[ABC310G] Takahashi And Pass-The-Ball Game

Problem Statement

There are $N$ Takahashi.

The $i$-th Takahashi has an integer $A_i$ and $B_i$ balls.

An integer $x$ between $1$ and $K$, inclusive, will be chosen uniformly at random, and they will repeat the following operation $x$ times.

For every $i$, the $i$-th Takahashi gives all his balls to the $A_i$-th Takahashi.

Beware that all $N$ Takahashi simultaneously perform this operation.

For each $i=1,2,\ldots,N$, find the expected value, modulo $998244353$, of the number of balls the $i$-th Takahashi has at the end of the operations.

How to find a expected value modulo $998244353$

It can be proved that the sought probability is always a rational number. Additionally, the constraints of this problem guarantee that if the sought probability is represented as an irreducible fraction $\frac{y}{x}$, then $x$ is not divisible by $998244353$.

Here, there is a unique $0\leq z\lt998244353$ such that $y\equiv xz\pmod{998244353}$, so report this $z$.

Constraints

$1\leq N\leq 2\times10^5$
$1\leq K\leq 10^{18}$
$K$ is not a multiple of $998244353$.
$1\leq A _ i\leq N\ (1\leq i\leq N)$
$0\leq B _ i\lt998244353\ (1\leq i\leq N)$
All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$ $K$

$A _ 1$ $A _ 2$ $\cdots$ $A _ N$

$B _ 1$ $B _ 2$ $\cdots$ $B _ N$

Output

Print the expected value of the number of balls the $i$-th Takahashi has at the end of the operations for $i=1,2,\ldots,N$, separated by spaces, in a single line.

Sample Input 1

Sample Output 1

3 0 499122179 499122178 5

During two operations, the five Takahashi have the following number of balls.

If $x=1$ is chosen, the five Takahashi have $4,0,1,2,5$ balls.

If $x=2$ is chosen, the five Takahashi have $2,0,4,1,5$ balls.

Thus, the sought expected values are $3,0,\dfrac52,\dfrac32,5$.
Print these values modulo $998244353$, that is, $3,0,499122179,499122178,5$, separated by spaces.

Sample Input 2

Sample Output 2

111 0 0

After one or more operations, the first Takahashi gets all balls.

Sample Input 3

16 1000007

16 12 6 12 1 8 14 14 5 7 6 5 9 6 10 9

719092922 77021920 539975779 254719514 967592487 476893866 368936979 465399362 342544824 540338192 42663741 165480608 616996494 16552706 590788849 221462860

Sample Output 3

817852305 0 0 0 711863206 253280203 896552049 935714838 409506220 592088114 0 413190742 0 363914270 0 14254803

Sample Input 4

24 100000000007

19 10 19 15 1 20 13 15 8 23 22 16 19 22 2 20 12 19 17 20 16 8 23 6

944071276 364842194 5376942 671161415 477159272 339665353 176192797 2729865 676292280 249875565 259803120 103398285 466932147 775082441 720192643 535473742 263795756 898670859 476980306 12045411 620291602 593937486 761132791 746546443

Sample Output 4

918566373 436241503 0 0 0 455245534 0 356196743 0 906000633 0 268983266 21918337 0 733763572 173816039 754920403 0 273067118 205350062 0 566217111 80141532 0

期望是假的，其实题目是让 $x$ 走 $i(i\le k)$ 步后的点加上 $a_i$

容易发现是一棵基环树，环上树上分开考虑。

树上的点长剖计算就好，复杂度 $O(n)$，关键是环上的点。

对每个树上的点，考虑他给环上的点带来的贡献，破环成链，差分计算即可。注意按照距离的不同，走的时候他有可能没走到。

#include<bits/stdc++.h>

using namespace std;

const int N=2e5+5,P=998244353;

typedef long long LL;

int n,q[N],l=1,r,a[N],b[N],e_num,hd[N],ans[N],inv,c[N<<1],in[N],s[N],dp[N],dfn[N],tme,son[N];

struct edge{

    int v,nxt;

}e[N];

LL k;

int read()

{

    char ch=getchar();

    int s=0;

    while(ch<'0'||ch>'9')

        ch=getchar();

    while(ch>='0'&&ch<='9')

        s=s*10+ch-48,ch=getchar();

    return s;

}

int pown(int x,int y)

{

    if(!y)

        return 1;

    int t=pown(x,y>>1);

    if(y&1)

        return 1LL*t*t%P*x%P;

    return 1LL*t*t%P;

}

void add_edge(int u,int v)

{

    e[++e_num]=(edge){v,hd[u]};

    hd[u]=e_num;

}

void doit(int x,int op)

{

    if(son[x])

        doit(son[x],1),ans[x]=(ans[son[x]]+b[son[x]])%P;

    for(int i=hd[x];i;i=e[i].nxt)

    {

        if(e[i].v==son[x])

            continue;

        doit(e[i].v,1);

        (ans[x]+=(ans[e[i].v]+b[e[i].v])%P)%=P;

        for(int j=0;j<dp[e[i].v];j++)

            (s[dfn[x]+j+1]+=s[dfn[e[i].v]+j])%=P;

    }

    if(k+1<dp[x])

        (ans[x]+=P-s[dfn[x]+k+1])%=P;

    if(!op)

        ans[x]=0;

}

void init(int x,int fa)

{

    s[dfn[x]=++tme]=b[x];

    if(son[x])

        init(son[x],x);

    for(int i=hd[x];i;i=e[i].nxt)

        if(e[i].v^son[x])

            init(e[i].v,x);

}

void dfs(int x,int dep,int fr)

{

    for(int i=hd[x];i;i=e[i].nxt)

        dfs(e[i].v,dep+1,fr);

    for(int i=hd[x];i;i=e[i].nxt)

        if(dp[e[i].v]>dp[son[x]])

            son[x]=e[i].v;

    dp[x]=dp[son[x]]+1;

    if(dep<=k)

    {

        LL p=(k-dep)%r,q=(k-dep)/r%P;

        (c[fr]+=1LL*(q+1)*b[x]%P)%=P;

        (c[fr+p+1]+=(P-b[x])%P)%=P;

        (c[fr+r]+=(P-1LL*q*b[x]%P)%P)%=P;

    }

}

void solve(int x)

{

    q[r=1]=x;

    int p=a[x];

    while(p^x)

    {

        q[++r]=p;

        p=a[p];

    }

    for(int i=1;i<=r;i++)

    {

        dfs(q[i],in[q[i]]=0,i);

        init(q[i],0);

        doit(q[i],0);

    }

    for(int i=1;i<=2*r;i++)

        (c[i]+=c[i-1])%=P,(ans[q[(i-1)%r+1]]+=c[i])%=P;

    for(int i=1;i<=r;i++)

        (ans[q[i]]+=(P-b[q[i]])%P)%=P;

    for(int i=1;i<=2*r;i++)

        c[i]=0;

}

int main()

{

    scanf("%d%lld",&n,&k),inv=pown(k%P,P-2);

    for(int i=1;i<=n;i++)

        a[i]=read(),in[a[i]]++;

    for(int i=1;i<=n;i++)

        b[i]=read();

    for(int i=1;i<=n;i++)

        if(!in[i])

            q[++r]=i;

    while(l<=r)

    {

        in[a[q[l]]]--;

        add_edge(a[q[l]],q[l]);

        if(!in[a[q[l]]])

            q[++r]=a[q[l]];

        l++;

    }

    for(int i=1;i<=n;i++)

        if(in[i])

            solve(i);

    for(int i=1;i<=n;i++)

        printf("%lld ",1LL*ans[i]*inv%P);

}