「联合省选 2020 A」组合数问题题解

非常显然的，我们展开 $f (k)$ ，于是有：

\begin{aligned} (1) & \sum_{k = 0}^{n} \sum_{i = 0}^{m} a_{i} k^{i} x^{k} (\binom{n}{k}) \\ (2) & = & \sum_{k = 0}^{n} \sum_{i = 0}^{m} a_{i} \sum_{j = 0}^{i} {\begin{array}{c} i \\ j \end{array}} (\binom{k}{j}) j! x^{k} (\binom{n}{k}) \\ (3) & = & \sum_{i = 0}^{m} a_{i} \sum_{j = 0}^{i} {\begin{array}{c} i \\ j \end{array}} j! \sum_{k = 0}^{n} x^{k} (\binom{n}{j}) (\binom{n - j}{k - j}) \\ (4) & = & \sum_{i = 0}^{m} a_{i} \sum_{j = 0}^{i} {\begin{array}{c} i \\ j \end{array}} j! (\binom{n}{j}) \sum_{k = 0}^{n - j} x^{k + j} (\binom{n - j}{k}) \\ (5) & = & \sum_{i = 0}^{m} a_{i} \sum_{j = 0}^{i} {\begin{array}{c} i \\ j \end{array}} n^{\underset{―}{j}} (x + 1)^{n - j} x^{j} \\ (6) & = & \sum_{i = 0}^{m} n^{\underset{―}{i}} (x + 1)^{n - i} x^{i} \sum_{j = i}^{m} a_{j} {\begin{array}{c} j \\ i \end{array}} \end{aligned}

然后就直接 $O (m^{2})$ 做出这道题。

1|0code

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using ci = const int;

int n, x, p, m, a[1005];
int S[1005][1005];

int qpow(int x, int y)
{
	int z = 1;
	while (y)
	{
		if (y & 1) z = 1ll * z * x % p;
		x = 1ll * x * x % p, y >>= 1;
	}
	return z;
}

int main()
{
	// freopen(".in", "r", stdin);
	// freopen(".out", "w", stdout);

	cin >> n >> x >> p >> m;
	for (int i = 0; i <= m; ++i) cin >> a[i];
	S[0][0] = 1;
	for (int i = 1; i <= m; ++i)
		for (int j = 1; j <= i; ++j)
			S[i][j] = (S[i - 1][j - 1] + 1ll * S[i - 1][j] * j) % p;
	int ans = 0;
	for (int i = 0; i <= m; ++i)
	{
		ll n1 = 1;
		for (int j = 0; j < i; ++j) (n1 *= n - j) %= p;
		ll n2 = n1 * qpow(x + 1, n - i) % p * qpow(x, i) % p;
		for (int j = i; j <= m; ++j)
			(ans += n2 * a[j] % p * S[j][i] % p) %= p;
	}
	cout << ans;

	return 0;
}

__EOF__

本文作者：TrueFalse
本文链接：https://www.cnblogs.com/cqbzljh/p/17794335.html
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