;i++)a[i].x=rdn(); for(int i=0;i<=m;i++)b[i].x=rdn(); len=1; while(len<=n+m)len<<=1;//<= for(int i=0;i<len;i++) r[i]=(r[i>>1]>>1)+((i&1)?len>>1:0); fft(a,0); fft(b,0); for(int i=0;i<len;i++) a[i]=a[i]*b[i]; fft(a,1); for(int i=0;i<=n+m;i++) printf("%d ",int(a[i].x/len+0.5));puts(""); return 0; }