「学习笔记」组合计数与中国剩余定理

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Combination

思路

Lucas 定理 \((6)\) 板子题.

Code

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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

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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

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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.

设 \(f(i)\) 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

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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

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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：

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

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

代码

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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

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

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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 \(k+1\) 与 \(k\) 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 \(\gcd\) 把 \(k^n\) 膜 \(k+1-n\) 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 & f1, std::vector & 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  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

排列组合
——OI-Wiki

——Rolling_star

浅谈Lucas定理应用及组合数建模
——BerryKanry

——GXZlegend

——Pycr

中国剩余定理
——OI-Wiki

扩展Lucas定理
——HorizonWind

学习笔记数学——中国剩余定理数学——组合数学总类——数学

「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

思路

Code

[SDOI2016]排列计数

思路

代码

[ZJOI2010]排列计数

思路

代码

BZOJ2839 集合计数

思路

代码

牡牛和牝牛

思路

代码

序列统计

思路

代码

[SDOI2009] 虔诚的墓主人

思路

代码

[SDOI2010]地精部落

思路

代码

[ZJOI2011]看电影

思路

代码

中国剩余定理

【模板】中国剩余定理（CRT）/ 曹冲养猪

思路

代码

Strange Way to Express Integers

思路

代码

礼物

思路

代码

[SDOI2010]古代猪文

思路

代码

Reference

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