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：

[egin{aligned}mathbf{Ans}&=sum_{a=1}^{infty}([x^a]G)cdotinom{a-1}{m-1}\&=sum_{a=0}^{infty}!left([x^{a}]dfrac{G}{x} ight)!cdotinom{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}} ight} ight]^{a_i})。

则 ([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。

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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 (displaystylesum_{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}} ight} ight]^{a_i}) 的 (hat m) 阶导：

• 对于 (displaystyle t=prod_{i=1}^{n}t_i) 的 (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}inom{k}{a_{1ldots n}}prod_{i=1}^{n}t_i^{(a_i)})，其中 (dbinom{k}{a_{1ldots 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。

令 (f_i=mathbf{OGF}left{0,underset{b_i}{underbrace{1,ldots,1}} ight}) ，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：

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

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

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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} ight)^{(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}} ight})，稍加推导可以得到：

• (f_i^{(0)}(1))，即 (f_i(1))，等于 (b_i)（显然）。
• 对于 (jge1)，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(nmlog 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
", tAns);
	return 0;
}