洛谷.3803.[模板]多项式乘法(FFT)
题目链接:洛谷、LOJ.
FFT相关:快速傅里叶变换(FFT)详解、FFT总结、从多项式乘法到快速傅里叶变换.
5.4 又看了一遍,这个也不错。
2019.3.7 叕看了一遍,推荐这个。
#include <cmath>
#include <cctype>
#include <cstdio>
#include <algorithm>
#define gc() getchar()
const int N=1e6+5;
const double PI=acos(-1);
int n,m;
struct Complex
{
double x,y;
Complex(double xx=0,double yy=0) {x=xx, y=yy;}
Complex operator + (const Complex &a) {return Complex(x+a.x, y+a.y);}
Complex operator - (const Complex &a) {return Complex(x-a.x, y-a.y);}
Complex operator * (const Complex &a) {return Complex(x*a.x-y*a.y, x*a.y+y*a.x);}
}A[N*3],B[N*3];//size!
void Fast_Fourier_Transform(Complex *a,int lim,int opt)
{
for(int j=0,i=0; i<lim; ++i)
{
if(i>j) std::swap(a[i],a[j]);
for(int l=lim>>1; (j^=l)<l; l>>=1);
}
for(int i=2; i<=lim; i<<=1)//最后等于lim即整个序列的合并
{
int mid=i>>1;
Complex Wn(cos(2.0*PI/i),opt*sin(2.0*PI/i)),t;
for(int j=0; j<lim; j+=i)
{
Complex w(1,0);
for(int k=0; k<mid; ++k,w=w*Wn)
a[j+mid+k]=a[j+k]-(t=w*a[j+mid+k]),
a[j+k]=a[j+k]+t;
}
}
}
inline int read()
{
int now=0,f=1;register char c=gc();
for(;!isdigit(c);c=gc()) if(c=='-') f=-1;
for(;isdigit(c);now=now*10+c-'0',c=gc());
return now*f;
}
int main()
{
n=read(),m=read();
for(int i=0; i<=n; ++i) A[i].x=(double)read();//scanf("%lf",&A[i].x);
for(int i=0; i<=m; ++i) B[i].x=(double)read();//scanf("%lf",&B[i].x);
int lim=1;
while(lim<=n+m) lim<<=1;
Fast_Fourier_Transform(A,lim,1);
Fast_Fourier_Transform(B,lim,1);
for(int i=0; i<=lim; ++i) A[i]=A[i]*B[i];//size!
Fast_Fourier_Transform(A,lim,-1);
for(int i=0; i<=n+m; ++i) printf("%d ",(int)(A[i].x/lim+0.5));
return 0;
}
递归实现:
#include <cmath>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#define gc() getchar()
const int N=2e6+5;
const double PI=acos(-1.0);
int n,m;
struct Complex
{
double x,y;
Complex(double xx=0,double yy=0) {x=xx, y=yy;}
Complex operator + (const Complex &a) {return Complex(x+a.x, y+a.y);}
Complex operator - (const Complex &a) {return Complex(x-a.x, y-a.y);}
Complex operator * (const Complex &a) {return Complex(x*a.x-y*a.y, x*a.y+y*a.x);}
}A[N],B[N];
inline int read()
{
int now=0,f=1;register char c=gc();
for(;!isdigit(c);c=gc()) if(c=='-') f=-1;
for(;isdigit(c);now=now*10+c-'0',c=gc());
return now*f;
}
void Fast_Fourier_Transform(Complex *a,int lim,int type)
{
if(lim==1) return;
Complex a1[lim>>1],a2[lim>>1];//爆栈
for(int i=0; i<lim; i+=2)
a1[i>>1]=a[i], a2[i>>1]=a[i+1];
Fast_Fourier_Transform(a1,lim>>1,type),
Fast_Fourier_Transform(a2,lim>>1,type);
Complex Wn(cos(2.0*PI/lim),type*sin(2.0*PI/lim)),w(1,0),t;//Wn:单位根 w:幂
for(int i=0; i<(lim>>1); ++i,w=w*Wn)
a[i]=a1[i]+(t=w*a2[i]),
a[i+(lim>>1)]=a1[i]-t;
}
int main()
{
n=read(),m=read();
for(int i=0; i<=n; ++i) A[i].x=read();
for(int i=0; i<=m; ++i) B[i].x=read();
int lim=1;
while(lim<=n+m) lim<<=1;
Fast_Fourier_Transform(A,lim,1);
Fast_Fourier_Transform(B,lim,1);
for(int i=0; i<=lim; ++i) A[i]=A[i]*B[i];
Fast_Fourier_Transform(A,lim,-1);
for(int i=0; i<=n+m; ++i) printf("%d ",(int)(A[i].x/lim+0.5));
return 0;
}
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很久以前的奇怪但现在依旧成立的签名
attack is our red sun $$\color{red}{\boxed{\color{red}{attack\ is\ our\ red\ sun}}}$$ ------------------------------------------------------------------------------------------------------------------------
很久以前的奇怪但现在依旧成立的签名
attack is our red sun $$\color{red}{\boxed{\color{red}{attack\ is\ our\ red\ sun}}}$$ ------------------------------------------------------------------------------------------------------------------------