Codeforces 1278F: Cards

题目传送门:CF1278F

题意简述:

\(n\) 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 \(0\)

\(\displaystyle \Sigma x = \sum_{i=1}^{n} x_i\),求 \(E({(\Sigma x)}^k)\)

题解:

\[\begin{aligned} \mathbf{Ans} &= \sum_{i=0}^{n} \binom{n}{i} p^i (1-p)^{n-i} i^k \\ &= \sum_{i=0}^{n} \binom{n}{i} p^i (1-p)^{n-i} \sum_{j=0}^{k} {k \brace j} i^{\underline{j}} \\ &= \sum_{j=0}^{k} {k \brace j} \sum_{i=0}^{n} \binom{n}{i} p^i (1-p)^{n-i} i^{\underline{j}} \\ &= \sum_{j=0}^{k} {k \brace j} n^{\underline{j}} \sum_{i=0}^{n} \binom{n-j}{i-j} p^i (1-p)^{n-i} \\ &= \sum_{j=0}^{k} {k \brace j} n^{\underline{j}} p^j \sum_{i=0}^{n-j} \binom{n-j}{i} p^i (1-p)^{n-j-i} \\ &= \sum_{j=0}^{k} {k \brace j} n^{\underline{j}} p^j \end{aligned} \]

通常幂转下降幂是常用套路。注意这个恒等式:\(\displaystyle \binom{n}{i} i^{\underline{j}} = \binom{n-j}{i-j} n^{\underline{j}}\)

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\n", Ans);
	return 0;
}
posted @ 2019-12-20 15:33  粉兔  阅读(...)  评论(...编辑  收藏