hdu6801.Game on a Circle

题目大意

n个石子围成一圈，从第一个开始顺时针考虑，每次有p的概率取走当前石子，问c号石子是第1...n个被取走的概率

n<=1e6

题解

记q=1-p，枚举石子c被取走之前的完整轮数t，设x^i表示是第i+1个被取走的概率

(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实际上是考虑可以取已经取走的石子，并且最后概率加起来不变且只记一次，还可以统一考虑

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})

(=psum_isum_jinom{c-1}{i}inom{n-c}{j}sum_{t=0}^{infty} q^{t+(t+1)i+tj}(1-x)^{i+j}x^{n-1-i-j})

(=psum_isum_jinom{c-1}{i}inom{n-c}{j}q^i(1-x)^{i+j}x^{n-1-i-j}sum_{t=0}^{infty} q^{t(1+i+j)})

(=psum_isum_jinom{c-1}{i}inom{n-c}{j}q^i(1-x)^{i+j}x^{n-1-i-j}frac{1}{1-q^{i+j+1}})

(=psum_{k=0}^{n-1}sum_{i+j=k}inom{c-1}{i}inom{n-c}{j}q^i(1-x)^{k}x^{n-1-k}frac{1}{1-q^{k+1}})

设(f_k=sum_{i+j=k}inom{c-1}{i}inom{n-c}{j}q^i)

(=psum_{k=0}^{n-1}f_k(1-x)^{k}x^{n-1-k}frac{1}{1-q^{k+1}})

(=psum_{i=0}^{n-1}x^isum_{k=0}^{n-1}f_k(1-x)^{k}frac{1}{1-q^{k+1}}[x^{i+k+1-n}])

(=psum_{i=0}^{n-1}x^isum_{k=0}^{n-1}f_k inom{k}{n-i-1}(-1)^{i+k+1-n}frac{1}{1-q^{k+1}})

求出f之后就可以卷了

f也可以卷，但是1e6很悬

(f_k=sum_{i+j=k}inom{c-1}{i}inom{n-c}{j}q^i)

设(F(x)=sum_i x^if_i)，则(F(x)=(1+qx)^{c-1}(1+x)^{n-c})

考虑推出f的递推式，有显然的sb推法：

---

((1+qx)^{c-1}=sum inom{c-1}{i}q^ix^i)

(((1+qx)^{c-1})'=sum inom{c-1}{i}q^iix^{i-1}=sum inom{c-2}{i-1}(c-1)q^ix^{i-1}=(c-1)qsum inom{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)qfrac{F(x)}{1+qx})

((1+x)^{n-c})同理

然后在群里被教育了一顿

---

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)是复合函数求导，(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})，根据x^i的系数可以列出等式：

(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}))

完结撒花

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
",(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;
}