Description：
\(n<=10,max(w)<=1e6\)
题解：
考虑暴力，相当于走多维格子图，不能走有些点。
套路就是设\(f[i]\)表示第一次走到i的方案数
\(f[i]=起点到点i的方案数-\sum_{j在i前面}f[j]*j到i的方案数\)
不妨把前缀和后缀的分开，设为f和g。
f上的点形如(i,i,…)
设\(m=min(w)\),\(w-=m\)
则g上的点形如(w+i,……)
这样就顺序了。
且\(i->j的方案数\)之和坐标差有关，那就可以分治NTT了。
还可以优化，设\(f->f的转移多项式是A,f->g是B，g->f是C\)
则有
\(F=A-F*A-G*C\)
\(G=B-G*A-F*B\)
最后答案就是\(G[m]\)
消元可得：
\(G={B \over (1+A)^2-BC}\)
所以求逆即可。
Code：
#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, B = y; i <= B; i ++)
#define ff(i, x, y) for(int i = x, B = y; i < B; i ++)
#define fd(i, x, y) for(int i = x, B = y; i >= B; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
using namespace std;
const int mo = 998244353;
ll ksm(ll x, ll y) {
ll s = 1;
for(; y; y /= 2, x = x * x % mo)
if(y & 1) s = s * x % mo;
return s;
}
typedef vector<ll> V;
#define pb push_back
#define si size()
const int nm = 1 << 21;
namespace ntt {
ll w[nm], a[nm], b[nm]; int r[nm];
void build() {
for(int i = 1; i < nm; i *= 2) {
ll v = ksm(3, (mo - 1) / 2 / i);
w[i] = 1; ff(j, 1, i) w[i + j] = w[i + j - 1] * v % mo;
}
}
void dft(ll *a, int n, int f) {
ff(i, 0, n) {
r[i] = r[i / 2] / 2 + (i & 1) * (n / 2);
if(i < r[i]) swap(a[i], a[r[i]]);
} ll b;
for(int i = 1; i < n; i *= 2) for(int j = 0; j < n; j += 2 * i)
ff(k, 0, i) b = a[i + j + k] * w[i + k], a[i + j + k] = (a[j + k] - b) % mo, a[j + k] = (a[j + k] + b) % mo;
if(f == -1) {
reverse(a + 1, a + n);
b = ksm(n, mo - 2);
ff(i, 0, n) a[i] = (a[i] + mo) * b % mo;
}
}
void fft(V &p, V &q) {
int p0 = p.si + q.si - 1, n = 1;
for(; n < p0; n *= 2);
ff(i, 0, n) a[i] = b[i] = 0;
ff(i, 0, p.si) a[i] = p[i];
ff(i, 0, q.si) b[i] = q[i];
dft(a, n, 1); dft(b, n, 1);
ff(i, 0, n) a[i] = a[i] * b[i] % mo;
dft(a, n, -1);
p.resize(p0);
ff(i, 0, p0) p[i] = a[i];
}
}
V operator * (V p, V q) {
ntt :: fft(p, q);
return p;
}
void dft(V &p, int f) {
ff(i, 0, p.si) ntt :: a[i] = p[i];
ntt :: dft(ntt :: a, p.si, f);
ff(i, 0, p.si) p[i] = ntt :: a[i];
}
V qni(V a) {
int a0 = a.si, n0 = 1;
while(n0 < a0) n0 *= 2;
V b; b.resize(1); b[0] = ksm(a[0], mo - 2);
for(int n = 2; n <= n0; n *= 2) {
V d = b; d.resize(n); b.resize(2 * n);
V c = a; c.resize(n); c.resize(2 * n);
dft(c, 1); dft(b, 1);
ff(i, 0, b.si) b[i] = c[i] * b[i] % mo * b[i] % mo;
dft(b, -1); b.resize(n);
ff(i, 0, b.si) b[i] = (2 * d[i] - b[i] + mo) % mo;
}
b.resize(a0);
return b;
}
const int N = 1e7 + 5;
ll fac[N], nf[N];
V a, b, c;
void build(int n) {
fac[0] = 1; fo(i, 1, n) fac[i] = fac[i - 1] * i % mo;
nf[n] = ksm(fac[n], mo - 2); fd(i, n, 1) nf[i - 1] = nf[i] * i % mo;
}
int n, m, w[11];
void build2() {
a.resize(m + 1); b.resize(m + 1); c.resize(m + 1);
fo(i, 1, m) a[i] = fac[n * i] * ksm(nf[i], n) % mo;
fo(i, 0, m) {
int s = 0; b[i] = 1;
fo(j, 1, n) s += w[j] + i, b[i] = b[i] * nf[w[j] + i] % mo;
b[i] = b[i] * fac[s] % mo;
}
fo(i, 0, m) {
int s = 0; c[i] = 1;
fo(j, 1, n) {
if(-w[j] + i < 0) c[i] = 0; else
s += -w[j] + i, c[i] = c[i] * nf[-w[j] + i] % mo;
}
c[i] = c[i] * fac[s] % mo;
}
}
ll f[N], g[N];
int main() {
freopen("queue.in", "r", stdin);
freopen("queue.out", "w", stdout);
ntt :: build();
scanf("%d", &n); m = 1e6;
fo(i, 1, n) scanf("%d", &w[i]), m = min(m, w[i]);
fo(i, 1, n) w[i] -= m;
{
int ye = 1;
fo(i, 1, n) if(w[i]) ye = 0;
if(ye) {
pp("0\n"); return 0;
}
}
build(1e7);
build2();
a[0] ++; a = a * a; a.resize(m + 1);
c = c * b; c.resize(m + 1);
ff(i, 0, a.si) a[i] = (a[i] - c[i] + mo) % mo;
a = qni(a); b = b * a;
pp("%lld\n", b[m]);
}