两个多项式系数的卷积

Posted 2023-04-23

MAtlab 多项式的乘法实际上是多项式系数向量间的卷积运算?
向量间的卷积运算是怎么回事,多项式函数间的卷积运算又是怎么回事呢?
向量间的卷积运算有计算公式吗?具体怎么运算的?

参考技术A 本质上是一样的.
向量卷积和多项式函数间的卷积从算法这个角度上,是一样的,所以都是conv函数.
多项式函数的系数,其实就是一个向量.
补充问题：
向量卷积就是离散信号的卷积,看任意一本《信号与系统》的书就可以.
y(n)=∑x(m)h(n-m) (求和变量为m)

模板FFT

题意简述

求两个多项式的卷积

题解思路

先将多项式转化为点值表示法，再相乘，最后转化为系数表示法

注意：用三次变两次优化会掉精

代码（递归）

#include <cmath>
#include <cstdio>
const int N=4000010;
const double Pi=acos(-1.0);
int n,m,len=1;
struct Complex 
    double x,y;
    Complex(double xx=0,double yy=0)  x=xx; y=yy; 
a[N],b[N];
Complex operator +(Complex x,Complex y)  return Complex(x.x+y.x,x.y+y.y); 
Complex operator -(Complex x,Complex y)  return Complex(x.x-y.x,x.y-y.y); 
Complex operator *(Complex x,Complex y)  return Complex(x.x*y.x-x.y*y.y,x.x*y.y+x.y*y.x); 
void FFT(Complex *x,const int& len,const int& type) 
    if (len==1) return;
    Complex x1[len>>1],x2[len>>1];
    for (register int i=0;i<len;i+=2) 
        x1[i>>1]=x[i]; x2[i>>1]=x[i+1];
    
    FFT(x1,len>>1,type); FFT(x2,len>>1,type);
    Complex UR=Complex(cos(2.0*Pi/len),type*sin(2.0*Pi/len)),w=Complex(1,0);
    for (register int i=0,_n=len>>1;i<_n;++i,w=w*UR) 
        x[i]=x1[i]+x2[i]*w;
        x[i+_n]=x1[i]-x2[i]*w;
    

int main() 
    scanf("%d%d",&n,&m);
    for (register int i=0;i<=n;++i) scanf("%lf",&a[i].x);
    for (register int i=0;i<=m;++i) scanf("%lf",&b[i].x);
    n+=m; for (;len<=n;len<<=1);
    FFT(a,len,1); FFT(b,len,1);
    for (register int i=0;i<=len;++i) a[i]=a[i]*b[i];
    FFT(a,len,-1);
    for (register int i=0;i<=n;++i) printf("%d%c",(int)(a[i].x/len+0.5)," \n"[i==n]);

代码（非递归）

#include <cmath>
#include <cctype>
#include <cstdio>
#include <algorithm>
const int N=4000010;
const double Pi=acos(-1.0);
int n,m,lim=1,l=-1,r[N];
char ch;
inline int read()  for (;!isdigit(ch=getchar());); return ch-48; 
struct Complex 
    double x,y;
    Complex(double xx=0,double yy=0)  x=xx; y=yy; 
a[N];
Complex operator +(Complex x,Complex y)  return Complex(x.x+y.x,x.y+y.y); 
Complex operator -(Complex x,Complex y)  return Complex(x.x-y.x,x.y-y.y); 
Complex operator *(Complex x,Complex y)  return Complex(x.x*y.x-x.y*y.y,x.x*y.y+x.y*y.x); 
inline void FFT(Complex *x,const int& lim,const int& type) 
    for (register int i=0;i<lim;++i) if (i<r[i]) std::swap(x[i],x[r[i]]);
    for (register int i=1;i<lim;i<<=1) 
        Complex UR(cos(Pi/i),sin(Pi/i)*type);
        for (register int j=0,I=i<<1;j<lim;j+=I) 
            Complex w(1,0);
            for (register int k=0;k<i;++k,w=w*UR) 
                Complex t1=x[j+k],t2=w*x[k+i+j];
                x[j+k]=t1+t2; x[j+k+i]=t1-t2;
            
        
    

int main() 
    scanf("%d%d",&n,&m);
    for (register int i=0;i<=n;++i) a[i].x=read();
    for (register int i=0;i<=m;++i) a[i].y=read();
    n+=m; for (;lim<=n;lim<<=1,++l);
    for (register int i=0;i<lim;++i) r[i]=(r[i>>1]>>1)|((i&1)<<l);
    FFT(a,lim,1);
    for (register int i=0;i<lim;++i) a[i]=a[i]*a[i];
    FFT(a,lim,-1);
    for (register int i=0;i<=n;++i) printf("%d%c",(int)(a[i].y/lim/2+0.5)," \n"[i==n]);