【LOJ6067】【2017 山东一轮集训 Day3】第三题 FFT
题目大意
给你 (n,b,c,d,e,a_0,a_1,ldots,a_{n-1})，定义
[egin{align}
x_k&=b imes c^{4k}+d imes c^{2k}+e\
f(x)&=sum_{i=0}^{n-1}a_ix^i
end{align}
]
求 (f(x_0),f(x_1),ldots,f(x_{n-1}))。
对 ({10}^6+3) 取模。
题解
直接多项式多点求值（显然）会TLE。
当 (c=0) 时：
[egin{align}
x_k&=
egin{cases}
b+d+e,&k=0\
e,&k
eq 0\
end{cases}
end{align}
]
直接求就好了。
当 (c eq 0,b=0) 时：
[egin{align}
f(x_k)&=sum_{i=0}^{n-1}a_i{(dc^{2k}+e)}^i\
&=sum_{i=0}^{n-1}a_isum_{j=0}^{i}{(dc^{2k})}^je^{i-j}inom{i}{j}\
&=sum_{j=0}^{n-1}frac{1}{j!}d^jc^{2kj}sum_{i=j}^{n-1}i!a_ie^{i-j}frac{1}{(i-j)!}\
&=sum_{j=0}^{n-1}frac{1}{j!}d^jc^{2kj}sum_{i=0}^{n-j-1}(n-i-1)!a_{n-i-1}e^{(n-i-1)-j}frac{1}{((n-i-1)-j)!}\
&=sum_{j=0}^{n-1}frac{1}{j!}d^jc^{2kj}g_j\
&=sum_{j=0}^{n-1}frac{1}{j!}d^jc^{k^2-({k-j)}^2+j^2}g_j\
&=c^{k^2}sum_{j=0}^{n-1}frac{1}{j!}d^jc^{j^2}g_jfrac{1}{c^{{(k-j)}^2}}
end{align}
]
当 (c eq 0,b eq 0) 时：
[egin{align}
bc^{4k}+dc^{2k}+e&=b(c^{4k}+frac{d}{b}c^{2k})+e\
&=b(c^{4k}+frac{d}{b}c^{2k}+{(frac{d}{2b})}^2-{(frac{d}{2b})}^2)+e\
&=b{(c^{2k}+frac{d}{2b})}^2-frac{d^2}{4b}+e\
&=b{(c^{2k}+d')}^2+e'\
end{align}
]
然后用 (d') 和 (e') 替换原来的 (d) 和 (e)：
[egin{align}
f(x_k)&=sum_{i=0}^{n-1}a_i{(b{(c^{2k}+d)}^2+e)}^i\
&=sum_{i=0}^{n-1}a_isum_{j=0}^{i}b^j{(c^{2k}+d)}^{2j}e^{i-j}inom{i}{j}\
&=sum_{j=0}^{n-1}b^j{(c^{2k}+d)}^{2j}frac{1}{j!}sum_{i=j}^{n-1}a_ii!e^{i-j}frac{1}{i-j}\
&=sum_{j=0}^{n-1}b^j{(c^{2k}+d)}^{2j}frac{1}{j!}g_j\
&=sum_{j=0}^{n-1}b^jfrac{1}{j!}g_jsum_{i=0}^{2j}c^{2ki}d^{2j-i}inom{2j}{i}\
&=sum_{i=0}^{2n-2}c^{2ki}frac{1}{i!}sum_{j=lceilfrac{i}{2}
ceil}^{n-1}frac{(2j)!}{j!}b^jg_jfrac{1}{(2j-i)!}d^{2j-i}\
&=sum_{i=0}^{2n-2}c^{2ki}frac{1}{i!}sum_{2jgeq i}^{2n-2}frac{(2j)!}{j!}b^jg_jfrac{1}{(2j-i)!}d^{2j-i}\
&=sum_{i=0}^{2n-2}c^{2ki}frac{1}{i!}h_i\
&=c^{k^2}sum_{i=0}^{2n-2}frac{1}{i!}c^{i^2}h_ifrac{1}{c^{{(i-k)}^2}}
end{align}
]
那些卷积都可以FFT（任意模数FFT）。
时间复杂度：(O(nlog n))
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<functional>
#include<cmath>
#include<vector>
#include<assert.h>
//using namespace std;
using std::min;
using std::max;
using std::swap;
using std::sort;
using std::reverse;
using std::random_shuffle;
using std::lower_bound;
using std::upper_bound;
using std::unique;
typedef long long ll;
typedef unsigned long long ull;
typedef std::pair<int,int> pii;
typedef std::pair<ll,ll> pll;
void open(const char *s){
#ifndef ONLINE_JUDGE
char str[100];sprintf(str,"%s.in",s);freopen(str,"r",stdin);sprintf(str,"%s.out",s);freopen(str,"w",stdout);
#endif
}
int rd(){int s=0,c,b=0;while(((c=getchar())<'0'||c>'9')&&c!='-');if(c=='-'){c=getchar();b=1;}do{s=s*10+c-'0';}while((c=getchar())>='0'&&c<='9');return b?-s:s;}
void put(int x){if(!x){putchar('0');return;}static int c[20];int t=0;while(x){c[++t]=x%10;x/=10;}while(t)putchar(c[t--]+'0');}
int upmin(int &a,int b){if(b<a){a=b;return 1;}return 0;}
int upmax(int &a,int b){if(b>a){a=b;return 1;}return 0;}
const ll p=1000003;
const int N=530000;
ll fp(ll a,ll b)
{
if(!b)
return 1;
b=(b%(p-1)+p-1)%(p-1);
ll s=1;
for(;b;b>>=1,a=a*a%p)
if(b&1)
s=s*a%p;
return s;
}
namespace fft
{
const int W=524288;
const ll M=1000;
typedef double db;
db pi=acos(-1);
struct cp
{
db x,y;
cp(db a=0,db b=0):x(a),y(b){}
};
cp operator +(cp a,cp b){return cp(a.x+b.x,a.y+b.y);}
cp operator -(cp a,cp b){return cp(a.x-b.x,a.y-b.y);}
cp operator *(cp a,cp b){return cp(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
cp operator /(cp a,int b){return cp(a.x/b,a.y/b);}
cp conj(cp a){return cp(a.x,-a.y);}
cp muli(cp a){return cp(-a.y,a.x);}
cp divi(cp a){return cp(a.y,-a.x);}
int rev[N];
cp *w[20];
void fft(cp *a,int n,int t)
{
for(int i=1;i<n;i++)
{
rev[i]=(rev[i>>1]>>1)|(i&1?n>>1:0);
if(rev[i]>i)
swap(a[rev[i]],a[i]);
}
for(int i=2,s=1;i<=n;i<<=1,s++)
for(int j=0;j<n;j+=i)
for(int k=0;k<i/2;k++)
{
cp u=a[j+k];
cp v=a[j+k+i/2]*w[s][k];
a[j+k]=u+v;
a[j+k+i/2]=u-v;
}
if(t==-1)
{
reverse(a+1,a+n);
for(int i=0;i<n;i++)
a[i]=a[i]/n;
}
}
void dft(db *a,db *b,cp *c,cp *d,int n)
{
static cp a1[N],a2[N];
for(int i=0;i<n;i++)
a1[i]=cp(a[i],b[i]);
fft(a1,n,1);
for(int i=0;i<n;i++)
a2[i]=conj(a1[i]);
reverse(a2+1,a2+n);
for(int i=0;i<n;i++)
{
c[i]=(a1[i]+a2[i])/2;
d[i]=divi(a1[i]-a2[i])/2;
}
}
void idft(db *a,db *b,cp *c,cp *d,int n)
{
static cp a1[N];
for(int i=0;i<n;i++)
a1[i]=c[i]+muli(d[i]);
fft(a1,n,-1);
for(int i=0;i<n;i++)
{
a[i]=a1[i].x;
b[i]=a1[i].y;
}
}
void init()
{
for(int i=1;i<=19;i++)
w[i]=new cp[1<<(i-1)];
for(int i=0;i<W/2;i++)
w[19][i]=cp(cos(2*pi/W*i),sin(2*pi/W*i));
for(int i=18;i>=1;i--)
for(int j=0;j<1<<(i-1);j++)
w[i][j]=w[i+1][j<<1];
}
void mul(ll *a,ll *b,ll *c,int n,int m,int l)
{
static db a1[N],a2[N],b1[N],b2[N],c1[N],c2[N],d1[N],d2[N];
static cp a3[N],a4[N],b3[N],b4[N],c3[N],c4[N],d3[N],d4[N];
int k=1;
while(k<=n+m)
k<<=1;
for(int i=0;i<k;i++)
a1[i]=a2[i]=b1[i]=b2[i]=0;
for(int i=0;i<=n;i++)
{
a[i]=(a[i]+p)%p;
a1[i]=a[i]/M;
a2[i]=a[i]%M;
}
for(int i=0;i<=m;i++)
{
b[i]=(b[i]+p)%p;
b1[i]=b[i]/M;
b2[i]=b[i]%M;
}
dft(a1,a2,a3,a4,k);
dft(b1,b2,b3,b4,k);
for(int i=0;i<k;i++)
{
c3[i]=a3[i]*b3[i];
c4[i]=a3[i]*b4[i];
d3[i]=a4[i]*b3[i];
d4[i]=a4[i]*b4[i];
}
idft(c1,c2,c3,c4,k);
idft(d1,d2,d3,d4,k);
for(int i=0;i<=l;i++)
c[i]=((ll)(c1[i]+0.5)%p*M%p*M%p+(ll)(c2[i]+0.5)%p*M%p+(ll)(d1[i]+0.5)%p*M%p+(ll)(d2[i]+0.5)%p)%p;
}
}
int n;
ll B,C,D,E;
ll a[N];
ll inv[N],fac[N],ifac[N];
void init()
{
inv[1]=fac[0]=fac[1]=ifac[0]=ifac[1]=1;
for(int i=2;i<=200000;i++)
{
inv[i]=-p/i*inv[p%i]%p;
inv[i]=(inv[i]+p)%p;
fac[i]=fac[i-1]*i%p;
ifac[i]=ifac[i-1]*inv[i]%p;
}
}
namespace pregao
{
ll b[N],c[N],s[N];
void gao()
{
for(int i=0;i<n;i++)
{
b[i]=a[n-i-1]*fac[n-i-1]%p;
c[i]=ifac[i]*fp(E,i)%p;
}
fft::mul(b,c,s,n-1,n-1,n-1);
reverse(s,s+n);
}
}
namespace gao1
{
ll d[N],e[N],f[N],g[N],h[N],ans[N];
void gao()
{
pregao::gao();
for(int i=0;i<n;i++)
{
d[i]=pregao::s[i]*fp(D,i)%p*fp(C,(ll)i*i)%p*ifac[i]%p;
e[i]=fp(C,-(ll)i*i);
}
fft::mul(d,e,f,n-1,n-1,n-1);
reverse(d,d+n);
e[0]=0;
fft::mul(d,e,g,n-1,n-1,n-1);
for(int i=0;i<n;i++)
{
ans[i]=((f[i]+g[n-i-1])%p+p)%p;
ans[i]=ans[i]*fp(C,(ll)i*i)%p;
}
for(int i=0;i<n;i++)
printf("%lld
",ans[i]);
}
}
namespace gao2
{
ll b[N],c[N],d[N],e[N],f[N],g[N],ans[N];
void gao()
{
E=(E-D*D%p*fp(4*B,p-2)%p+p)%p;
D=D*fp(2*B,p-2)%p;
pregao::gao();
for(int i=0;i<n;i++)
b[2*n-2*i-2]=fac[2*i]*ifac[i]%p*pregao::s[i]%p*fp(B,i)%p;
for(int i=0;i<2*n-1;i++)
c[i]=fp(D,i)*ifac[i]%p;
fft::mul(b,c,d,2*n-2,2*n-2,2*n-2);
reverse(d,d+2*n-1);
for(int i=0;i<2*n-1;i++)
{
d[i]=d[i]*ifac[i]%p*fp(C,(ll)i*i)%p;
e[i]=fp(C,-(ll)i*i);
}
fft::mul(d,e,f,n-1,n-1,n-1);
reverse(d,d+2*n-1);
e[0]=0;
fft::mul(d,e,g,2*n-2,2*n-2,2*n-2);
for(int i=0;i<n;i++)
ans[i]=(fp(C,(ll)i*i)*(f[i]+g[2*n-i-2])%p+p)%p;
for(int i=0;i<n;i++)
printf("%lld
",ans[i]);
}
}
namespace gao0
{
void gao()
{
ll ans=0;
for(int i=n-1;i>=0;i--)
{
ans=ans*(E+B+D)%p;
ans=(ans+a[i])%p;
}
printf("%lld
",ans);
ans=0;
for(int i=n-1;i>=0;i--)
{
ans=ans*E%p;
ans=(ans+a[i])%p;
}
for(int i=1;i<n;i++)
printf("%lld
",ans);
}
}
int main()
{
open("loj6067");
fft::init();
init();
scanf("%d%lld%lld%lld%lld",&n,&B,&C,&D,&E);
for(int i=0;i<n;i++)
a[i]=rd();
if(!C)
gao0::gao();
else if(!B)
gao1::gao();
else
gao2::gao();
return 0;
}