【模板】FFT

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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]);
}

【模板】FFT

标签：operator   题意   递归   str   模板   turn   getc   scan   print   

原文地址：https://www.cnblogs.com/xuyixuan/p/11326532.html