题目传送门:CF1278F。
题意简述:
有 (n) 个独立随机变量 (x_i),每个随机变量都有 (p = 1/m) 的概率取 (1),有 ((1-p)) 的概率取 (0)。
令 (displaystyle Sigma x = sum_{i=1}^{n} x_i),求 (E({(Sigma x)}^k))。
题解:
[egin{aligned} mathbf{Ans} &= sum_{i=0}^{n} inom{n}{i} p^i (1-p)^{n-i} i^k \ &= sum_{i=0}^{n} inom{n}{i} p^i (1-p)^{n-i} sum_{j=0}^{k} {k race j} i^{underline{j}} \ &= sum_{j=0}^{k} {k race j} sum_{i=0}^{n} inom{n}{i} p^i (1-p)^{n-i} i^{underline{j}} \ &= sum_{j=0}^{k} {k race j} n^{underline{j}} sum_{i=0}^{n} inom{n-j}{i-j} p^i (1-p)^{n-i} \ &= sum_{j=0}^{k} {k race j} n^{underline{j}} p^j sum_{i=0}^{n-j} inom{n-j}{i} p^i (1-p)^{n-j-i} \ &= sum_{j=0}^{k} {k race j} n^{underline{j}} p^j end{aligned}
]
通常幂转下降幂是常用套路。注意这个恒等式:(displaystyle inom{n}{i} i^{underline{j}} = inom{n-j}{i-j} n^{underline{j}})。
下面是代码,时间复杂度为 (mathcal O (k^2)):
#include <cstdio>
typedef long long LL;
const int Mod = 998244353;
const int MK = 5005;
inline int qPow(int b, int e) {
int a = 1;
for (; e; e >>= 1, b = (LL)b * b % Mod)
if (e & 1) a = (LL)a * b % Mod;
return a;
}
int N, M, K;
int S[MK][MK], Ans;
int main() {
scanf("%d%d%d", &N, &M, &K);
M = qPow(M, Mod - 2);
S[0][0] = 1;
for (int i = 1; i <= K; ++i)
for (int j = 1; j <= i; ++j)
S[i][j] = (S[i - 1][j - 1] + (LL)j * S[i - 1][j]) % Mod;
for (int i = 1, C = 1; i <= K; ++i)
C = (LL)C * (N - i + 1) % Mod * M % Mod,
Ans = (Ans + (LL)S[K][i] * C) % Mod;
printf("%d
", Ans);
return 0;
}