首先 \(m\) 中奇数个数是 \(\left\lceil\frac{m}{2}\right\rceil\)，偶数个数是 \(\left\lfloor\frac{m}{2}\right\rfloor\)。下文为了方便记 \(p=\left\lceil\frac{m}{2}\right\rceil,q=\left\lfloor\frac{m}{2}\right\rfloor\)。
考虑先写出暴力的式子，枚举取奇数的盒子个数：
\[\sum_{i=1}^{n}\binom{n}{i}p^iq^{n-i}i^k \]依旧把 \(i^k\) 展开：
\[=\sum_{i=1}^{n}\binom{n}{i}p^iq^{n-i}\sum_{j=1}^{k}\begin{Bmatrix}k\\ j\end{Bmatrix}j!\binom{i}{j} \]\[=\sum_{j=1}^{k}\begin{Bmatrix}k\\ j\end{Bmatrix}j!\sum_{i=1}^{n}\binom{n}{i}\binom{i}{j}p^iq^{n-i} \]\[=\sum_{j=1}^{k}\begin{Bmatrix}k\\ j\end{Bmatrix}j!\binom{n}{j}\sum_{i=j}^{n}\binom{n-j}{i-j}p^iq^{n-i} \]\[=\sum_{j=1}^{k}\begin{Bmatrix}k\\ j\end{Bmatrix}n^{\underline{j}}\sum_{i=0}^{n-j}\binom{n-j}{i}p^{i+j}q^{n-j-i} \]\[=\sum_{j=1}^{k}\begin{Bmatrix}k\\ j\end{Bmatrix}n^{\underline{j}}p^j\sum_{i=0}^{n-j}\binom{n-j}{i}p^{i}q^{n-j-i} \]\[=\sum_{j=1}^{k}\begin{Bmatrix}k\\ j\end{Bmatrix}n^{\underline{j}}p^j(p+q)^{n-j} \]\[=\sum_{j=1}^{k}\begin{Bmatrix}k\\ j\end{Bmatrix}n^{\underline{j}}p^jm^{n-j} \]然后 \(\mathcal O(k^2)\) 预处理斯特林数啥的就好了。
Code：
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2005, mod = 998244353;
int T;
int n, m, k;
int S[N][N], f[N];
int qpow(int x, int y) {
int res = 1;
while (y) {
if (y & 1) res = 1ll * res * x % mod;
x = 1ll * x * x % mod;
y >>= 1;
}
return res;
}
void init(int n) {
S[0][0] = 1;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= i; ++j)
S[i][j] = (S[i - 1][j - 1] + 1ll * S[i - 1][j] * j % mod) % mod;
}
void solve() {
scanf("%d%d%d", &n, &m, &k);
f[0] = 1;
for (int i = 1; i <= min(n, k); ++i) f[i] = 1ll * f[i - 1] * (n - i + 1) % mod;
int cnt = (m + 1) / 2;
int ans = 0;
int x = cnt, y = qpow(m, n - 1), invm = qpow(m, mod - 2);
for (int i = 1; i <= min(n, k); ++i) {
int tmp = 1ll * S[k][i] * f[i] % mod * x % mod * y % mod;
ans = (ans + tmp) % mod, x = 1ll * x * cnt % mod, y = 1ll * y * invm % mod;
}
printf("%d\n", ans);
}
int main() {
init(2000);
scanf("%d", &T);
while (T--) solve();
return 0;
}