NOI2019 斗主地
题解
发现随机归并后每种可能的状态概率是一样的。
把操作反着看，发现相当于每次等概率随机一些牌放到最前面。
对于type=1，答案相当于第i张卡做完第m..1次随机，在它前面的卡数期望+1，相当于每张卡经过m次随机在它前面的概率和+1
于是我们可以只考虑一张卡在它前面或者在它后面。
那么f[x][0/1]表示顺序做完第m..x次随机，某张卡在它前面/后面，最后在它前面的概率。
对于type=1，答案相当于在它前面的（卡数平方+2卡数）期望+1，也就相当于$2\times$每对卡（无先后顺序）经过m次随机都在它前面的概率和+$3\times$ 每张卡经过m次随机在它前面的概率和+1
再记g[x][0/1/2]表示顺序做完第m..x次随机，某对卡都在它前面/一张在前一张在后/都在它后面，最后都在它前面的概率。
/*
Author: QAQ Automaton
Lang: C++
Prog: landlords.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
#define int long long
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define inf 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const int SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// print the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed integer
inline bool read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (char &x) {
x=gc();
return x!=EOF;
}
inline bool read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
*x=0;
return 1;
}
template<typename A,typename ...B>
inline bool read(A &x,B &...y){
return read(x)&&read(y...);
}
// print a signed integer
inline bool write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (char x) {
putc(x);
return 0;
}
inline bool write(const char *x){
while(*x){putc(*x);++x;}
return 0;
}
inline bool write(char *x){
while(*x){putc(*x);++x;}
return 0;
}
template<typename A,typename ...B>
inline bool write(A x,B ...y){
return write(x)||write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int a[105],b[105];
int fac[10000005],inv[10000005],invf[10000005]; const int p=998244353;
int down(int a,int b){return b<0||a<b?0:fac[a]*invf[a-b]%p;}
int C(int a,int b){return b<0||a<b?0:fac[a]*invf[a-b]%p*invf[b]%p;}
int fpm(int a,int b){
int c=1;
for(;b;b>>=1,a=a*a%p)if(b&1)c=c*a%p;
return c;
}
int n,m,type;
int f[500005][2],g[500005][3];
signed main(){
freopen("landlords.in","r",stdin);
freopen("landlords.out","w",stdout);
read(n,m,type);
fac[0]=fac[1]=inv[1]=invf[0]=invf[1]=1;
for(int i=2;i<=n;++i){
fac[i]=fac[i-1]*i%p;
inv[i]=(p-p/i)*inv[p%i]%p;
invf[i]=inv[i]*invf[i-1]%p;
}
if(type==1){
f[0][0]=1;
f[0][1]=0;
for(int i=1;i<=m;++i){
int x;
read(x);
int w=C(n-2,x-1)*fpm(C(n,x),p-2)%p;
f[i][0]=(f[i-1][0]*(p+1-w)+f[i-1][1]*w)%p;
f[i][1]=(f[i-1][0]*w+f[i-1][1]*(p+1-w))%p;
}
}
else{
f[0][0]=1;
f[0][1]=0;
g[0][0]=1;
g[0][1]=g[0][2]=0;
for(int i=1;i<=m;++i){
int x;
read(x);
int qaq=fpm(C(n,x),p-2);
int w=C(n-2,x-1)*qaq%p;
f[i][0]=(f[i-1][0]*(p+1-w)+f[i-1][1]*w)%p;
f[i][1]=(f[i-1][0]*w+f[i-1][1]*(p+1-w))%p;
int c0=C(n-3,x)*qaq%p,c1=C(n-3,x-1)*qaq%p,c2=C(n-3,x-2)*qaq%p,c3=C(n-3,x-3)*qaq%p;
g[i][0]=((c0+c1+c1+c2+c3)*g[i-1][0]+(c2+c2)*g[i-1][1]+(c1)*g[i-1][2])%p;
g[i][1]=((c1+c2)*g[i-1][0]+(c0+c1+c2+c3)*g[i-1][1]+(c1+c2)*g[i-1][2])%p;
g[i][2]=((c2)*g[i-1][0]+(c1+c1)*g[i-1][1]+(c0+c1+c2+c2+c3)*g[i-1][2])%p;
}
}
int q;
read(q);
for(;q;--q){
int x;
read(x);
if(type==1)write(((x-1)*f[m][0]+(n-x)*f[m][1]+1)%p,'\n');
else{
write((1+3*f[m][0]*(x-1)+3*f[m][1]*(n-x)+(x-1)*(x-2)%p*g[m][0]+2*(x-1)*(n-x)%p*g[m][1]+(n-x)*(n-x-1)%p*g[m][2])%p,'\n');
}
}
return 0;
}