【BZOJ4555】【TJOI2016】【HEOI2016】求和

题目

传送门

解法

我们可以用容斥来求第二类斯特林数

我们知道, 第二类斯特林数\(S(n, k)\)\(n\)个元素放进\(k\)个无标号的盒子里, 不可以含有空的。 于是我们可以考虑可以含有空的,且盒子有标号, 情况下的数量, 这明显是\(\sum\limits_{j = 0}^{k}{k \choose j}(k-j)^n\)

于是, 根据容斥原理可得:\(S(n, k) = \frac{1}{k!}\sum_{j = 0}^{k}(k-j)^n{k \choose j}(-1)^i\)

于是

\[\begin{aligned} &\sum_{i = 0}^{n}\sum_{j = 0}^{i}S(i, j)\\ &=\sum_{i = 0}^{n}\sum_{j = 0}^{i}S(i, j)\\ &=\sum_{i = 0}^{n}\sum_{j = 0}^{n}2^jj!{\frac{1}{j!}}\sum_{k = 0}^{j}(j-k)^i(-1)^j{j \choose k}\\ &= \sum_{i = 0}^{n}\sum_{j = 0}^{n}2^j\sum_{k = 0}^{j}(j-k)^i(-1)^j\frac{j!}{k!(j-k)!}\\ &= \sum_{i = 0}^{n}\sum_{j = 0}^{n}2^j\sum_{k = 0}^{j}(j-k)^i(-1)^j\frac{j!}{k!(j-k)!}\\ &= \sum_{i = 0}^{n}\sum_{j = 0}^{n}2^jj!\sum_{k = 0}^{j}(j-k)^i(-1)^j\frac{1}{k!(j-k)!}\\ &= \sum_{i = 0}^{n}\sum_{j = 0}^{n}2^jj!\sum_{k = 0}^{j}(-1)^j\frac{(j-k)^i}{k!(j-k)!}\\ &= \sum_{i = 0}^{n}\sum_{j = 0}^{n}2^jj!\sum_{k = 0}^{j}(-1)^j\frac{1}{k!}\frac{(j-k)^i}{(j-k)!} \\ &= \sum_{j = 0}^{n}2^jj!\sum_{k = 0}^{j}(-1)^j\frac{1}{k!}\sum_{i = 0}^{n}\frac{(j-k)^i}{(j-k)!} \end{aligned} \]

于是, 这里出现了一个感人肺腑的卷积

我们设\(a(x) = \frac{1}{x!}(-1)^x\), \(b(x) = \sum_{k = 0}^{n}{k^{x}\over k!}\)

于是答案是\(\sum\limits_{j = 0}^{n}\sum\limits_{k = 0}^{j}a(k)b(j-k)\)

\(b\)可以用等比数列求和公式求出

代码

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstdlib>
#include <algorithm>

using namespace std;

typedef long long LL;

const LL mod = 998244353LL;

const int N = 400010;

inline LL power(LL a, LL n, LL mod)
{	LL Ans = 1;
	a %= mod;
	while (n)
	{	if (n & 1) Ans = (Ans * a) % mod;
		a = (a * a) % mod;
		n >>= 1;
	}
	return Ans;
}

inline LL Plus(LL a, LL b) { return a + b > mod ? a + b - mod : a + b; }

inline LL Minus(LL a, LL b) { return a - b < 0 ? a - b + mod : a - b; }

struct Mul
{	int Len, Bit;

	LL wn[N];

	int rev[N];

	void getReverse()
	{	for (int i = 0; i < Len; i++)
			rev[i] = (rev[i>>1] >> 1) | ((i&1) * (Len >> 1));
	}

	void NTT(LL * a, int opt)
	{	getReverse();
		for (int i = 0; i < Len; i++)
			if (i < rev[i]) swap(a[i], a[rev[i]]);
		int cnt = 0;
		for (int i = 2; i <= Len; i <<= 1)
		{	cnt++;
			for (int j = 0; j < Len; j += i)
			{	LL w = 1LL;
				for (int k = 0; k < (i>>1); k++)
				{	LL x = a[j + k];
					LL y = (w * a[j + k + (i>>1)]) % mod;
					a[j + k] = Plus(x, y);
					a[j + k + (i>>1)] = Minus(x, y);
					w = (w * wn[cnt]) % mod;
				}
			}
		}
		if (opt == -1)
		{	reverse(a + 1, a + Len);
			LL num = power(Len, mod-2, mod);
			for (int i = 0; i < Len; i++)
				a[i] = (a[i] * num) % mod;
		}
	}

	void getLen(int l)
	{	Len = 1, Bit = 0;
		for (; Len <= l; Len <<= 1) Bit++;
	}

	void init()
	{	for (int i = 0; i < 23; i++)
			wn[i] = power(3, (mod-1) / (1LL << i), mod);
	}
} Calc;

LL fac[N], ifac[N];

LL A[N], B[N], C[N];

int main()
{	int n;
	scanf("%d", &n);
	fac[0] = 1;
	for (int i = 1; i <= n; i++)
		fac[i] = fac[i-1] * i % mod;
	ifac[n] = power(fac[n], mod-2, mod);
	for (int i = n-1; i >= 0; i--)
		ifac[i] = ifac[i+1] * (i+1) % mod;
	for (int i = 0; i <= n; i++)
		A[i] = (i & 1 ? Minus(mod, 1) : 1) * ifac[i] % mod;
	B[0] = 1;
	B[1] = n + 1;
	for (int i = 2; i <= n; i++)
		B[i] = (power(i, n+1, mod) + mod - 1) % mod * power(i-1, mod-2, mod) % mod * ifac[i] % mod;
	Calc.init();
	Calc.getLen(n * 2 + 1);
	Calc.NTT(A, 1);
	Calc.NTT(B, 1);
	for (int i = 0; i < Calc.Len; i++)
		C[i] = A[i] * B[i] % mod;
	Calc.NTT(C, -1);
	LL Ans = 0;
	for (int i = 0; i <= n; i++)
		Ans = Plus(Ans, (power(2LL, i, mod) * fac[i] % mod * C[i] % mod));
	printf("%lld\n", Ans);
	return 0;
}
posted @ 2018-08-05 11:16  EZ_WYC  阅读(162)  评论(0编辑  收藏  举报