https://vjudge.net/contest/591700#problem/G
看到排列，先考虑置换换，题意转化为置换环相邻的不能再最终序列上相邻
而这个过程看起来很容斥，所以我们容斥：至少要 x x x 个相邻
我们发现每个置换环的所有边不能全部同时被选，所以我们每个置换环要分开考虑，最后再乘起来
然而这样的复杂度会炸。但是我们可以写成生成函数的形式，然后用启发式合并起来
cpp
#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else
#define debug(...) void(0)
#endif
#define int long long
inline int read(){int x=0,f=1;char ch=getchar(); while(ch<'0'||
ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
//srand(time(0));
#define N 2000010
//#define M
#define mo 998244353
#define G 3
inline int pw(int a, int b) {
int ans=1;
while(b) {
if(b&1) ans*=a;
a*=a; b>>=1;
ans%=mo; a%=mo;
}
return ans;
}
int fac[N], ifac[N];
void init(int n) {
int i;
for(i=fac[0]=1; i<=n; ++i) fac[i]=fac[i-1]*i%mo;
ifac[n]=pw(fac[n], mo-2);
for(i=n-1; i>=0; --i) ifac[i]=ifac[i+1]*(i+1)%mo;
}
int C(int n, int m) {
if(m>n) return 0;
return fac[n]*ifac[m]%mo*ifac[n-m]%mo;
}
const int Gi=pw(G, mo-2);
int Mod(int a) { if(a>=mo || a<=-mo) a%=mo; if(a<0) a+=mo; return a; }
void Add(int &a, int b) { a+=b; Mod(a); }
void Mul(int &a, int b) { Mod(b); a*=b; Mod(a); }
int rev[1<<22];
inline void revese(int n, int l) {
for(int i=0; i<n; ++i) rev[i]=((rev[i>>1]>>1)|((i&1)<<l-1));
}
inline void NTT(int *P, int n, int op) {
int i, j, k, w, W, X, Y;
for(i=0; i<n; ++i) if(i<rev[i]) swap(P[i], P[rev[i]]);
for(i=1; i<n; i<<=1) {
W=pw(op==1 ? G : Gi, (mo-1)/(i<<1)); //*****
for(j=0; j<n; j+=(i<<1)) {
for(k=0, w=1; k<i; ++k, w=Mod(w*W)) {
X=P[j+k], Y=Mod(w*P[j+k+i]);
P[j+k]=Mod(X+Y), P[j+k+i]=Mod(X-Y);
}
}
}
if(op==1) return ;
int inv=pw(n, mo-2);
for(i=0; i<n; ++i) P[i]=P[i]*inv%mo;
}
struct node {
int x, len;
bool operator <(const node &A) const {
return len>A.len;
}
};
priority_queue<node>q;
int u, ve;
int n, m, i, j, k, T;
int f[N], g[N], rt;
int a[N];
int F[N], w[N], ans;
vector<int>v[N];
int fa(int x) {
if(F[x]==x) return x;
return F[x]=fa(F[x]);
}
signed main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
// T=read();
// while(T--) {
//
// }
srand(time(0));
m=read(); init(m);
for(i=1; i<=m; ++i) F[i]=i;
for(i=1; i<=m; ++i) k=read(), F[fa(i)]=fa(k);
for(i=1; i<=m; ++i) w[fa(i)]++;
for(i=1; i<=m; ++i) if(fa(i)==i) {
debug("%lld %lld\n", i, w[i]);
++n;
for(j=0; j<w[i]; ++j) v[n].pb(C(w[i], j)); v[n].pb(0);
q.push({n, w[i]+1});
}
while(!q.empty()) {
u=q.top().x; q.pop();
if(q.empty()) break;
ve=q.top().x; q.pop();
k=++n;
int len1=v[u].size()-1, len2=v[ve].size()-1;
int len=len1+len2+1, m=len, n, le; //new n!
for(n=1, le=0; n<=m; n<<=1) ++le;
revese(n, le);
for(i=0; i<v[u].size() && i<n; ++i) f[i]=v[u][i]; for(; i<n; ++i) f[i]=0;
for(i=0; i<v[ve].size() && i<n; ++i) g[i]=v[ve][i]; for(; i<n; ++i) g[i]=0;
for(i=0; i<n; ++i) debug("%lld ", f[i]); debug("\n");
for(i=0; i<n; ++i) debug("%lld ", g[i]); debug("\n");
NTT(f, n, 1); NTT(g, n, 1);
for(i=0; i<n; ++i) f[i]=Mod(f[i]*g[i]);
NTT(f, n, -1);
for(i=0; i<n; ++i) v[k].pb(f[i]);
q.push({k, n});
}
debug("%lld\n", n);
for(i=0; i<m; ++i) debug("%lld ", v[n][i]); debug("\n");
for(i=0, k=1; i<m; ++i, k=-k) Add(ans, v[n][i]*k%mo*fac[m-i]%mo);
ans=Mod(ans);
printf("%lld\n", ans);
return 0;
}