是个水题？
重点是得发现一个结论：若排列 \(a\) 不合法，则一定存在 \(1 \le i < n\)，满足 \(a_{1\ldots i}\) 是 \(1\ldots i\) 的一个排列。
则此时 \(i+1\) 是第一个使得序列不完美的位置。
设 \(f_n\) 表示 \(n\) 的答案，则取补集得 \[ f_n = n! - \sum\limits_{i=1}^{n-1} f_i (n-i)! \]
移项并令 \(f_0 = 0\) 得 \[ \sum\limits_{i=0}^n f_i(n-i)! + [n=0] = n! \]
设 \(F(x) = \sum\limits_{i=0}^{\infty} f_ix^i,G(x) = \sum\limits_{i=0}^{\infty} i!x^i\)，则根据上式有 \[ \begin{align*} F(x)G(x)+1 &= G(x) \\ (1-F(x))G(x) &= 1 \\ F(x) &= 1 - \frac1{G(x)} \end{align*} \]
多项式求逆即可。
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using namespace std;
const int N = 1 << 18;
const int mod = 998244353;
const int G = 3;
inline int fpow(int a,int b)
{
int ret = 1;
for(;b;b >>= 1)
(b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
return ret;
}
struct poly
{
int a[N + 5];
inline const int &operator[](int x) const
{
return a[x];
}
inline int &operator[](int x)
{
return a[x];
}
inline void clear(int x = 0)
{
memset(a + x,0,(N - x + 1) << 2);
}
} f;
int T,len,k,n,lg2[N + 5];
int cnt[N + 5];
int rev[N + 5],fac[N + 5],ifac[N + 5],inv[N + 5];
int rt[N + 5],irt[N + 5];
inline void init(int len)
{
for(n = 1;n < len;n <<= 1);
for(register int i = 2;i <= n;++i)
lg2[i] = lg2[i >> 1] + 1;
int w = fpow(G,(mod - 1) / n);
rt[n >> 1] = 1;
for(register int i = (n >> 1) + 1;i <= n;++i)
rt[i] = (long long)rt[i - 1] * w % mod;
for(register int i = (n >> 1) - 1;i;--i)
rt[i] = rt[i << 1];
fac[0] = 1;
for(register int i = 1;i <= n;++i)
fac[i] = (long long)fac[i - 1] * i % mod;
ifac[n] = fpow(fac[n],mod - 2);
for(register int i = n;i;--i)
ifac[i - 1] = (long long)ifac[i] * i % mod;
for(register int i = 1;i <= n;++i)
inv[i] = (long long)ifac[i] * fac[i - 1] % mod;
}
inline void ntt(poly &a,int type,int n)
{
type == -1 && (reverse(a.a + 1,a.a + n),1);
int lg = lg2[n] - 1;
for(register int i = 0;i < n;++i)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg),
i < rev[i] && (swap(a[i],a[rev[i]]),1);
for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
for(register int i = 0;i < n;i += w)
for(register int j = 0;j < m;++j)
{
int t = (long long)rt[m | j] * a[i | j | m] % mod;
a[i | j | m] = dec(a[i | j],t),a[i | j] = add(a[i | j],t);
}
if(type == -1)
for(register int i = 0;i < n;++i)
a[i] = (long long)a[i] * inv[n] % mod;
}
inline void mul(poly &a,const poly &b,int n)
{
static poly x,y;
int lim = 1;
x.clear(),y.clear();
for(;lim < (n << 1);lim <<= 1);
x = a,y = b;
x.clear(n),y.clear(n);
ntt(x,1,lim),ntt(y,1,lim);
for(register int i = 0;i < lim;++i)
x[i] = (long long)x[i] * y[i] % mod;
ntt(x,-1,lim);
x.clear(n),a = x;
}
inline poly inverse(const poly &f,int n)
{
static int s[30];
static poly g,h,q;
int lim = 1,top = 0;
g.clear();
for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
g[0] = fpow(f[0],mod - 2);
for(;top;--top)
{
n = s[top];
for(;lim < (n << 1);lim <<= 1);
q = g,h = f,h.clear(n);
ntt(g,1,lim),ntt(h,1,lim);
for(register int i = 0;i < lim;++i)
g[i] = (long long)g[i] * g[i] % mod * h[i] % mod;
ntt(g,-1,lim);
for(register int i = 0;i < n;++i)
g[i] = dec(add(q[i],q[i]),g[i]);
g.clear(n);
}
return g;
}
inline void derivative(poly &f,int n)
{
for(register int i = 1;i < n;++i)
f[i - 1] = (long long)f[i] * i % mod;
f[n - 1] = 0;
}
inline void integral(poly &f,int n)
{
for(register int i = n - 1;~i;--i)
f[i + 1] = (long long)f[i] * inv[i + 1] % mod;
f[0] = 0;
}
inline poly ln(const poly &f,int n)
{
static poly g;
g = f,derivative(g,n),mul(g,inverse(f,n),n),integral(g,n);
return g;
}
inline poly exp(const poly &f,int n)
{
static int s[30];
static poly g,h;
int lim = 1,top = 0;
g.clear();
for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
g[0] = 1;
for(;top;--top)
{
n = s[top];
for(;lim < (n << 1);lim <<= 1);
h = g,g = ln(g,n);
for(register int i = 0;i < n;++i)
g[i] = dec(f[i],g[i]);
g[0] = add(g[0],1);
ntt(g,1,lim),ntt(h,1,lim);
for(register int i = 0;i < lim;++i)
g[i] = (long long)g[i] * h[i] % mod;
ntt(g,-1,lim);
g.clear(n);
}
return g;
}
inline poly power(const poly &f,int k,int n)
{
static poly g;
g = ln(f,n);
for(register int i = 0;i < n;++i)
g[i] = (long long)g[i] * k % mod;
g = exp(g,n);
return g;
}
namespace Mod_sqrt
{
typedef pair<int,int> cp;
int w;
inline cp operator*(const cp &a,const cp &b)
{
return cp(((long long)a.first * b.first % mod + (long long)a.second * b.second % mod * w % mod) % mod,((long long)a.first * b.second % mod + (long long)a.second * b.first % mod) % mod);
}
inline cp pow(cp a,int b)
{
cp ret(1,0);
for(;b;b >>= 1)
(b & 1) && (ret = ret * a,1),a = a * a;
return ret;
}
inline int mod_sqrt(int x)
{
int y = rand() % mod;
for(;fpow(w = ((long long)y * y % mod - x + mod) % mod,mod - 1 >> 1) <= 1;y = rand() % mod);
cp ret = pow(cp(y,1),mod + 1 >> 1);
return min(ret.first,mod - ret.first);
}
}
using Mod_sqrt::mod_sqrt;
inline poly sqrt(const poly &f,int n)
{
static int s[30];
static poly g,h;
int top = 0;
g.clear();
for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
g[0] = mod_sqrt(f[0]);
for(;top;--top)
{
n = s[top];
for(register int i = 0;i < n;++i)
h[i] = add(g[i],g[i]);
h = inverse(h,n),mul(g,g,n);
for(register int i = 0;i < n;++i)
g[i] = add(g[i],f[i]);
mul(g,h,n);
}
return g;
}
int main()
{
init((len = 1e5 + 1) << 1);
for(register int i = 0;i < len;++i)
f[i] = fac[i];
f = inverse(f,len);
for(register int i = 0;i < len;++i)
f[i] = dec(0,f[i]);
f[0] = add(f[0],1),scanf("%d",&T);
for(int x;T;--T)
scanf("%d",&x),printf("%d\n",f[x]);
}