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  • 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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  • 原文地址:https://www.cnblogs.com/SovietPower/p/8444779.html
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