近日膜你赛做到了和这个核心思路类似的题……
于是就回来写了（？
考虑把图按最短路分层，则能连的边只有相邻两层的和同一层内的。
对于前一种边，对于第 \(i\) 层的任意一个点，往 \(i-1\) 层连边的生成函数为 \((1+x)^{t_i}-1\)。
第 \(i\) 层所有点往上一层连边的生成函数即 \([(1+x)^{t_i}-1]^{ t_{i-1} }\)。
分治 NTT / 手动 ln exp 胡乱分析一下可以做到 \(O(n \log^2 n)\) / \(O(n \log n)\)（视 \(n,m,\sum t_i t_{i-1}\) 同阶）。
考虑后一种边，则设可能的这样的边的总数为 \(s\)，有 \[ s = \sum\limits_{i=1}^{n-1} \binom{t_i}2 \]
则这部分边的生成函数是 \[ \sum\limits_{i\ge 0} \binom si x^i = (1+x)^s \]
直接多项式快速幂即可。
不过这样会慢一点，也可以像官方题解那样观察到在 \(s \bmod p < i < p\) 时有 \(\binom si \equiv 0 \pmod p\)，所以直接按照套路维护 \(s^{\underline i}\) 即可。
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using namespace std;
const int N = 2e5;
const int mod = 998244353;
int n,m;
int c[N + 5],s;
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;
}
namespace Poly
{
const int N = 1 << 19;
const int G = 3;
int lg2[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()
{
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;
}
struct poly
{
vector<int> a;
inline poly(int x = 0)
{
x && (a.push_back(x),1);
}
inline poly(const vector<int> &o)
{
a = o,shrink();
}
inline poly(const poly &o)
{
a = o.a,shrink();
}
inline void shrink()
{
for(;!a.empty() && !a.back();a.pop_back());
}
inline int size() const
{
return a.size();
}
inline void resize(int x)
{
a.resize(x);
}
inline int operator[](int x) const
{
if(x < 0 || x >= size())
return 0;
return a[x];
}
inline void clear()
{
vector<int>().swap(a);
}
inline poly rever() const
{
return poly(vector<int>(a.rbegin(),a.rend()));
}
inline void ntt(int type = 1)
{
int n = size();
type == -1 && (reverse(a.begin() + 1,a.end()),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;
}
friend inline poly operator+(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = add(a[i],b[i]);
return poly(ret);
}
friend inline poly operator-(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = dec(a[i],b[i]);
return poly(ret);
}
friend inline poly operator*(poly a,poly b)
{
if(a.a.empty() || b.a.empty())
return poly();
int lim = 1,tot = a.size() + b.size() - 1;
for(;lim < tot;lim <<= 1);
a.resize(lim),b.resize(lim);
a.ntt(),b.ntt();
for(register int i = 0;i < lim;++i)
a.a[i] = (long long)a[i] * b[i] % mod;
a.ntt(-1),a.shrink();
return a;
}
poly &operator+=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = add(a[i],o[i]);
return *this;
}
poly &operator-=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = dec(a[i],o[i]);
return *this;
}
poly &operator*=(poly o)
{
return (*this) = (*this) * o;
}
poly deriv() const
{
if(a.empty())
return poly();
vector<int> ret(size() - 1);
for(register int i = 0;i < size() - 1;++i)
ret[i] = (long long)(i + 1) * a[i + 1] % mod;
return poly(ret);
}
poly integ() const
{
if(a.empty())
return poly();
vector<int> ret(size() + 1);
for(register int i = 0;i < size();++i)
ret[i + 1] = (long long)a[i] * inv[i + 1] % mod;
return poly(ret);
}
inline poly modxn(int n) const
{
if(a.empty())
return poly();
n = min(n,size());
return poly(vector<int>(a.begin(),a.begin() + n));
}
inline poly inver(int m) const
{
poly ret(fpow(a[0],mod - 2));
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (2 - modxn(k) * ret)).modxn(k);
return ret.modxn(m);
}
inline pair<poly,poly> div(poly o) const
{
if(size() < o.size())
return make_pair(poly(),*this);
poly f,g;
f = (rever().modxn(size() - o.size() + 1) * o.rever().inver(size() - o.size() + 1)).modxn(size() - o.size() + 1).rever();
g = (modxn(o.size() - 1) - o.modxn(o.size() - 1) * f.modxn(o.size() - 1)).modxn(o.size() - 1);
return make_pair(f,g);
}
inline poly log(int m) const
{
return (deriv() * inver(m)).integ().modxn(m);
}
inline poly exp(int m) const
{
poly ret(1);
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (1 - ret.log(k) + modxn(k))).modxn(k);
return ret.modxn(m);
}
inline poly pow(int m,int k1,int k2 = -1) const
{
if(a.empty())
return poly();
if(k2 == -1)
k2 = k1;
int t = 0;
for(;t < size() && !a[t];++t);
if((long long)t * k1 >= m)
return poly();
poly ret;
ret.resize(m);
int u = fpow(a[t],mod - 2),v = fpow(a[t],k2);
for(register int i = 0;i < m - t * k1;++i)
ret.a[i] = (long long)operator[](i + t) * u % mod;
ret = ret.log(m - t * k1);
for(register int i = 0;i < ret.size();++i)
ret.a[i] = (long long)ret[i] * k1 % mod;
ret = ret.exp(m - t * k1),t *= k1,ret.resize(m);
for(register int i = m - 1;i >= t;--i)
ret.a[i] = (long long)ret[i - t] * v % mod;
for(register int i = 0;i < t;++i)
ret.a[i] = 0;
return ret;
}
};
}
using Poly::init;
using Poly::poly;
inline int C(int n,int m)
{
return n < m ? 0 : (long long)Poly::fac[n] * Poly::ifac[m] % mod * Poly::ifac[n - m] % mod;
}
poly f[N + 5],g,ans;
inline poly solve(int l,int r)
{
if(l == r)
return f[l];
int mid = l + r >> 1;
return solve(l,mid) * solve(mid + 1,r);
}
int main()
{
init();
scanf("%d%d",&n,&m);
int d;
for(register int i = 1;i <= n;++i)
scanf("%d",&d),++c[d];
for(register int i = 1;i < n;++i)
{
f[i].resize(c[i - 1] + 1);
for(register int j = 1;j <= c[i - 1];++j)
f[i].a[j] = C(c[i - 1],j);
f[i] = f[i].pow(c[i] * c[i - 1] + 1,c[i]);
}
ans = solve(1,n - 1).modxn(m + 1);
for(register int i = 1;i < n;++i)
s = (s + (long long)c[i] * (c[i] - 1) / 2) % mod;
g.resize(min(m,s) + 1);
for(register int i = 0,p = 1;i <= min(m,s);++i,p = (long long)p * (s - i + 1) % mod)
g.a[i] = (long long)p * Poly::ifac[i] % mod;
ans = (ans * g).modxn(m + 1);
printf("%d\n",ans[m]);
}