首先，这种形式肯定是把组合数消掉一点，然后后面再二项式定理处理一下。但是怎么搞呢？
开始尝试了拉格朗日插值，但是有一项非常毒瘤。（我甚至少抄一项推出了 (O(n)) 的式子……）
要消掉组合数一定是与阶乘有关的形式。连续点值启发着我们使用下降幂。众所周知，点值转下降幂系数只需要卷上一个 (e^{-x}) 即可。
这时候考虑只用式子里代入下降幂：
[egin{align*}
& sum_{k=0}^n k^{underline{m}} inom{n}{k} x^k left( 1 - x
ight) ^ {n - k} \
=& sum_{k=0}^n frac{k!}{left(k-m
ight)!} frac{n!}{k!left(n-k
ight)!} x^k left( 1 - x
ight) ^ {n - k} \
=& sum_{k=0}^n frac{n!}{left(n-m
ight)!} frac{left(n - m
ight)!}{left(k-m
ight)!left(n-k
ight)!} x^k left( 1 - x
ight) ^ {n - k} \
=& n^{underline{m}} x^m sum_{k=0}^n inom{n - m}{n-k} x^{k-m} left( 1 - x
ight) ^ {n - k} \
=& n^{underline{m}} x^m sum_{k=0}^n left(1 - x + x
ight) ^ {n - m} \
=& n^{underline{m}} x^m
end{align*}
]
容易计算。所以瓶颈就在转下降幂。如果实现好的话 (O(n^2)) 也是可以的。
#include <bits/stdc++.h>
const int mod = 998244353;
const int MAXN = 20010;
typedef long long LL;
void reduce(int & x) { x += x >> 31 & mod; }
int mul(int a, int b) { return (LL) a * b % mod; }
int fastpow(int a, int b) {
int res = 1; a %= mod;
for (; b; b >>= 1, a = mul(a, a)) if (b & 1) res = mul(res, a);
return res;
}
int A[MAXN], inv[MAXN], fac[MAXN], B[MAXN];
int n, m, X;
int inv2[MAXN];
int main() {
inv[0] = inv[1] = fac[0] = fac[1] = 1;
for (int i = 2; i != MAXN; ++i) {
fac[i] = mul(fac[i - 1], i);
inv[i] = mul(inv[mod % i], mod - mod / i);
}
inv2[0] = 1, inv2[1] = mod - 1;
for (int i = 2; i != MAXN; ++i) {
inv[i] = mul(inv[i - 1], inv[i]);
reduce(inv2[i] = i & 1 ? mod - inv[i] : inv[i]);
}
std::cin >> n >> m >> X; ++m;
for (int i = 0; i < m; ++i) std::cin >> A[i], A[i] = mul(A[i], inv[i]);
for (int i = 0, t; i < m; ++i)
for (int j = 0; i + j < m; ++j)
reduce(B[i + j] += (LL) A[i] * inv2[j] % mod - mod);
for (int i = 0; i < m; ++i) A[i] = mul(A[i], fac[i]);
int now = 1, ans = 0;
for (int i = 0; i < m; ++i) {
reduce(ans += (LL) fastpow(X, i) * now % mod * B[i] % mod - mod);
now = mul(now, n - i);
}
std::cout << ans << std::endl;
return 0;
}