HDU 4656 Evaluation（MTT）

题意

(x_k=bc^{2k}+d)

(displaystyle F(x)=sum_{i=0}^{n-1}a_ix^i)

给定 ({a},b,c,d,n) ，求 (F(x_0),F(x_1),cdots,F(x_{n-1}))

思路

设 (ans_k=F(x_k))

[ans_k=sum_{i=0}^{n-1}a_i(bcdot c^{2k}+d)^i ]

二项展开得

[ans_k=sum_{i=0}^{n-1}sum_{j=0}^{i}a_i{ichoose j}(bcdot c^{2k})^jd^{i-j} ]

设 (displaystyle f(i)=a_id^ii!,g(j)={b^jover {d^jj!}})

[ans_k=sum_{i=0}^{n-1}sum_{j=0}^{i}f(i)g(j)c^{2kj}{1over(i-j)!}\ ans_k=sum_{j=0}^{n-1}g(j)c^{2kj}sum_{i=j}^{n-1}f(i){1over(i-j)!} ]

不难发现，后面那个 $ sum$ 与 (k) 无关，可以预处理出来。

我们设 (displaystyle h(j)=sum_{i=j}^{n-1}f(i){1over(i-j)!})

这个可以直接用多项式乘出来。

有能力 $ O(1)$ 求解 $ h(j)$ 后，得到

[ans_k=sum_{j=0}^{n-1}g(j)h(j)c^{2jk}\ ans_k=sum_{j=0}^{n-1}g(j)h(j)c^{j^2+k^2-(k-j)^2}\ ans_k=sum_{j=0}^{n-1}g(j)h(j){c^{j^2}c^{k^2}over {c^{{(k-j)}^2}}}\ ans_k=c^{k^2}sum_{j=0}^{n-1}g(j)h(j)c^{j^2}{1over {c^{{(k-j)}^2}}}\ ]

再进行一次多项式乘法就可以了。

代码

#include<bits/stdc++.h>
#define FOR(i,x,y) for(int i=(x),i##END=(y);i<=i##END;++i)
#define DOR(i,x,y) for(int i=(x),i##END=(y);i>=i##END;--i)
typedef long long ll;
using namespace std;
const double PI=acos(-1.0);
const int N=1<<18|5;
const int P=1e6+3;
namespace _Maths
{
	ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
	void exgcd(ll a,ll b,ll &x,ll &y)
	{
		if(!b){x=1,y=0;return;}
		exgcd(b,a%b,y,x);y-=a/b*x;
	}
	ll Pow(ll a,ll p,ll P)
	{
		ll res=1;
		for(;p>0;(a*=a)%=P,p>>=1)if(p&1)(res*=a)%=P;
		return res;
	}
	ll inv(ll a,ll P){ll x,y;exgcd(a,P,x,y);return (x%P+P)%P;}
};
using namespace _Maths;
struct Complex
{
	double x,y;
	Complex operator +(const Complex &_)const{return (Complex){x+_.x,y+_.y};}
	Complex operator -(const Complex &_)const{return (Complex){x-_.x,y-_.y};}
	Complex operator *(const Complex &_)const{return (Complex){x*_.x-y*_.y,x*_.y+y*_.x};}
	Complex operator /(const int &_)const{return (Complex){x/_,y/_};}
};
namespace _Polynomial
{
    const int K=(1<<15)-1,L=15;
    Complex A[N<<1],B[N<<1],C[N<<1],D[N<<1];
    Complex w[N<<1];int r[N<<1];
    void DFT(Complex *a,int op,int n)
    {
        FOR(i,0,n-1)if(i<r[i])swap(a[i],a[r[i]]);
        for(int i=2;i<=n;i<<=1)
            for(int j=0;j<n;j+=i)
                for(int k=0;k<i/2;k++)
                {
                    Complex u=a[j+k],t=w[(op==1?n/i*k:n-n/i*k)&(n-1)]*a[j+k+i/2];
                    a[j+k]=u+t,a[j+k+i/2]=u-t;
                }
        if(op==-1)FOR(i,0,n-1)a[i]=a[i]/n;
    }
    void multiply(const int *a,const int *b,int *c,int n1,int n2)
    {
        int n=1;
		while(n<n1+n2-1)n<<=1;
		FOR(i,0,n1-1)A[i].x=a[i]&K,A[i].y=a[i]>>L;
		FOR(i,0,n2-1)B[i].x=b[i]&K,B[i].y=b[i]>>L;
		FOR(i,n1,n-1)A[i].x=A[i].y=0;
		FOR(i,n2,n-1)B[i].x=B[i].y=0;
		FOR(i,0,n-1)r[i]=(r[i>>1]>>1)|((i&1)*(n>>1));
		FOR(i,0,n-1)w[i]=(Complex){cos(2*PI*i/n),sin(2*PI*i/n)};
		
		DFT(A,1,n),DFT(B,1,n);
		FOR(i,0,n-1)
		{
			int j=(n-i)&(n-1);
			C[i]=(Complex){0.5*(A[i].x+A[j].x),0.5*(A[i].y-A[j].y)}*B[i];
			D[i]=(Complex){0.5*(A[i].y+A[j].y),0.5*(A[j].x-A[i].x)}*B[i];
		}
		DFT(C,-1,n),DFT(D,-1,n);
		FOR(i,0,n-1)
		{
			ll s=C[i].x+0.5,t=C[i].y+0.5,u=D[i].x+0.5,v=D[i].y+0.5;
			c[i]=(s%P+((t+u)%P<<L)+(v%P<<L<<L))%P;
		}
    }
};
int fac[N],ifac[N],h[N];
int A[N],B[N],C[N<<1];
int a[N],b,c,d;
int n;

ll f(int i){return (ll)a[i]*Pow(d,i,P)%P*fac[i]%P;}
ll g(int i){return (ll)Pow(b,i,P)*inv(Pow(d,i,P)%P*fac[i]%P,P)%P;}

int main()
{
	fac[0]=fac[1]=1;FOR(i,2,N-1)fac[i]=(ll)fac[i-1]*i%P;
	ifac[0]=ifac[1]=1;FOR(i,2,N-1)ifac[i]=(ll)(P-P/i)*ifac[P%i]%P;
	FOR(i,2,N-1)ifac[i]=(ll)ifac[i-1]*ifac[i]%P;
	scanf("%d%d%d%d",&n,&b,&c,&d);
	FOR(i,0,n-1)scanf("%d",&a[i]);
	
	FOR(i,0,n-1)A[i]=f(i);
	FOR(i,1-n,0)B[(n-1)+i]=ifac[-i];
	_Polynomial::multiply(A,B,C,n,n);
	FOR(i,0,n-1)h[i]=C[(n-1)+i];
	
	FOR(i,0,n-1)A[i]=g(i)*h[i]%P*Pow(c,(ll)i*i,P)%P;
	FOR(i,1-n,n-1)B[(n-1)+i]=inv(Pow(c,(ll)i*i,P),P);
	_Polynomial::multiply(A,B,C,n,2*n-1);
	FOR(i,0,n-1)printf("%lld
",Pow(c,(ll)i*i,P)*C[(n-1)+i]%P);
    return 0;
}