对于询问，我们可以分两种情况讨论：
1.第x株植物不变种且排名不变，当且仅当对于所有的ai<ax且kai>=ax的植物不变种，设这样的植物有p株
2.第x株植物变种且排名不变，当且仅当对于所有的ax<=ai<kax的植物变种，设这样的植物有q株
则x排名不变的概率为(1-pr)^(p+1)+pr^(q+1)，用数据结构维护即可，这里使用的是线段树
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=200005;
const long long mod=998244353;
int n,m,pr,k,left,right,a[N],tree[N<<2],t[N*2],len,op[N],x[N],y[N];
long long ans,p[N],np[N];
void modify(int k,int l,int r,int x,int v){
if (l==r){
tree[k]+=v;
return ;
}
int mid=l+r>>1;
if (x<=mid) modify(k<<1,l,mid,x,v);
else modify(k<<1|1,mid+1,r,x,v);
tree[k]=tree[k<<1]+tree[k<<1|1];
}
int query(int k,int l,int r,int x,int y){
if (x<=l&&r<=y){
return tree[k];
}
int mid=l+r>>1,s=0;
if (x<=mid) s+=query(k<<1,l,mid,x,y);
if (mid<y) s+=query(k<<1|1,mid+1,r,x,y);
return s;
}
int main(){
freopen ("lycoris.in","r",stdin);
freopen ("lycoris.out","w",stdout);
scanf ("%d%d%d%d",&n,&m,&pr,&k);
p[0]=np[0]=1;
for (register int i=1;i<=n;++i){
p[i]=1ll*p[i-1]*pr%mod;
np[i]=1ll*np[i-1]*(1-pr)%mod;
}
for (register int i=1;i<=n;++i){
scanf ("%d",&a[i]);
t[++len]=a[i];
}
for (register int i=1;i<=m;++i){
scanf ("%d%d",&op[i],&x[i]);
if (op[i]==1){
scanf ("%d",&y[i]);
t[++len]=y[i];
}
}
sort (t+1,t+1+len);
len=unique(t+1,t+1+len)-t-1;
for (register int i=1;i<=n;++i){
modify(1,1,len,lower_bound(t+1,t+1+len,a[i])-t,1);
}
for (register int i=1;i<=m;++i){
if (op[i]==1){
modify(1,1,len,lower_bound(t+1,t+1+len,a[x[i]])-t,-1);
a[x[i]]=y[i];
modify(1,1,len,lower_bound(t+1,t+1+len,a[x[i]])-t,1);
}
else{
ans=0ll;
right=lower_bound(t+1,t+1+len,a[x[i]])-t-1;
left=lower_bound(t+1,t+1+len,(a[x[i]]-1)/k+1)-t;
if (left<=right) ans=np[query(1,1,len,left,right)];
else ans=1;
ans=ans*(1-pr)%mod;
left=lower_bound(t+1,t+1+len,a[x[i]])-t;
right=lower_bound(t+1,t+1+len,a[x[i]]*k)-t-1;
if (t[len]<a[x[i]]*k) right=len;
int num=0;
if (left<=right) num=query(1,1,len,left,right)-1;
ans=(ans+pr*p[num])%mod;
printf ("%lld\n",(ans+mod)%mod);
}
}
return 0;
}