解题过程:
(答案=sum^n_{i=0}*C^i_n*{frac{1}{m}}^i*{frac{m-1}{m}}^{n-i}*i^k)
根据第二类斯特林数的性质(n^k=sum^k_{i=0}S^i_k*i!*C^i_n=sum^k_{i=0}S^i_k*n^underline{i})将普通幂转为下降幂
(=sum^n_{i=0}C^i_n*{frac{1}{m}}^i*{frac{m-1}{m}}^{n-i}sum^k_{j=0}S^j_k*i^underline{j})
(=sum^k_{j=0}S^j_ksum^n_{i=0}C^i_n{frac{1}{m}}^i{frac{m-1}{m}}^{n-i}i^underline{j})
把(C^i_n)化出来就是(frac{n!}{(n-i)!i!})
所以(frac{n!}{(n-i)!i!}i^underline{j}=frac{n!}{(n-i)!(i-j)!}=frac{(n-j)!n^underline{j}}{(n-i)!(i-j)!}=C^{i-j}_{n-j}n^underline{j})
答案(=sum^k_{j=0}S^j_kn^underline{j}sum^n_{i=j}C^{i-j}_{n-j}*{frac{1}{m}}^i*{frac{m-1}{m}}^{n-i})
通过变换积分上下限有
(=sum^k_{j=0}S^j_kn^underline{j}{frac{1}{m}}^{j}sum^{n-j}_{i=0}C^{i}_{n-j}*{frac{1}{m}}^i*{frac{m-1}{m}}^{n-i-j})
(=sum^k_{j=0}S^j_kn^underline{j}{frac{1}{m}}^{j})
(n^2)预处理第二类斯特林数，其他的可以过程中求
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
// clock_t c1 = clock();
// std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define ls ((x) << 1)
#define rs ((x) << 1 | 1)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<ll, ll> PLL;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 5e5 + 7;
const ll MAXM = 1e5 + 7;
const ll MOD = 998244353;
const double eps = 1e-6;
const double pi = acos(-1.0);
ll quick_pow(ll a, ll b)
{
ll ans = 1;
while (b)
{
if (b & 1)
ans = (1LL * ans * a) % MOD;
a = (1LL * a * a) % MOD;
b >>= 1;
}
return ans;
}
ll s[5005][5005];
void go()
{
s[0][0] = 1;
for (int i = 1; i <= 5000; i++)
for (int j = 1; j <= i; j++)
s[i][j] = (s[i - 1][j - 1] + j * s[i - 1][j]) % MOD;
}
int main()
{
ll n, m, k;
go();
scanf("%lld%lld%lld", &n, &m, &k);
ll inv = quick_pow(m, MOD - 2);
ll ans = 0;
ll up = 1;
for (int i = 1; i <= k; i++)
{
(((up *= (n - i + 1)) %= MOD) *= inv) %= MOD;
(ans += s[k][i] * up) %= MOD;
}
printf("%lld
", ans);
return 0;
}