Comet 67E: ffort

题目传送门:Comet 67E

用了个傻逼做法 A 了这题,欢迎观赏睿智做法!

题意简述:

题目说得很清楚了(这次是我不想写了)。

题解:

为了方便,令 \(m\) 为敌人数,\(n\) 为己方士兵种数。设 \(\mathbf{Ans}\) 为答案,则有:

\[\begin{aligned}\mathbf{Ans}&=\sum_{a=1}^{\infty}([x^a]G)\cdot\binom{a-1}{m-1}\\&=\sum_{a=0}^{\infty}\!\left([x^{a}]\dfrac{G}{x}\right)\!\cdot\binom{a}{\hat m}\\&=\frac{1}{\hat m!}\sum_{i=0}^{\infty}([x^i]F)i^{\underline{\hat m}}\\\end{aligned} \]

意即枚举 \(a\) 为打出的伤害数,则将这些伤害分配给敌人的方法数为 \(\dbinom{a-1}{m-1}\)

其中 \(G\) 为给己方士兵分配伤害的方案数的 \(\mathbf{OGF}\),即 \(\displaystyle G=\prod_{i=1}^{n}\left[\mathbf{OGF}\left\{0,\underset{b_i}{\underbrace{1,\ldots,1}}\right\}\right]^{a_i}\)

\([x^a]G\) 就为将伤害分配给己方士兵的方案数。

接下来令 \(\hat m=m-1\)\(F=\dfrac{G}{x}\),变换求和指标并提出 \(\dfrac{1}{\hat m!}\),留下下降幂形式。


下降幂经常出现在多次求导后的多项式中,即 \(\displaystyle f^{(k)}(x)=\sum_{i=k}^{\infty}f_ii^{\underline{k}}\cdot x^{i-k}\)

那么 \(\displaystyle\sum_{i=0}^{\infty}f_ii^{\underline{\hat m}}=f^{(\hat m)}(1)\)

据此,则有:

\[\mathbf{Ans}=\frac{1}{\hat m!}F^{(m)}(1) \]

考虑 \(\displaystyle F=\frac{1}{x}\prod_{i=1}^{n}\left[\mathbf{OGF}\left\{0,\underset{b_i}{\underbrace{1,\ldots,1}}\right\}\right]^{a_i}\)\(\hat m\) 阶导:

  • 对于 \(\displaystyle t=\prod_{i=1}^{n}t_i\)\(k\) 阶导,重复使用乘法法则 \((fg)'=f'g+fg'\) 即可得出:
  • \(\displaystyle t^{(k)}=\sum_{a_1+a_2+\cdots+a_n=k}\binom{k}{a_{1\ldots n}}\prod_{i=1}^{n}t_i^{(a_i)}\),其中 \(\dbinom{k}{a_{1\ldots n}}\) 即为多重组合数。
  • 从生成函数的角度看来,即 \(\displaystyle t^{(k)}=k![z^k]\prod_{i=1}^{n}\sum_{j=0}^{\infty}\frac{t_i^{(j)}}{j!}z^j\),即每个 \(\{t_i^{(0)},t_i^{(1)},\ldots\}\) 的二项卷积。

\(f_i=\mathbf{OGF}\left\{0,\underset{b_i}{\underbrace{1,\ldots,1}}\right\}\) ,特别地 \(f_0=\dfrac{1}{x}\)\(a_0=1\),那么有 \(\displaystyle F=\prod_{i=0}^{n}f_i^{a_i}\)

于是:

\[\begin{aligned}\mathbf{Ans}&=\frac{1}{\hat m}F^{(m)}(1)\\&=\frac{1}{\hat m}\left(\hat m!\left[z^{\hat m}\right]\prod_{i=0}^{n}\left(\sum_{j=0}^{\infty}\frac{f_i^{(j)}}{j!}z^{j}\right)^{a_i}\right)(1)\\&=[z^m]\prod_{i=0}^{n}\left(\sum_{j=0}^{\infty}\frac{f_i^{(j)}(1)}{j!}z^j\right)^{a_i}\end{aligned} \]

因为 \(n\times m\le 10^5\),所以后面只要算出 \(\dfrac{f_i^{(j)}(1)}{j!}\)\(0\le i\le n\)\(0\le j\le m\)),然后多项式快速幂暴力乘即可。


接下来考虑如何计算 \(\dfrac{f_i^{(j)}(1)}{j!}\)

对于 \(f_0=\dfrac{1}{x}\),有 \(\left(\dfrac{1}{x}\right)^{(k)}(1)=(-1)^kk!\),所以 \(\dfrac{f_0^{(j)}(1)}{j!}=(-1)^j\)

对于 \(f_i=\mathbf{OGF}\left\{0,\underset{b_i}{\underbrace{1,\ldots,1}}\right\}\),稍加推导可以得到:

  • \(f_i^{(0)}(1)\),即 \(f_i(1)\),等于 \(b_i\)(显然)。
  • 对于 \(j\ge1\),有 \(f_i^{(j)}(1)=\dfrac{(b_i+1)^{\underline{j+1}}}{j+1}\)(证明略),于是 \(\dfrac{f_i^{(j)}(1)}{j!}=\dfrac{(b_i+1)^{\underline{j+1}}}{(j+1)!}\) 可以递推求出。

那么这题就做完了,代码如下,复杂度 \(\mathcal O(nm\log m)\)

#include <cstdio>
#include <algorithm>

typedef long long LL;
const int Mod = 998244353;
const int G = 3, iG = 332748118;
const int MS = 1 << 19;

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;
}

inline int gInv(int b) { return qPow(b, Mod - 2); }

int Inv[MS], Fac[MS], iFac[MS];

inline void Init(int N) {
	Fac[0] = 1;
	for (int i = 1; i < N; ++i) Fac[i] = (LL)Fac[i - 1] * i % Mod;
	iFac[N - 1] = gInv(Fac[N - 1]);
	for (int i = N - 1; i >= 1; --i) iFac[i - 1] = (LL)iFac[i] * i % Mod;
	for (int i = 1; i < N; ++i) Inv[i] = (LL)Fac[i - 1] * iFac[i] % Mod;
}

inline int Binom(int N, int M) {
	if (M < 0 || M > N) return 0;
	return (LL)Fac[N] * iFac[M] % Mod * iFac[N - M] % Mod;
}

int Sz, InvSz, R[MS];

inline int getB(int N) { int Bt = 0; while (1 << Bt < N) ++Bt; return Bt; }

inline void InitFNTT(int N) {
	int Bt = getB(N);
	if (Sz == (1 << Bt)) return ;
	Sz = 1 << Bt, InvSz = Mod - (Mod - 1) / Sz;
	for (int i = 1; i < Sz; ++i) R[i] = R[i >> 1] >> 1 | (i & 1) << (Bt - 1);
}

inline void FNTT(int *A, int Ty) {
	for (int i = 0; i < Sz; ++i) if (R[i] < i) std::swap(A[R[i]], A[i]);
	for (int j = 1, j2 = 2; j < Sz; j <<= 1, j2 <<= 1) {
		int wn = qPow(~Ty ? G : iG, (Mod - 1) / j2), w, X, Y;
		for (int i = 0, k; i < Sz; i += j2) {
			for (k = 0, w = 1; k < j; ++k, w = (LL)w * wn % Mod) {
				X = A[i + k], Y = (LL)w * A[i + j + k] % Mod;
				A[i + k] -= (A[i + k] = X + Y) >= Mod ? Mod : 0;
				A[i + j + k] += (A[i + j + k] = X - Y) < 0 ? Mod : 0;
			}
		}
	}
	if (!~Ty) for (int i = 0; i < Sz; ++i) A[i] = (LL)A[i] * InvSz % Mod;
}

inline void PolyConv(int *_A, int N, int *_B, int M, int *_C) {
	static int A[MS], B[MS];
	InitFNTT(N + M - 1);
	for (int i = 0; i < N; ++i) A[i] = _A[i];
	for (int i = N; i < Sz; ++i) A[i] = 0;
	for (int i = 0; i < M; ++i) B[i] = _B[i];
	for (int i = M; i < Sz; ++i) B[i] = 0;
	FNTT(A, 1), FNTT(B, 1);
	for (int i = 0; i < Sz; ++i) A[i] = (LL)A[i] * B[i] % Mod;
	FNTT(A, -1);
	for (int i = 0; i < N + M - 1; ++i) _C[i] = A[i];
}

inline void PolyInv(int *_A, int N, int *_B) {
	static int A[MS], B[MS], tA[MS], tB[MS];
	for (int i = 0; i < N; ++i) A[i] = _A[i];
	for (int i = N, B = getB(N); i < 1 << B; ++i) A[i] = 0;
	B[0] = gInv(A[0]);
	for (int L = 1; L < N; L <<= 1) {
		int L2 = L << 1, L4 = L << 2;
		InitFNTT(L4);
		for (int i = 0; i < L2; ++i) tA[i] = A[i];
		for (int i = L2; i < Sz; ++i) tA[i] = 0;
		for (int i = 0; i < L; ++i) tB[i] = B[i];
		for (int i = L; i < Sz; ++i) tB[i] = 0;
		FNTT(tA, 1), FNTT(tB, 1);
		for (int i = 0; i < Sz; ++i) tB[i] = tB[i] * (2 - (LL)tA[i] * tB[i] % Mod + Mod) % Mod;
		FNTT(tB, -1);
		for (int i = 0; i < L2; ++i) B[i] = tB[i];
	}
	for (int i = 0; i < N; ++i) _B[i] = B[i];
}

inline void PolyLn(int *_A, int N, int *_B) {
	static int tA[MS], tB[MS];
	for (int i = 1; i < N; ++i) tA[i - 1] = (LL)_A[i] * i % Mod;
	PolyInv(_A, N - 1, tB);
	InitFNTT(N + N - 3);
	for (int i = N - 1; i < Sz; ++i) tA[i] = 0;
	for (int i = N - 1; i < Sz; ++i) tB[i] = 0;
	FNTT(tA, 1), FNTT(tB, 1);
	for (int i = 0; i < Sz; ++i) tA[i] = (LL)tA[i] * tB[i] % Mod;
	FNTT(tA, -1);
	_B[0] = 0;
	for (int i = 1; i < N; ++i) _B[i] = (LL)tA[i - 1] * Inv[i] % Mod;
}

inline void PolyExp(int *_A, int N, int *_B) {
	static int A[MS], B[MS], tA[MS], tB[MS];
	for (int i = 0; i < N; ++i) A[i] = _A[i];
	for (int i = N, B = getB(N); i < 1 << B; ++i) A[i] = 0;
	B[0] = 1;
	for (int L = 1; L < N; L <<= 1) {
		int L2 = L << 1, L4 = L << 2;
		for (int i = L; i < L2; ++i) B[i] = 0;
		PolyLn(B, L2, tA);
		InitFNTT(L4);
		for (int i = 0; i < L2; ++i) tA[i] = (!i + A[i] - tA[i] + Mod) % Mod;
		for (int i = L2; i < Sz; ++i) tA[i] = 0;
		for (int i = 0; i < L; ++i) tB[i] = B[i];
		for (int i = L; i < Sz; ++i) tB[i] = 0;
		FNTT(tA, 1), FNTT(tB, 1);
		for (int i = 0; i < Sz; ++i) tA[i] = (LL)tA[i] * tB[i] % Mod;
		FNTT(tA, -1);
		for (int i = 0; i < L2; ++i) B[i] = tA[i];
	}
	for (int i = 0; i < N; ++i) _B[i] = B[i];
}

int M, N;
int Ans[MS], Tmp[MS];

int main() {
	scanf("%d%d", &M, &N), --M;
	Init(M + 2);
	for (int i = 0; i <= M; ++i) Ans[i] = i & 1 ? Mod - 1 : 1;
	while (N--) {
		int a, b;
		scanf("%d%d", &a, &b);
		Tmp[0] = 1;
		int coef = (LL)(b + 1) * gInv(b) % Mod;
		for (int i = 1; i <= M; ++i) {
			coef = (LL)coef * (b - i + 1) % Mod * Inv[i + 1] % Mod;
			Tmp[i] = coef;
		}
		PolyLn(Tmp, M + 1, Tmp);
		for (int i = 0; i <= M; ++i) Tmp[i] = (LL)Tmp[i] * a % Mod;
		PolyExp(Tmp, M + 1, Tmp);
		int qwq = qPow(b, a);
		for (int i = 0; i <= M; ++i) Tmp[i] = (LL)Tmp[i] * qwq % Mod;
		PolyConv(Ans, M + 1, Tmp, M + 1, Ans);
	}
	int tAns = Ans[M];
	printf("%d\n", tAns);
	return 0;
}
posted @ 2019-09-21 23:25  粉兔  阅读(443)  评论(1编辑  收藏