CF960G Bandit Blues 第一类斯特林数+分治+FFT

题目传送门

https://codeforces.com/contest/960/problem/G

题解

首先整个排列的最大值一定是 (A) 个前缀最大值的最后一个，也是 (B) 个后缀最大值的最后一个。

那么枚举一下最大值的位置为 (i)，那么左右两边各选一些数的方案数为 (inom {n-1}{i-1})。

然后，左边有 (i-1) 个数，要分成 (A-1) 个部分，每一个部分的第一个数是所有数中最大的，并且每一个部分之间的最大值要递增。

可以发现这个问题等价于把 (i-1) 个数分成 (A-1) 个环——不是排列是因为第一个数的位置是固定的。因此可以很容易地写出答案

[sum_{i=1}^{n-1} inom{n-1}{i-1}egin{bmatrix}i-1\A-1end{bmatrix}egin{bmatrix}n-i\B-1end{bmatrix} ]

发现既然是先把一堆数分成两份再每一份分成 (A-1) 和 (B-1) 个环，那么也可以看成先分成 (A+B-2) 个环然后再把环分成两份。

因此方案数等价于

[egin{bmatrix}n-1\A+B-2end{bmatrix}inom{A+B-2}{A-1} ]

求第一类斯特林数可以用分治+FFT。

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b , 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b , 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I>
inline void read(I &x) {
	int f = 0, c;
	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
	x = c & 15;
	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
	f ? x = -x : 0;
}

const int N = 4e5 + 7;
const int P = 998244353;
const int Gi = 332748118;
const int G = 3;

int n, Aa, Bb, nlg;
int A[N], B[N], pw[20][N], cc[20];

int fac[N], inv[N], ifac[N];
inline void ycl(const int &n = ::n) {
	fac[0] = 1; for (int i = 1; i <= n; ++i) fac[i] = (ll)fac[i - 1] * i % P;
	inv[1] = 1; for (int i = 2; i <= n; ++i) inv[i] = (ll)(P - P / i) * inv[P % i] % P;
	ifac[0] = 1; for (int i = 1; i <= n; ++i) ifac[i] = (ll)ifac[i - 1] * inv[i] % P;
}
inline int C(int x, int y) {
	if (x < y) return 0;
	return (ll)fac[x] * ifac[y] % P * ifac[x - y] % P;
}

inline int smod(int x) { return x >= P ? x - P : x; }
inline void sadd(int &x, const int &y) { x += y; x >= P ? x -= P : x; }
inline int fpow(int x, int y) {
	int ans = 1;
	for (; y; y >>= 1, x = (ll)x * x % P) if (y & 1) ans = (ll)ans * x % P;
	return ans;
}

inline void NTT(int *a, int n, int f) {
	for (int i = 0, j = 0; i < n; ++i) {
		if (i > j) std::swap(a[i], a[j]);
		for (int l = n >> 1; (j ^= l) < l; l >>= 1) ;
	}
	for (int i = 1; i < n; i <<= 1) {
		int w = fpow(f > 0 ? G : Gi, (P - 1) / (i << 1));
		for (int j = 0; j < n; j += i << 1)
			for (int k = 0, e = 1; k < i; ++k, e = (ll)e * w % P) {
				int x = a[j + k], y = (ll)e * a[i + j + k] % P;
				a[j + k] = smod(x + y), a[i + j + k] = smod(x + P - y);
			}
	}
	if (f < 0) for (int i = 0, p = fpow(n, P - 2); i < n; ++i) a[i] = (ll)a[i] * p % P;
}
inline void Mul(int *a, int *b, int *c, int n, int m) {
	int l = 1;
	while (l <= n + m) l <<= 1;
	for (int i = 0; i <= n; ++i) A[i] = a[i];
	for (int i = n + 1; i < l; ++i) A[i] = 0;
	for (int i = 0; i <= m; ++i) B[i] = b[i];
	for (int i = m + 1; i < l; ++i) B[i] = 0;
	NTT(A, l, 1), NTT(B, l, 1);
	for (int i = 0; i < l; ++i) A[i] = (ll)A[i] * B[i] % P;
	NTT(A, l, -1);
	for (int i = 0; i <= n + m; ++i) c[i] = A[i];
}

int a[N], b[N], c[N];
inline int Calc_Strling1(int n, int m) {
	if (!n) return m == 0;
	if (m > n || m <= 0) return 0;
	a[1] = 1;
	for (int k = 0; k < nlg; ++k) {
		int l = cc[k];
		for (int i = 0; i <= l; ++i) b[i] = (ll)a[l - i] * fac[l - i] % P
		for (int i = 0; i <= l; ++i) c[i] = (ll)pw[k][i] * ifac[i] % P;
		Mul(b, c, b, l, l);
		for (int i = 0; i <= l; ++i) c[i] = (ll)b[l - i] * ifac[i] % P;
		Mul(a, c, a, l, l);
		if (cc[k + 1] & 1) {
			l = cc[k + 1];
			for (int i = l; i; --i) a[i] = (a[i - 1] + (l - 1ll) * a[i]) % P;
			a[0] = (l - 1ll) * a[0] % P;
		}
	}
	return a[m];
}

inline void work() {
	ycl();
	cc[0] = n - 1;
	while (cc[nlg] >> 1) cc[nlg + 1] = cc[nlg] >> 1, ++nlg;
	std::reverse(cc, cc + nlg + 1);
	for (int i = 0; i < nlg; ++i) {
		pw[i][0] = 1;
		for (int j = 1; j <= n; ++j) pw[i][j] = (ll)pw[i][j - 1] * cc[i] % P;
	}
	printf("%I64d
", (ll)Calc_Strling1(n - 1, Aa + Bb - 2) * C(Aa + Bb - 2, Aa - 1) % P);
}

inline void init() {
	read(n), read(Aa), read(Bb);
}

int main() {
#ifdef hzhkk
	freopen("hkk.in", "r", stdin);
#endif
	init();
	work();
	fclose(stdin), fclose(stdout);
	return 0;
}