解题思路

由于本蒟蒻不太会用 $L^A{T}_{E}X$ ，本题解题思路改用手写。

代码实现

不开 long long 见祖宗，十年 OI 一场空。

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 5005, P = 998244353;
int a[N], dp[N][N];
int qpow(int x, int y) {
	int res = 1;
	while (y) {
		if (y & 1) {
			(res *= x) %= P;
		}
		(x *= x) %= P;
		y >>= 1;
	}
	return res;
}
signed main() {
	ios :: sync_with_stdio(0);
	int n, m, k;
	cin >> n >> m >> k;
	a[0] = 1;
	for (int i = 1; i <= k; ++i) {
		a[i] = a[i - 1] * qpow(i, P - 2) % P;
	}
	dp[1][1] = 1;
	for (int i = 2; i <= k; ++i) {
		for (int j = 1; j <= i; ++j) {
			dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1]) % P * j % P;
		}
	}
	int res = 0, mul = qpow(m, P - 2), base = mul, C = n;
	for (int i = 1; i <= k; ++i) {
		(res += base * a[i] % P * dp[k][i] % P * C % P) %= P;
		(base *= mul) %= P;
		(C *= n - i) %= P;
	}
	cout << res << '\n';
	return 0;
}