hdu6801.Game on a Circle
題目大意
n個石子圍成一圈，從第一個開始順時針考慮，每次有p的概率取走當前石子，問c號石子是第1...n個被取走的概率
n<=1e6
題解
記q=1-p，列舉石子c被取走之前的完整輪數t，設x^i表示是第i+1個被取走的概率
\(\sum_{t=0}^{\infty} q^tp(q^{t+1}+(1-q^{t+1})x)^{c-1}(q^{t}+(1-q^{t})x)^{n-c}\)
其中1-q^t實際上是考慮可以取已經取走的石子，並且最後概率加起來不變且只記一次，還可以統一考慮
1-q^t不好搞，所以拆一下
\(=\sum_{t=0}^{\infty} q^tp(q^{t+1}(1-x)+x)^{c-1}(q^{t}(1-x)+x)^{n-c}\)
\(=p\sum_i\sum_j\binom{c-1}{i}\binom{n-c}{j}\sum_{t=0}^{\infty} q^{t+(t+1)i+tj}(1-x)^{i+j}x^{n-1-i-j}\)
\(=p\sum_i\sum_j\binom{c-1}{i}\binom{n-c}{j}q^i(1-x)^{i+j}x^{n-1-i-j}\sum_{t=0}^{\infty} q^{t(1+i+j)}\)
\(=p\sum_i\sum_j\binom{c-1}{i}\binom{n-c}{j}q^i(1-x)^{i+j}x^{n-1-i-j}\frac{1}{1-q^{i+j+1}}\)
\(=p\sum_{k=0}^{n-1}\sum_{i+j=k}\binom{c-1}{i}\binom{n-c}{j}q^i(1-x)^{k}x^{n-1-k}\frac{1}{1-q^{k+1}}\)
設\(f_k=\sum_{i+j=k}\binom{c-1}{i}\binom{n-c}{j}q^i\)
\(=p\sum_{k=0}^{n-1}f_k(1-x)^{k}x^{n-1-k}\frac{1}{1-q^{k+1}}\)
\(=p\sum_{i=0}^{n-1}x^i\sum_{k=0}^{n-1}f_k(1-x)^{k}\frac{1}{1-q^{k+1}}[x^{i+k+1-n}]\)
\(=p\sum_{i=0}^{n-1}x^i\sum_{k=0}^{n-1}f_k \binom{k}{n-i-1}(-1)^{i+k+1-n}\frac{1}{1-q^{k+1}}\)
求出f之後就可以捲了
f也可以卷，但是1e6很懸
\(f_k=\sum_{i+j=k}\binom{c-1}{i}\binom{n-c}{j}q^i\)
設\(F(x)=\sum_i x^if_i\)，則\(F(x)=(1+qx)^{c-1}(1+x)^{n-c}\)
考慮推出f的遞推式，有顯然的sb推法：
\((1+qx)^{c-1}=\sum \binom{c-1}{i}q^ix^i\)
\(((1+qx)^{c-1})'=\sum \binom{c-1}{i}q^iix^{i-1}=\sum \binom{c-2}{i-1}(c-1)q^ix^{i-1}=(c-1)q\sum \binom{c-2}{i-1}q^{i-1}x^{i-1}\)
\(=(c-1)q(1+qx)^{c-2}\)
\(F(x)=(1+qx)^{c-1}(1+x)^{n-c}\)
\(((1+qx)^{c-1})'(1+x)^{n-c}=(c-1)q(1+qx)^{c-2}(1+x)^{n-c}=(c-1)q\frac{F(x)}{1+qx}\)
\((1+x)^{n-c}\)同理
然後在群裡被教育了一頓
正常推法：
\(\ln F(x)=\frac{F'(x)}{F(x)}=(\ln (1+qx)^{c-1}(1+x)^{n-c})'\)
\(=((c-1)\ln (1+qx)+(n-c)\ln (1+x))'\)
注意其中的ln(1+x)和ln(1+qx)是複合函式求導，(1+qx)'=q
\(=\frac{(c-1)q}{1+qx}+\frac{n-c}{1+x}\)
所以\(F'(x)=\frac{(c-1)qF(x)}{1+qx}+\frac{(n-c)F(x)}{1+x}\)
有\([x^i]F(x)=(i+1)f_{i+1}\)，根據x^i的係數可以列出等式：
\(q(i-1)f_{i-1}+(q+1)if_i+(i+1)f_{i+1}=((c-1)q+n-c)f_i+((c-1)q+(n-c)q)f_{i-1}\)
整理可得
\(f_{i+1}=\frac{1}{i+1}(((c-1)q+n-c-(q+1)i)f_i+((c-1)q+(n-c)q-q(i-1))f_{i-1})\)
\(f_{i+1}=\frac{1}{i+1}(((c-1)q+n-c-(q+1)i)f_i+q(n-i)f_{i-1})\)
完結撒花
code
#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define mod 998244353
#define Mod 998244351
#define ll long long
#define G 3
//#define file
using namespace std;
ll a[2097152],b[2097152],f[2097152],jc[1000001],Jc[1000001],w[1000001],p,q,T,ans,N2;
int N,len,n,i,j,k,l,A,B,c;
ll qpower(ll a,int b) {ll ans=1; while (b) {if (b&1) ans=ans*a%mod;a=a*a%mod;b>>=1;} return ans;}
void dft(ll *a,int tp)
{
static ll A[2097152];
int i,j,k,l,S=N,s1=2,s2=1;
ll u,v,w,W;
fo(i,0,N-1)
{
j=i,k=0;
fo(l,1,len)
k=k*2+(j&1),j>>=1;
A[i]=a[k];
}
memcpy(a,A,N*8);
fo(i,1,len)
{
w=(tp==1)?qpower(G,(mod-1)/s1):qpower(G,(mod-1)-(mod-1)/s1);S>>=1;
fo(j,0,S-1)
{
W=1;
fo(k,0,s2-1)
{
u=a[j*s1+k],v=a[j*s1+k+s2]*W%mod;
a[j*s1+k]=(u+v)%mod;
a[j*s1+k+s2]=(u-v)%mod;
W=W*w%mod;
}
}
s1<<=1,s2<<=1;
}
}
ll C(int n,int m) {if (n<m) return 0;return jc[n]*Jc[m]%mod*Jc[n-m]%mod;}
void init()
{
int i,j,k,l;
f[0]=1;f[1]=(C(c-1,1)*q+C(n-c,1))%mod;
fo(i,2,n-1) f[i]=((q*(c-1)+n-c-(q+1)*(i-1))%mod*f[i-1]+(q*(n-(i-1)))%mod*f[i-2])%mod*w[i]%mod;
}
int main()
{
#ifdef file
freopen("hdu6801.in","r",stdin);
#endif
jc[0]=jc[1]=Jc[0]=Jc[1]=w[1]=1;
fo(i,2,1000000) w[i]=mod-w[mod%i]*(mod/i)%mod,jc[i]=jc[i-1]*i%mod,Jc[i]=Jc[i-1]*w[i]%mod;
scanf("%d",&T);
for (;T;--T)
{
scanf("%d%d%d%d",&n,&A,&B,&c);len=ceil(log2(n))+1;N=qpower(2,len);N2=qpower(N,Mod);
p=1ll*A*qpower(B,Mod)%mod,q=1-p;
memset(f,0,N*8);
memset(a,0,N*8);
memset(b,0,N*8);
init();
fo(i,0,n-1) a[i]=qpower(1-qpower(q,i+1),Mod)*f[i]%mod*jc[i]%mod*(((i+1-n)&1)?-1:1);
fo(i,n-1,2*(n-1)) b[(n-1)*2-i]=Jc[i-(n-1)];
dft(a,1),dft(b,1);
fo(i,0,N-1) a[i]=a[i]*b[i]%mod;
dft(a,-1);
fo(i,0,n-1) printf("%lld\n",(a[2*(n-1)-i]*N2%mod*((i&1)?-1:1)+mod)%mod*p%mod*Jc[n-1-i]%mod);
}
fclose(stdin);
fclose(stdout);
return 0;
}