“学习笔记”组合计数与中国剩余定理

“学习笔记”组合计数与中国剩余定理

点击查看目录


Warning:

本文内含有大量 $\LaTeX$ 公式,可能会引起不适与页面卡顿.

知识点

排列

\(n\) 个元素里选出 \(m\) 个元素排成一列的方案数.

计算公式:

\[A_{n}^{m}=n(n-1)(n-2)···(n-m+1)=\frac{n!}{(n-m)!}\tag{1}
\]

还有一种特殊情况:全排列

\[A_{n}^{n}=n!\tag{2}
\]

错排列

求对于一个排列 \(1\sim n\),满足任意 \(i\) 都不在第 \(i\) 位上的排列有多少个.

递推公式为:

\[D_{n}=(n-1)(D_{n-1}+D_{n-2})\tag{3}
\]

证明点这里

我们先把 \(n\) 放在第 \(n\) 位,然后对于任意一个有 \(n-1\) 个数的排列,我们分情况讨论:

  • 这个排列满足要求:随便找一个数和 \(n\) 换,方案数为 \((n-1)D_{n-1}\).
  • 这个排列有且只有一位 \(k(1\le k\le n-1)\) 不满足要求:把 \(k\)\(n\) 交换过来,方案数为 \((n-1)D_{n-2}\).
  • 这个排列有 \(k(2\le k\le n-1)\) 位不满足要求:不可能一次换完,应该在计算 \(D_{n-1}\) 时就已经换完了.

合并一下,总方案数为:

\[D_{n}=(n-1)(D_{n-1}+D_{n-2})
\]

组合数

式子

\(n\) 个元素里选出一个 \(m\) 个元素的集合的方案数.

组合与排列的区别:组合没有顺序

计算公式:

\[\dbinom{n}{m}=\frac{A^{m}_{n}}{A^{m}_{m}}=\frac{n!}{m!(n-m)!}\tag{4}
\]

可以理解为排列数去掉顺序.

一些性质

\[\dbinom{n}{m}=\dbinom{n}{n-m}=\dbinom{n-1}{m}+\dbinom{n-1}{m-1}\tag{5}
\]

卢卡斯定理

用于求较大的且模数 \(p\in\mathbb{P}\) 的组合数.

公式:

\[\dbinom{n}{m}\bmod p=\dbinom{\lfloor n/p\rfloor}{\lfloor m/p\rfloor}\cdot\dbinom{n\bmod p}{m\bmod p}\bmod p\tag{6}
\]

谔项式定理

公式:

\[(a+b)^n=\sum_{i=0}^{n}\dbinom{n}{i}a^{i}b^{n-i}\tag{7}
\]

点击查看证明

数学归纳法.

\[\begin{aligned}
(a+b)^1&=a^0b^1+a^1b^0=a+b\\
(a+b)^{n+1}
&=(a+b)\sum_{i=0}^{n}\dbinom{n}{i}a^{i}b^{n-i}\\
&=\sum_{i=0}^{n}\dbinom{n}{i}a^{i+1}b^{n-i}+\sum_{i=0}^{n}\dbinom{n}{i}a^{i}b^{n+1-i}\\
&=\sum_{i=1}^{n+1}\dbinom{n}{i-1}a^{i}b^{n+1-i}+\sum_{i=0}^{n}\dbinom{n}{i}a^{i}b^{n+1-i}\\
&=a^{n+1}b^{0}+a^{0}b^{n+1}+\sum_{i=1}^{n}\left(\dbinom{n}{i-1}+\dbinom{n}{i}\right)a^{i}b^{n+1-i}\\
&=\dbinom{n+1}{n+1}a^{n+1}b^{0}+\dbinom{n+1}{0}a^{0}b^{n+1}+\sum_{i=1}^{n}\dbinom{n+1}{i}a^{i}b^{n+1-i}\\
&=\sum_{i=0}^{n+1}\dbinom{n+1}{i}a^{i}b^{n+1-i}\\
\end{aligned}
\]

组合意义

显然展开后每一项都是 \(n\) 次的.

那么考虑构成 \(a\)\(i\) 次的项的方案数.

显然为从 \(n\)\(a\) 里选出 \(i\)\(a\) 的方案数,即 \(\dbinom{n}{i}\).

那么这一项为 \(\dbinom{n}{i}a^{i}b^{n-i}\).

全部展开后就是 \(\sum_{i=0}^{n}\dbinom{n}{i}a^{i}b^{n-i}\).

谔项式反演

相当有趣但相当长.

这里挂几个式子,以后有机会单独整理一下学习笔记并挂在这里.

Good Reference

形式零

\[f(n)=\sum_{i=0}^{n}(-1)^i\dbinom{n}{i}g(i)\Longleftrightarrow g(n)=\sum_{i=0}^{n}(-1)^i\dbinom{n}{i}f(i)\tag{8}
\]

形式一

\[f(n)=\sum_{i=0}^{n}\dbinom{n}{i}g(i)\Longleftrightarrow g(n)=\sum_{i=0}^{n}(-1)^{n-i}\dbinom{n}{i}f(i)\tag{9}
\]

形式谔

\[f(n)=\sum_{i=n}^{m}\dbinom{i}{n}g(i)\Longleftrightarrow g(n)=\sum_{i=n}^{m}(-1)^{i-n}\dbinom{i}{n}f(i)\tag{10}
\]

小技巧:线性推阶乘逆元

好像是 SoyTony 教的我,%%%

求完阶乘后求出最后一个的逆元,然后用一点阶乘的逆元的性质,反推回去即可.

在组合题中及其常用,可以线性预处理后直接 \(\Theta(1)\) 求组合数.

点击查看代码
inline ll FastPow (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a % P;
		a = a * a % P, b >>= 1;
	}
	return ans;
}
inline void Pre () {
	fac[0] = 1;
	_for (i, 1, n) fac[i] = fac[i - 1] * i % P;
	inv[n] = FastPow (fac[n], P - 2);
	for_ (i, n - 1, 0) inv[i] = inv[i + 1] * (i + 1) % P;
	return;
}

中国剩余定理(CRT)

解决如下形式的方程组:

\[\begin{cases}
x \equiv a_1 \pmod{m_1}\\
x \equiv a_2 \pmod{m_2}\\
\dots\\
x \equiv a_n \pmod{m_n}\\
\end{cases}
\]

其中,\(x,a_i,m_i\) 均为正整数,保证 \(m_i\) 互质.

做法

\(M=\prod_{i=1}^{n}m_i\)\(n_i=\frac{M}{m_i}\)\(n_i^{-1}\) 表示 \(n_i\) 在膜 \(m_i\) 意义下的逆元.

答案是:

\[\sum_{i=1}^{n}a_i n_i n_i^{-1}\pmod{M}
\]

注意:\(m_i\) 互质但 \(m_i\) 不一定是质数,求逆元不能用费马小定理,只能用扩欧!

证明

首先解这样一组方程:

\[\begin{cases}
x_i \equiv 0 \pmod{m_1}\\
\dots\\
x_i \equiv a_i \pmod{m_i}\\
\dots\\
x_i \equiv 0 \pmod{m_n}\\
\end{cases}
\]

显然,\(x_i=a_i n_i n_i^{-1}\) 是一组合法解:

  • 对于第 \(i\) 组方程,由于在膜 \(m_i\) 意义下 \(n_i n_i^{-1}=1\),所以 \(a_i n_i n_i^{-1} \equiv a_i \pmod{m_i}\).

  • 对于第 \(j(j\neq i)\) 组方程,由于 \(n_i\)\(m_j\) 倍数,所以 \(a_j n_j n_j^{-1} \equiv 0 \pmod{m_j}\).

那么如何合并?可以发现 \(x_i+x_j\) 仍是原方程的解,因此答案就是 \(\sum_{i=1}^{n}x_i\pmod{M}=\sum_{i=1}^{n}a_i n_i n_i^{-1}\pmod{M}\).

EXCRT

解决问题没有变,但 \(m_i\) 不互质.

该算法主要运用合并的思想.

首先解这个方程:

\[\begin{cases}
x \equiv a_1 \pmod{m_1}\\
x \equiv a_2 \pmod{m_2}\\
\end{cases}
\]

其中 \(m_1,m_2\) 不互质.

转化为不定方程:

\[\begin{aligned}
x &= a_1 + p \cdot m_1\\
&= a_2 + q \cdot m_2\\
\end{aligned}
\]

进行一个移项:

\[p \cdot m_1 – q \cdot m_2 = a_2 – a_1
\]

\(a_2-a_1\) 不能被 \(\gcd(m_1,m_2)\) 整除时,原方程无解,否则一定可以解出来一组 \(p,q\).

那么最后可以合并得到一个方程:

\[x \equiv p \cdot m_1 + a_1 \pmod{\operatorname{lcm}(m_1,m_2)}
\]

按这种方法两两合并,最后得到答案.

ExLucas

因为前置比较多,所以放到最后写.

问题

求:

\[\dbinom{n}{m} \bmod{P} (P\in \mathbb{N^*})
\]

拆为 CRT

\(P\) 不一定是质数,那么我们考虑将它拆成质数解决.

\(P=p_1^{k_1}\cdot p_2^{k_2}\cdot p_3^{k_3}\cdots p_t^{k_t}(p_i\in \mathbb{P})\).

那么可以列出方程:

\[\begin{cases}
x \equiv \dbinom{n}{m} \pmod{p_1^{k_1}}\\
x \equiv \dbinom{n}{m} \pmod{p_2^{k_2}}\\
\dots\\
x \equiv \dbinom{n}{m} \pmod{p_t^{k_t}}\\
\end{cases}
\]

可以发现 \(x\) 就是最终结果,这里可以使用 CRT 解决.

那么现在把焦点放在这个方程上:

\[x \equiv \dbinom{n}{m} \pmod{p^k}\\
\]

构造余数

我们现在需要解决的是计算 \(\dbinom{n}{m} \bmod{p^k}\),即 \(\dfrac{n!}{m!(n-m)!} \bmod{p^k}\).

然而 \(n!,m!\)\((n-m)!\) 可能与 \(p^k\) 不互质,不能直接求逆元,那么我们就把它们的因数中的 \(p\) 提前去掉.

我们设 \(g(n)\) 表示 \(n\) 的因数里有多少个 \(p\)\(f(n)\) 表示 \(\dfrac{n!}{p^{g(n)}}\).

那么原式就是:

\[\dfrac{f(n)}{f(m)f(n-m)} \times P^{g(n)-g(m)-g(n-m)} \bmod{p^k}
\]

构造函数

现在我们只需要解决函数 \(f(n)\)\(g(n)\) 即可,那么如何计算?

首先我们拆分一下 \(n!\):

\[\begin{aligned}

n!

&=1\cdot2\cdot3\cdots n&\pmod{p^k}\\

&=(p\cdot2p\cdot3p\cdots\left\lfloor\frac{n}{p}\right\rfloor p)(1\cdot2\cdot3\cdots n)&\pmod{p^k}\\

&=p^{\left\lfloor\frac{n}{p}\right\rfloor}(\left\lfloor\frac{n}{p}\right\rfloor)!(\prod_{i=1,i\bmod p\neq0}^{n}i)&\pmod{p^k}\\

&=p^{\left\lfloor\frac{n}{p}\right\rfloor}(\left\lfloor\frac{n}{p}\right\rfloor)!(\prod_{i=1,i\bmod p\neq0}^{p^k}i)(\prod_{i=p^k+1,i\bmod p\neq0}^{2p^k}i)\cdots(\prod_{i=p^k(\left\lfloor\frac{n}{p^k}\right\rfloor-1)+1,i\bmod{p}\neq0}^{p^k\left\lfloor\frac{n}{p^k}\right\rfloor}i)(\prod_{i=p^k\left\lfloor\frac{n}{p^k}\right\rfloor+1,i\bmod{p}\neq0}^{n}i)&\pmod{p^k}\\

&=p^{\left\lfloor\frac{n}{p}\right\rfloor}(\left\lfloor\frac{n}{p}\right\rfloor)!(\prod_{i=1,i\bmod p\neq0}^{p^k}i)(\prod_{i=1,i\bmod{p}\neq0}^{p^k}i)\cdots(\prod_{i=1,i\bmod{p}\neq0}^{p^k}i)(\prod_{i=p^k\left\lfloor\frac{n}{p^k}\right\rfloor+1,i\bmod{p}\neq0}^{n}i)&\pmod{p^k}\\

&=p^{\left\lfloor\frac{n}{p}\right\rfloor}(\left\lfloor\frac{n}{p}\right\rfloor)!(\prod_{i=1,i\bmod{p}\neq0}^{p^k}i)^{\left\lfloor\frac{n}{p^k}\right\rfloor}(\prod_{i=1,i\bmod{p}\neq0}^{n\bmod{p^k}}i)&\pmod{p^k}

\end{aligned}
\]

我们的目的是除去 \(p\),因此 \(p^{\left\lfloor\frac{n}{p}\right\rfloor}\) 应去除.\((\left\lfloor\frac{n}{p}\right\rfloor)!\) 里还可能会有 \(p\),所以最后式子为:

\[f(n)=f(\left\lfloor\frac{n}{p}\right\rfloor)(\prod_{i=1,i\bmod{p}\neq0}^{p^k}i)^{\left\lfloor\frac{n}{p^k}\right\rfloor}(\prod_{i=1,i\bmod{p}\neq0}^{n\bmod{p^k}}i)\pmod{p^k}
\]

边界为 \(f(0)=1\).

从刚才的式子也可以看出来,每次递推会诞生 \(\left\lfloor\frac{n}{p}\right\rfloor\)\(p\),由于它还在往下递推,所以还会产生 \(g(\left\lfloor\frac{n}{p}\right\rfloor)\)\(p\).

那么递推式为:

\[g(n)=\left\lfloor\frac{n}{p}\right\rfloor+g(\left\lfloor\frac{n}{p}\right\rfloor)
\]

代码

点击查看代码
const ll N = 1e5 + 10, INF = 1ll << 40;

namespace MathBasic {
	inline void GetFactor (ll x, std::vector <ll>& f1, std::vector <ll>& f2) {
		_for (i, 2, x) {
			if (!(x % i)) {
				f1.push_back (i), f2.push_back (0);
				while (!(x % i)) ++f2[f2.size () - 1], x /= i;
			}
		}
		return;
	}
	inline ll FastPow (ll a, ll b, ll MOD = INF) {
		ll ans = 1;
		while (b) {
			if (b & 1) ans = ans * a % MOD;
			a = a * a % MOD, b >>= 1;
		}
		return ans;
	}
	ll ExGcd (ll a, ll b, ll& x, ll& y) {
		if (!b) { x = 1, y = 0;return a; }
		ll g = ExGcd (b, a % b, x, y), _x = x;
		x = y, y = _x - y * (a / b);
		return g;
	}
	inline ll Inv (ll a, ll P) {
		ll x, y; ExGcd (a, P, x, y);
		return (x % P + P) % P;
	}
}

namespace EXLUCAS {
	using namespace MathBasic;
	ll FDP (ll x, ll P, ll pk) { //FacDivP
		if (x == 0) return 1;
		ll ans = 1;
		_for (i, 1, pk) if (i % P) ans = ans * i % pk;
		ans = FastPow (ans, x / pk, pk);
		_for (i, 1, x % pk) if (i % P) ans = ans * (i % pk) % pk;
		return FDP (x / P, P, pk) * ans % pk;
	}
	ll Index (ll x, ll P) { 
		if (x < P) return 0;
		return (x / P) + Index (x / P, P);
	}
	ll a[N], md[N];
	inline ll ExLucas (ll n, ll m, ll P) {
		std::vector <ll> p, k;
		p.push_back (0), k.push_back (0);
		GetFactor (P, p, k);
		ll len = p.size () - 1, ans = 0;
		_for (i, 1, len) {
			md[i] = FastPow (p[i], k[i]);
			a[i] = FDP (n, p[i], md[i]) * Inv (FDP (m, p[i], md[i]), md[i]) % md[i] * Inv (FDP (n - m, p[i], md[i]), md[i]) % md[i];
			a[i] = a[i] * FastPow (p[i], Index (n, p[i]) - Index (n - m, p[i]) - Index (m, p[i]), md[i]) % md[i];
		}
		_for (i, 1, len) {
			ll q = P / md[i], x, y;
			ExGcd (q, md[i], x, y);
			ans = (ans + a[i] * q % P * ((x % P + P) % P) % P) % P;
		}
		return ans;
	}
}

例题

排列组合

排队

题意

\(n\) 名男同学,\(m\) 名女同学和两名老师要排成一条直线,并且任意两名女同学不能相邻,两名老师也不能相邻,求一共有多少种排法?

\(n,m\le 2000\)

思路

要分类讨论.

如果两名老师被男生隔开,则方案数为 男生的排列乘老师的排列乘女生的排列

即:

\[n!\times A_{n+1}^{2}\times A_{n+3}^{m}
\]

如果两名老师被女生隔开,则 用来隔开老师的女生 乘上 老师的排列数 再乘上 这两名老师与这名女生可插的空 的方案数为 \(2m(n+1)\),再乘上男生的排列数和剩余女生的排列数得到:

\[2m(n+1)\times n!\times A_{n+2}^{m-1}
\]

合并一下可得:

\[n!(A_{n+1}^{2}\times A_{n+3}^{m}+2m(n+1)\times A_{n+2}^{m-1})
\]

但是由于 \(n,m\le 2000\),要手写高精才能过……

Code
点击查看代码
const ll N = 10000, k = 1000000000;
ll n, m;
class BigNum {
public:
	ll len = 0, a[N] = {0};
	BigNum () { memset(a, 0, sizeof(a));}
	inline void In (ll num){
		while (num) {
			a[++ len] = num % k;
			num /= k;
		}
		return;
	}
	inline void Out () {
		for_(i, len, 1)
			printf((i == len ? "%lld" : "%.9lld"), a[i]);
		if (!len) puts("0");
		puts("");
		return;
	}
	void operator = (BigNum another) {
		len = another.len;
		_for(i, 1, len) a[i] = another.a[i];
		return;
	}
	BigNum operator + (BigNum another) {
		BigNum answer;
		answer.len = max(len, another.len);
		_for(i, 1, answer.len) {
			answer.a[i] += a[i] + another.a[i];
			answer.a[i + 1] += answer.a[i] / k;
			answer.a[i] %= k;
		}
		while (answer.a[answer.len + 1]) ++ answer.len;
		return answer;
	}
	BigNum operator * (BigNum another) {
		BigNum answer;
		answer.len = len + another.len - 1;
		_for(i, 1, answer.len) {
			_for(j, 1, another.len) {
				answer.a[i + j - 1] += a[i] * another.a[j];
				answer.a[i + j] += answer.a[i + j - 1] / k;
				answer.a[i + j - 1] %= k;
			}
		}
		while (answer.a[answer.len + 1]) ++ answer.len;
		return answer;
	}
} mm, nn, ans;
namespace SOLVE {
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar();
		while (!isdigit(c)) { if (c == '-') w = -1; c = getchar();}
		while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
		return x * w;
	}
	inline void In () {
		n = rnt(), m = rnt();
		nn.In(n), mm.In(m);
		return;
	}
	inline void Solve () {
		BigNum a, b, one;
		one.In(1), a.In(2), b.In(1);
		a = a * (nn + one) * mm;
		for_(i, n + 2, n - m + 4) {
			BigNum ii;
			ii.In(i);
			a = a * ii;
		}
		b = nn * (nn + one);
		for_(i, n + 3, n - m + 4) {
			BigNum ii;
			ii.In(i);
			b = b * ii;
		}
		ans = a + b;
		_for(i, 2, n) {
			BigNum ii;
			ii.In(i);
			ans = ans * ii;
		}
		return;
	}
	inline void Out () {
		ans.Out();
		return;
	}
}

Combination

思路

Lucas 定理 \((6)\) 板子题.

Code
点击查看代码
namespace SOLVE {
	const ll P = 1e4 + 7, N = 1e4 + 10;
	ll T, x, y, fac[N], inv[N];
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar();
		while (!isdigit(c)) { if (c == '-') w = -1; c = getchar();}
		while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
		return x * w;
	}
	inline ll FastPow (ll a, ll b) {
		ll ans = 1;
		while (b) {
			if (b & 1) ans = ans * a % P;
			a = a * a % P, b >>= 1;
		}
		return ans;
	}
	inline void Pre () {
		fac[0] = 1;
		_for (i, 1, P)
			fac[i] = fac[i - 1] * i % P;
		inv[P - 1] = FastPow(fac[P - 1], P - 2);
		for_ (i, P - 2, 0)
			inv[i] = inv[i + 1] * (i + 1) % P;
		return ;
	}
	inline ll C (ll n, ll m) {
		if (m > n) return 0;
		return fac[n] * inv[n - m] % P * inv[m] % P;
	}
	inline ll Lucas (ll n, ll m) {
		if (m == 0) return 1;
		return C(n % P, m % P) * Lucas(n / P, m / P) % P;
	}
	inline void In () {
		x = rnt(), y = rnt();
		return ;
	}
	inline void Out () {
		printf("%lld\n", Lucas(x, y));
		return ;
	}
}

[SDOI2016]排列计数

思路

我们钦定 \(m\) 个数为稳定的,方案数为 \(\dbinom{n}{m}\).

在剩下的 \(n-m\) 个位置里要保证每个数不稳定.

欸那不就是错排列 \((3)\) 吗?

那么方案数就是 \(D_{n-m}\).

总方案数就是 \(\dbinom{n}{m}D_{n-m}\)\(\Theta(n)\) 预处理一下错排列,阶乘与逆元可用 \(\Theta(1)\) 求出单次询问.

代码
点击查看代码
namespace SOLVE {
	const ll P = 1e9 + 7, N = 1e6 + 10, M = 1e6;
	ll T, n, m, d[N], fac[N], inv[N];
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar();
		while (!isdigit(c)) { if (c == '-') w = -1; c = getchar();}
		while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
		return x * w;
	}
	inline ll FastPow(ll a, ll b) {
		ll ans = 1;
		while (b) {
			if(b & 1) ans = ans * a % P;
			a = a * a % P, b >>= 1;
		}
		return ans;
	}
	inline void Pre () {
		d[0] = 1, d[1] = 0, fac[0] = 1;
		_for (i, 2, M) d[i] = (i - 1) * ((d[i - 1] + d[i - 2]) % P) % P;
		_for (i, 1, M) fac[i] = fac[i - 1] * i % P;
		inv[M] = FastPow(fac[M], P - 2);
		for_ (i, M - 1, 0) inv[i] = inv[i + 1] * (i + 1) % P;
		return;
	}
	inline ll C(ll n, ll m) {
		return fac[n] * inv[n - m] % P * inv[m] % P;
	}
	inline void In () {
		n = rnt(), m = rnt();
		return ;
	}
	inline void Out () {
		printf("%lld\n", C(n, m) * d[n-m] % P);
		return ;
	}
}

[ZJOI2010]排列计数

思路

观察一下可以发现满足性质的序列是一个小根堆.

那么设 \(s_i\) 表示以 \(i\) 为根的堆的大小,\(f_i\) 表示以 \(i\) 为根的堆的可行方案数(此时该子堆里的序号不是最终序号,而是在子堆内大小的排名,因为归并到父堆时要算分配给子堆不同序号的方案数).

那么转移方程就是:

\[s_{i}=s_{i*2}+s_{i*2+1}+1\\
f_{i}=\dbinom{s_{i}-1}{s_{i*2}}f_{i*2}f_{i*2+1}
\]

(自己必须是最小的所以只能从 \(s_{i}-1\) 个序号选 \(s_{i*2}\) 分配给左儿子,剩下的全给右儿子)

代码
点击查看代码
namespace SOLVE {
	const ll N = 4e6 + 10;
	ll T, n, P, sz[N], f[N], fac[N];
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar();
		while (!isdigit(c)) { if (c == '-') w = -1; c = getchar();}
		while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
		return x * w;
	}
	inline ll FastPow (ll a, ll b) {
		ll ans = 1;
		while (b) {
			if(b & 1) ans = ans * a % P;
			a = a * a % P, b >>= 1;
		}
		return ans;
	}
	inline void Pre () {
		fac[0] = 1;
		_for (i, 1, std::min(P, n)) fac[i] = fac[i - 1] * i % P;
		_for (i, 1, n * 2 + 1) f[i] = 1;
		return;
	}
	inline ll Inv (ll n) {
		return FastPow(fac[n], P - 2);
	}
	inline ll C (ll n, ll m) {
		if(!n || !m) return 1;
		return fac[n] * Inv(n - m) % P * Inv(m) % P;
	}
	inline ll Lucas (ll n, ll m) {
		if(!n || !m) return 1;
		return C(n % P, m % P) * Lucas(n / P, m / P) % P;
	}
	inline void In () {
		n = rnt(), P = rnt();
		return ;
	}
	inline void Solve () {
		for_ (i, n, 1) {
			sz[i] = sz[i << 1] + sz[(i << 1) + 1] + 1;
			f[i] = f[i << 1] * f[(i << 1) + 1] % P * Lucas(sz[i] - 1, sz[i << 1]) % P;
		}
		return ;
	}
	inline void Out () {
		printf("%lld\n", f[1]);
		return ;
	}
}

BZOJ2839 集合计数

思路

谔项式反演.

\(f(i)\) 表示交集数量 \(\ge i\) 的方案数,\(g(i)\) 表示交集个数恰好为 \(i\) 个的方案数,那么答案为 \(g(k)\).

那么:

\[f(i)=\dbinom{n}{i}(2^{2^{n-i}}-1)
\]

即先确定 \(i\) 个必选,包含这 \(i\) 个的集合数为 \(2^{n-k}\) 个,每个集合都可以选或不选但不能一个不选,即 \(2^{2^{n-i}}-1\).

同时:

\[f(k)=\sum_{i=k}^{n}\dbinom{i}{k}g(i)
\]

等一下这式子是不是在哪里见过?

这不是 \((10)\) 吗?!

那么愉快的套一个谔项式反演:

\[\begin{aligned}
g(k)
&=\sum_{i=k}^{n}(-1)^{i-k}\dbinom{i}{k}f(i)\\
&=\sum_{i=k}^{n}(-1)^{i-k}\dbinom{i}{k}\dbinom{n}{i}(2^{2^{n-i}}-1)\\
\end{aligned}
\]

再加上一点预处理,就可以解决了.

代码
点击查看代码
namespace SOLVE {
	typedef long double ldb;
	typedef long long ll;
	typedef double db;
	const ll N = 1e6 + 10, P = 1e9 + 7;
	ll T, n, k, er[N], fac[N], inv[N], ans;
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar ();
		while (!isdigit (c)) { if (c == '-') w = -1; c = getchar (); }
		while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
		return x * w;
	}
	inline ll FastPow (ll a, ll b) {
		ll ans = 1;
		while (b) {
			if (b & 1) ans = ans * a % P;
			a = a * a % P, b >>= 1;
		}
		return ans;
	}
	inline void Pre () {
		fac[0] = 1, er[0] = 2;
		_for (i, 1, n) {
			fac[i] = fac[i - 1] * i % P;
			er[i] = (er[i - 1] * er[i - 1]) % P;
		}
		inv[n] = FastPow (fac[n], P - 2);
		for_ (i, n - 1, 0) inv[i] = inv[i + 1] * (i + 1) % P;
		return;
	}
	inline ll C (ll n, ll m) {
		if (!m) return 1;
		return fac[n] * inv[n - m] % P * inv[m] % P;
	}
	inline void In () {
		n = rnt (), k = rnt ();
		return;
	}
	inline void Solve () {
		_for (i, k, n) {
			ll w = ((i - k) & 1) ? -1 : 1;
			ans = (ans + (er[n - i] - 1 + P) % P * C (n, i) % P * C (i, k) % P * w + P) % P;
		}
		return;
	}
	inline void Out () {
		printf ("%lld\n", ans);
		return;
	}
}

牡牛和牝牛

思路

我们枚举牝牛的数量 \(i\),那么一定会有 \(k\times(i-1)\) 只牡牛被固定住,此时剩下 \(w(i)=(n-i-k\times(i-1))\times[i>0]+n\times[i=0]\) 只牡牛可以随便选位置.

观察一下,看上去是只有 \(k+1\) 个地方可以插空,然而两只牝牛之间可以放多只牡牛,如何解决这个问题?

既然可以重复放,那我们就把重复放的位置 \(\text{new}\) 出来!

即把空的个数改为 \(k+1+(w(i)-1)=k+i\).

这样会不会导致选的全都是 \(\text{new}\) 出来的呢?不会,因为我们只 \(\text{new}\) 出来了 \(w(i)-1\) 个空,剩下的一只牛必然会被放在原有的位置.

那么答案就是:

\[\sum_{i=0}^{n}[w(i)\ge0]\dbinom{i+w(i)}{w(i)}
\]

代码
点击查看代码
namespace SOLVE {
	typedef long double ldb;
	typedef long long ll;
	typedef double db;
	const ll N = 1e5 + 10, P = 5e6 + 11;
	ll n, k, fac[N], inv[N], ans;
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar();
		while (!isdigit(c)) { if (c == '-') w = -1; c = getchar();}
		while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
		return x * w;
	}
	inline ll FastPow (ll a, ll b) {
		ll ans = 1;
		while (b) {
			if (b & 1) ans = ans * a % P;
			a = a * a % P, b >>= 1;
		}
		return ans;
	}
	inline void Pre () {
		fac[0] = 1;
		_for (i, 1, n) fac[i] = fac[i - 1] * i % P;
		inv[n] = FastPow (fac[n], P - 2);
		for_ (i, n - 1, 0) inv[i] = inv[i + 1] * (i + 1) % P;
		return;
	}
	inline ll C (ll n, ll m) {
		return fac[n] * inv[n - m] % P * inv[m] % P;
	}
	inline void In () {
		n = rnt (), k = rnt ();
		return;
	}
	inline void Solve () {
		_for (i, 0, n) {
			ll w = i ? (n - i - k * (i - 1)) : n;
			if (w < 0) break;
			ans = (ans + C (i + w, w)) % P;
		}
		return ;
	}
	inline void Out () {
		printf ("%lld\n", ans);
		return ;
	}
}

序列统计

思路

本题和上一题有些类似,每个数也是可以重复选的.

那么设 \(m=r-l+1\),长度为 \(i\) 的序列的方案数为 \(\dbinom{m+i-1}{i}\).

然后推式子:

\[\begin{aligned}
\sum_{i=1}^{n}\dbinom{m+i-1}{i}
&=\sum_{i=1}^{n}\dbinom{m+i-1}{m-1}+\dbinom{m}{m}-1\\
&=\sum_{i=2}^{n}\dbinom{m+i-1}{m-1}+\dbinom{m}{m-1}+\dbinom{m}{m}-1\\
&=\sum_{i=2}^{n}\dbinom{m+i-1}{m-1}+\dbinom{m+1}{m}-1\\
&=\sum_{i=3}^{n}\dbinom{m+i-1}{m-1}+\dbinom{m+2}{m}-1\\
&=\sum_{i=4}^{n}\dbinom{m+i-1}{m-1}+\dbinom{m+3}{m}-1\\
&=\cdots\\
&=\sum_{i=n}^{n}\dbinom{m+i-1}{m-1}+\dbinom{m+n-1}{m}-1\\
&=\dbinom{m+n}{m}-1\\
\end{aligned}
\]

\(n,m\) 过大,需要用到 \(\text{Lucas}\) 定理 \((6)\).

代码
点击查看代码
namespace SOLVE {
	typedef long double ldb;
	typedef long long ll;
	typedef double db;
	const ll N = 1e6 + 10, P = 1e6 + 3;
	ll T, n, m, l, r, fac[N], inv[N];
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar ();
		while (!isdigit (c)) { if (c == '-') w = -1; c = getchar (); }
		while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
		return x * w;
	}
	inline ll FastPow (ll a, ll b) {
		ll ans = 1;
		while (b) {
			if (b & 1) ans = ans * a % P;
			a = a * a % P, b >>= 1;
		}
		return ans;
	}
	inline void Pre () {
		fac[0] = 1;
		_for (i, 1, P - 1) fac[i] = fac[i - 1] * i % P;
		inv[P - 1] = FastPow (fac[P - 1], P - 2);
		for_ (i, P - 2, 0) inv[i] = inv[i + 1] * (i + 1) % P;
		return;
	}
	inline ll C (ll n, ll m) {
		if (n < m) return 0;
		if (!n || !m) return 1;
		return fac[n] * inv[n - m] % P * inv[m] % P;
	}
	inline ll Lucas (ll n, ll m) {
		if (n < m) return 0;
		if (!n || !m) return 1;
		return C (n % P, m % P) * Lucas (n / P, m / P) % P;
	}
	inline void In () {
		n = rnt (), l = rnt (), r = rnt ();
		m = r - l + 1;
		return;
	}
	inline void Out () {
		printf ("%lld\n", (Lucas (m + n, m) + P - 1) % P);
		return;
	}
}

[SDOI2009] 虔诚的墓主人

思路

题解

代码

感觉以前写的代码太丑了.

于是又写了一份.

点击查看代码
namespace SOLVE {
	typedef long double ldb;
	typedef long long ll;
	typedef double db;
	const ll N = 1e5 + 10, P = 2147483648;
	ll n, m, w, k, C[N][20], ans;
	ll cx[N], cy[N], nx[N], ny;
	class TREE {
	public:
		ll x, y;
		inline bool operator < (TREE another) {
			return (y == another.y) ? (x < another.x) : (y < another.y);
		}
	} tr[N];
	class TreeArray {
	public:
		ll b[N];
		inline ll lowbit (ll x) { return x & -x; }
		inline void Update (ll x, ll y) {
			while (x <= n) {
				b[x] = (b[x] + y) % P;
				x += lowbit (x);
			}
			return;
		}
		inline ll Query (ll x) {
			ll ans = 0;
			while (x) {
				ans = (ans + b[x]) % P;
				x -= lowbit (x);
			}
			return ans;
		}
	} ta;

	namespace LISAN {
		ll ls1[N], ls2[N];
		inline void lisan () {
			_for (i, 1, w) ls1[i] = tr[i].x;
			_for (i, 1, w) ls2[i] = tr[i].y;
			std::sort (ls1 + 1, ls1 + w + 1);
			std::sort (ls2 + 1, ls2 + w + 1);
			n = std::unique (ls1 + 1, ls1 + w + 1) - ls1;
			m = std::unique (ls2 + 1, ls2 + w + 1) - ls2;
			_for (i, 1, w) {
				tr[i].x = std::lower_bound (ls1 + 1, ls1 + n + 1, tr[i].x) - ls1;
				tr[i].y = std::lower_bound (ls2 + 1, ls2 + m + 1, tr[i].y) - ls2;
			}
			return;
		}
	}

	inline void Pre () {
		C[0][0] = 1;
		_for (i, 1, w) {
			C[i][0] = 1;
			_for (j, 1, k) C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % P;
		}
		return;
	}

	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar ();
		while (!isdigit (c)) { if (c == '-') w = -1; c = getchar (); }
		while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
		return x * w;
	}
	inline void In () {
		n = rnt (), m = rnt (), w = rnt ();
		_for (i, 1, w) tr[i].x = rnt (), tr[i].y = rnt ();
		k = rnt ();
		return;
	}
	inline void Solve () {
		LISAN::lisan ();
		std::sort (tr + 1, tr + w + 1);
		_for (i, 1, w) ++cx[tr[i].x], ++cy[tr[i].y];
		_for (i, 1, w - 1) {
			++ny, ++nx[tr[i].x];
			ll last = (ta.Query (tr[i].x) - ta.Query (tr[i].x - 1) + P) % P;
			if (tr[i].y == tr[i + 1].y) {
				ll up_down = C[ny][k] * C[cy[tr[i].y] - ny][k] % P;
				ll left_right = (ta.Query (tr[i + 1].x - 1) - ta.Query (tr[i].x) + P) % P;
				ans = (ans + up_down * left_right % P) % P;
			}
			else ny = 0;
			ta.Update (tr[i].x, (C [nx[tr[i].x]][k] * C [cx[tr[i].x] - nx[tr[i].x]][k] % P - last + P) % P);
		}
		return;
	}
	inline void Out () {
		printf ("%lld\n", ans);
		return;
	}
}

[SDOI2010]地精部落

思路

\(f_{i,0}\) 表示长度为 \(i\) 且第一段山为山谷的序列数量.

\(f_{i,1}\) 表示长度为 \(i\) 且第一段山为山峰的序列数量.

\[f_{i,k}=
\sum_{j=1}^{i}[j\bmod{2}=k]\dbinom{i-1}{j-1}f_{j-1,k}f_{i-j,0}
\]

代码
点击查看代码
namespace SOLVE {
	typedef long double ldb;
	typedef long long ll;
	typedef double db;
	const ll N = 4200 + 10;
	int n, P, f[N][2], C[N][N], ans;
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar ();
		while (!isdigit (c)) { if (c == '-') w = -1; c = getchar (); }
		while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
		return x * w;
	}
	inline void In () {
		n = rnt (), P = rnt ();
		return;
	}
	inline void Solve () {
		f[0][0] = f[0][1] = f[1][0] = 1, C[0][0] = 1;
		_for (i, 1, n) {
			C[i][0] = 1;
			_for (j, 1, i) {
				f[i][j & 1] = ((ll)(f[i][j & 1]) + (ll)(f[j - 1][j & 1]) * (ll)(f[i - j][0]) % P * C[i - 1][j - 1] % P) % P;
				C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
			}
		}
		return;
	}
	inline void Out () {
		printf ("%lld\n", (f[n][0] + f[n][1]) % P);
		return;
	}
}

[ZJOI2011]看电影

思路

这个式子还是蛮有意思的,但要用高精.

首先答案 \(=\frac{合法情况数}{总情况数}\),总情况数显然是 \(k^n\),难点在于如何算出合法情况数.

首先我们在 \(k\) 后面新增一个座位 \(k+1\) 然后拉链为环,让没有座位的人从头开始往后坐,这样一定所有人都会有座位,那么这样的环一共会有 \((k+1)^{n-1}\) 个(即圆排列).注意:这里暂时不考虑标号.

但是我们怎么判断做法是否合法呢?如果 \(k+1\) 这个位置最后有人,那么在没有环与新座位时就一定会有人站在这里,否则没人站着(即合法情况),也就是说 我们在环中找到空的位置当成 \(k+1\) 即可,空的位置一共有 \(k+1-n\) 个,所以最后答案为:

\[\frac{(k+1)^{n-1}(k+1-n)}{k^n}
\]

这就是拉链为环前莫名其妙地新增一个座位 \(k+1\) 的原因.

然后这个题非常恶心,要用高精,化简分数时要用高精除,这里考虑一种简单的方法:

显然 \(k+1\)\(k\) 互质,只能化简 \(\dfrac{k+1-n}{k^n}\).

这里 \(k+1-n\) 为低精,我们提前做一次 \(\gcd\)\(k^n\)\(k+1-n\) 转换为低精,就可以低精求 \(\gcd\) 了.

也就是说,最后我们只需要高精乘,高精除低精和高精膜低精即可.

代码
点击查看代码
namespace SOLVE {
	typedef long double ldb;
	typedef long long ll;
	typedef double db;
	const ll N = 110, B = 10000000; // Base
	ll T, n, k;
	class BigNum {
	public:
		ll num[N];
		inline void Print () {
			printf ("%lld", num[num[0]]);
			for_ (i, num[0] - 1, 1) printf ("%07lld", num[i]);
			return;
		}
		inline void Clear () {
			memset (num, 0, sizeof (num));
			num[0] = 1;
			return;
		}
		inline void In (ll number) { num[1] = number; }
		BigNum operator * (ll ano) {
			BigNum ans;
			ans.Clear ();
			ans.num[0] = num[0];
			_for (i, 1, num[0]) {
				ans.num[i] += num[i] * ano;
				ans.num[i + 1] += ans.num[i] / B;
				ans.num[i] %= B;
			}
			while (ans.num[ans.num[0] + 1]) ++ans.num[0];
			return ans;
		}
		BigNum operator * (BigNum ano) {
			BigNum ans;ans.Clear ();
			ans.num[0] = num[0] + ano.num[0] - 1;
			_for (i, 1, num[0]) {
				_for (j, 1, ano.num[0]) {
					ans.num[i + j - 1] += num[i] * ano.num[j];
					ans.num[i + j] += ans.num[i + j - 1] / B;
					ans.num[i + j - 1] %= B;
				}
			}
			while (ans.num[ans.num[0] + 1]) ++ans.num[0];
			return ans;
		}
		inline BigNum operator / (ll ano) {
			BigNum ans (*this);
			for_ (i, num[0], 1) {
				if (i > 1) ans.num[i - 1] += (ans.num[i] % ano) * B;
				ans.num[i] /= ano;
			}
			while (!ans.num[ans.num[0]] && ans.num[0] > 1) --ans.num[0];
			return ans;
		}
		inline ll operator % (ll ano) {
			ll ans = 0;
			for_ (i, num[0], 1) {
				ans = ans * B % ano;
				ans += num[i] % ano;
			}
			return ans;
		}
	} a, b;
	inline ll Gcd (ll a, ll b) {
		if (!b) return a;
		return Gcd (b, a % b);
	}
	inline BigNum FastPow (BigNum a, ll b) {
		BigNum ans;ans.Clear ();
		ans.num[0] = ans.num[1] = 1;
		while (b) {
			if (b & 1) ans = ans * a;
			a = a * a, b >>= 1;
		}
		return ans;
	}
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar ();
		while (!isdigit (c)) { if (c == '-') w = -1; c = getchar (); }
		while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
		return x * w;
	}
	inline void In () {
		n = rnt (), k = rnt ();
		return;
	}
	inline void Solve () {
		a.Clear (), b.Clear ();
		a.num[1] = k + 1, b.num[1] = k;
		a = FastPow (a, n - 1) * (k + 1 - n);
		b = FastPow (b, n);
		ll c = b % (k + 1 - n);
		ll g = Gcd (k + 1 - n, c);
		a = a / g, b = b / g;
		return;
	}
	inline void Out () {
		a.Print (), putchar (' ');
		b.Print (), puts ("");
		return;
	}
}

中国剩余定理

【模板】中国剩余定理(CRT)/ 曹冲养猪

思路

模板题.

代码
点击查看代码
namespace SOLVE {
	typedef long double ldb;
	typedef long long ll;
	typedef double db;
	const ll N = 20;
	ll n, a[N], m[N], M = 1, ans;
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar ();
		while (!isdigit (c)) { if (c == '-') w = -1; c = getchar (); }
		while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
		return x * w;
	}
	inline void exgcd (ll a, ll b, ll& x, ll& y) {
		if (!b) {
			x = 1, y = 0;
			return;
		}
		exgcd (b, a % b, x, y);
		ll _x = x;
		x = y, y = _x - (a / b) * y;
		return;
	}
	inline void In () {
		n = rnt ();
		_for (i, 1, n) {
			m[i] = rnt (), a[i] = rnt ();
			M *= m[i];
		}
		return;
	}
	inline void Solve () {
		_for (i, 1, n) {
			ll Mi = M / m[i], inv, y;
			exgcd (Mi, m[i], inv, y);
			ans = (ans + a[i] * Mi % M * (inv + m[i]) % M) % M;
		}
		return;
	}
	inline void Out () {
		printf ("%lld\n", ans);
		return;
	}
}

Strange Way to Express Integers

思路

EXCRT 模板题.

代码
点击查看代码
namespace SOLVE {
	typedef long double ldb;
	typedef long long ll;
	typedef double db;
	const ll N = 1e5 + 10;
	ll n, a[N], m[N], b, M;
	
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar ();
		while (!isdigit (c)) { if (c == '-') w = -1; c = getchar (); }
		while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
		return x * w;
	}
	ll Exgcd (ll a, ll b, ll& x, ll& y) {
		if (!b) { x = 1, y = 0; return a; }
		ll g = Exgcd (b, a % b, x, y), _x = x;
		x = y, y = _x - (a / b) * y;
		return g;
	}
	inline ll Lcm (ll a, ll b) {
		return a * b / std::__gcd (a, b);
	}
	inline ll FastMul (ll a, ll b, ll MOD) {
		ll ans = 0;
		while (b) {
			if (b & 1) ans = (ans + a) % MOD;
			a = (a + a) % MOD, b >>= 1;
		}
		return (ans + MOD) % MOD;
	}
	inline void In () {
		n = rnt ();
		_for (i, 1, n) m[i] = rnt (), a[i] = rnt ();
		return;
	}
	inline ll EXCRT () {
		b = a[1], M = m[1];
		_for (i, 2, n) {
			ll x, y, num = (a[i] - b % m[i] + m[i]) % m[i];
			ll g = Exgcd (M, m[i], x, y);
			if (num % g) return -1;
			b += M * FastMul(x, num / g, m[i] / g);
			M *= m[i] / g, b = (b % M + M) % M;
		}
		return (b % M + M) % M;
	}
	inline void Out () {
		printf ("%lld\n", EXCRT());
		return;
	}
}

礼物

思路

式子显然是:

\[\prod_{i=1}^{m}\dbinom{n-sum_{i-1}}{w_i}\bmod{P}
\]

直接扩卢即可.

代码
点击查看代码
const ll N = 1e5 + 10, INF = 1ll << 40;

namespace MathBasic {
	inline void GetFactor (ll x, std::vector <ll>& f1, std::vector <ll>& f2) {
		f1.push_back (0), f2.push_back (0);
		_for (i, 2, x) {
			if (!(x % i)) {
				f1.push_back (i), f2.push_back (0);
				while (!(x % i)) ++f2[f2.size () - 1], x /= i;
			}
		}
		return;
	}
	inline ll FastPow (ll a, ll b, ll Mod = INF) {
		ll ans = 1;
		while (b) {
			if (b & 1) ans = ans * a % Mod;
			a = a * a % Mod, b >>= 1;
		}
		return ans;
	}
	ll ExGcd (ll a, ll b, ll& x, ll& y) {
		if (!b) { x = 1, y = 0; return a; }
		ll g = ExGcd (b, a % b, x, y), _x = x;
		x = y, y = _x - (a / b) * y;
		return g;
	}
	inline ll Inv (ll a, ll P) {
		ll x, y;
		ExGcd (a, P, x, y);
		return (x % P + P) % P;
	}
}

namespace EXLUCAS {
	using namespace MathBasic;
	inline ll FDP (ll x, ll P, ll pk) {
		if (!x) return 1;
		ll ans = 1;
		_for (i, 1, pk) if (i % P) ans = ans * i % pk;
		ans = FastPow (ans, x / pk, pk);
		_for (i, 1, x % pk) if (i % P) ans = ans * i % pk;
		return ans * FDP (x / P, P, pk) % pk;
	}
	inline ll Index (ll x, ll P) {
		if (x < P) return 0;
		return (x / P) + Index (x / P, P);
	}
	ll a[N], md[N], P;
	std::vector <ll> p, k;
	inline void Pre (ll _P) {
		GetFactor (_P, p, k);
		P = _P;
		return;
	}
	inline ll ExLucas (ll n, ll m) {
		ll ans = 0, sz = p.size () - 1;
		_for (i, 1, sz) {
			md[i] = FastPow (p[i], k[i]);
			a[i] = FDP (n, p[i], md[i]) * Inv (FDP (m, p[i], md[i]), md[i]) % md[i] * Inv (FDP (n - m, p[i], md[i]), md[i]) % md[i];
			a[i] = a[i] * FastPow (p[i], Index (n, p[i]) - Index (m, p[i]) - Index (n - m, p[i]), md[i]) % md[i];
			ans = (ans + a[i] * (P / md[i]) % P * Inv (P / md[i], md[i]) % P) % P;
		}
		return ans;
	}
}

namespace SOLVE {
	ll P, n, m, w[10], sum[10], ans = 1;
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar ();
		while (!isdigit (c)) { if (c == '-') w = -1; c = getchar (); }
		while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
		return x * w;
	}
	inline void In () {
		P = rnt (), n = rnt (), m = rnt ();
		_for (i, 1, m) {
			w[i] = rnt ();
			sum[i] = sum[i - 1] + w[i];
		}
		return;
	}
	inline void Solve () {
		EXLUCAS::Pre (P);
		_for (i, 1, m) {
			if (n - sum[i - 1] < w[i]) { ans = -1; return; }
			ans = ans * EXLUCAS::ExLucas (n - sum[i - 1], w[i]) % P;
		}
		return;
	}
	inline void Out () {
		printf ((ans == -1) ? "Impossible\n" : "%lld\n", ans);
		return;
	}
}

[SDOI2010]古代猪文

思路

(为啥 SDOI2010 经典题这么多啊,猪国杀和 \(k\) 短路模板也是这里出的)

数论全家桶.

显然式子为:

\[g^{\sum_{n|k}\binom{n}{k}}\bmod{999911659}
\]

用一个费马小定理:

\[g^{\sum_{n|k}\binom{n}{k}\bmod{999911658}}\bmod{999911659}
\]

那么现在的问题就是:如何求出 \(\sum_{n|k}\binom{n}{k}\bmod{999911658}\)

仔细看两眼发现就是个扩卢.

然后就出来了.

(其实 \(999911658\) 就是 \(2, 3, 4679, 35617\) 这几个质数之积,可以简化一点写.)

代码
点击查看代码
const ll N = 50000, INF = 1ll << 40;

namespace MathBasic {
	inline ll FastPow (ll a, ll b, ll MOD = INF) {
		ll ans = 1;
		while (b) {
			if (b & 1) ans = ans * a % MOD;
			a = a * a % MOD, b >>= 1;
		}
		return ans;
	}
	inline ll ExGcd (ll a, ll b, ll& x, ll& y) {
		if (!b) { x = 1, y = 0; return a; }
		ll g = ExGcd (b, a % b, x, y), _x = x;
		x = y, y = _x - y * (a / b);
		return g;
	}
	inline ll Inv (ll a, ll P) {
		ll x, y;
		ExGcd (a, P, x, y);
		return (x % P + P) % P;
	}
}

namespace EXLUCAS {
	using namespace MathBasic;
	ll a[5], p[5] = { 0, 2, 3, 4679, 35617 }, fac[5][N], q[5];
	ll FDP (ll x, ll P, ll qwq) {
		if (!x) return 1;
		ll ans = FastPow (fac[qwq][P - 1], x / P, P);
		ans = ans * fac[qwq][x % P] % P;
		return ans * FDP (x / P, P, qwq) % P;
	}
	ll Index (ll x, ll P) {
		if (x < P) return 0;
		return (x / P) + Index (x / P, P);
	}
	inline void Pre () {
		fac[1][0] = fac[2][0] = fac[3][0] = fac[4][0] = 1;
		_for (k, 1, 4) {
			_for (i, 1, 36000) fac[k][i] = fac[k][i - 1] * i % p[k];
			q[k] = (999911658 / p[k]) * Inv (999911658 / p[k], p[k]);
		}
		return;
	}

	inline ll ExLucas (ll n, ll m, ll P) {
		ll ans = 0;
		_for (i, 1, 4) {
			a[i] = FDP (n, p[i], i) * Inv (FDP (m, p[i], i), p[i]) % P * Inv (FDP (n - m, p[i], i), p[i]) % p[i];
			a[i] = a[i] * FastPow (p[i], Index (n, p[i]) - Index (m, p[i]) - Index (n - m, p[i])) % p[i];
			ans = (ans + a[i] * q[i] % P) % P;
		}
		return ans;
	}
}

namespace SOLVE {
	ll n, g, idx, P = 999911659;
	inline ll rnt () {
		ll x = 0, w = 1; char c = getchar ();
		while (!isdigit (c)) { if (c == '-') w = -1; c = getchar (); }
		while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
		return x * w;
	}
	inline void In () {
		n = rnt (), g = rnt ();
		EXLUCAS::Pre ();
		return;
	}
	inline void Solve () {
		for (ll i = 1; i * i <= n; ++i) {
			if (n % i) continue;
			idx = (idx + EXLUCAS::ExLucas (n, i, P - 1)) % (P - 1);
			if (i * i != n) idx = (idx + EXLUCAS::ExLucas (n, n / i, P - 1)) % (P - 1);
		}
		return;
	}
	inline void Out () {
		printf ("%lld\n", g == P ? 0 : MathBasic::FastPow (g, idx, P));
		return;
	}
}

Reference