【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))是(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(sumlimits_{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

[egin{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^jsum_{k = 0}^{j}(j-k)^i(-1)^jfrac{j!}{k!(j-k)!}\ &= sum_{i = 0}^{n}sum_{j = 0}^{n}2^jsum_{k = 0}^{j}(j-k)^i(-1)^jfrac{j!}{k!(j-k)!}\ &= sum_{i = 0}^{n}sum_{j = 0}^{n}2^jj!sum_{k = 0}^{j}(j-k)^i(-1)^jfrac{1}{k!(j-k)!}\ &= sum_{i = 0}^{n}sum_{j = 0}^{n}2^jj!sum_{k = 0}^{j}(-1)^jfrac{(j-k)^i}{k!(j-k)!}\ &= sum_{i = 0}^{n}sum_{j = 0}^{n}2^jj!sum_{k = 0}^{j}(-1)^jfrac{1}{k!}frac{(j-k)^i}{(j-k)!} \ &= sum_{j = 0}^{n}2^jj!sum_{k = 0}^{j}(-1)^jfrac{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(sumlimits_{j = 0}^{n}sumlimits_{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
", Ans);
	return 0;
}