xgzc— math 专题训练（一）

Lucas定理

当\(p\)是质数时，有\((^n_m)\equiv(^{n/p}_{m/p}) * (^{n\%p}_{m\%p}) \pmod{p}\)

狄利克雷卷积

定义：\((f*g)(n)=\sum_{d|n}f(d)g(\frac{n}{d})\)

然后满足交换律，结合律，分配律

单位元：\(e=[n=1]\)，即\(f*e=e*f=f\)

逆元：对于每一个\(f(1)\ne0\)的函数\(f\)，存在逆元\(g\)使得\(f*g=e\)

那么\(g(n)\)满足递推式：

\(g(n)=\frac{1}{f(1)}([n=1]-\sum_{i|n,i\ne1}f(i)g(\frac{n}{i}))\)

Extra

定义1：\((f\oplus g)(x)=\sum_{x|d}f(\frac{d}{x})g(d)\)

那么有，\((f*g)\oplus h =f \oplus (g \oplus h)\)

定义2:\((f\cdot g)(x)=f(x)g(x)\)

那么当\(f\)是完全积性函数时，有\((f\cdot g)*(f\cdot h)=f\cdot (g*h)\)

常见数论函数

\(1(x)=1\)

\(id^k(x)=x^k\)

\(\phi(x)=\sum_{i=1}^{x}[gcd(i,x)=1]\)

\(d(x)=\sum_{d|x}1\)

\(\sigma(x)=\sum_{d|x}d\)

\(\mu：1\)的逆元

其中的一些关系：

\(d = 1*1\)

\(id = 1*\phi\)

\(\sigma=1*id=1*1*\phi=d*\phi\)

\(\phi=id*\mu\)

\(1=d*\mu\)

\(id=\sigma*\mu\)

莫比乌斯反演

如果\(g=f*1\)，则\(g*\mu=f\)

如果\(g=1\oplus f\),则\(f=\mu \oplus g\)

证明：\(\mu \oplus g=\mu \oplus (1 \oplus f)=(\mu * 1)\oplus f = f\)

例题

[SDOI2015] 约数个数和

求\(\sum_{i=1}^n\sum_{i=1}^md(ij)\)

\(d(ij)=\sum_{a|i}\sum_{b|j}[gcd(a,b)=1]\)

那么，原式等于

\[\sum_{i=1}^n\sum_{i=1}^m\sum_{a|i}\sum_{b|j}[gcd(a,b)=1]=
\sum_{i=1}^n\sum_{j=1}^m\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{j}\rfloor[gcd(i,j)=1]
\]

\[\sum_{d=1}^n\mu(d)\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\lfloor\frac{n}{id}\rfloor\sum_{i=1}^{\lfloor\frac{m}{d}\rfloor}\lfloor\frac{m}{id}\rfloor
\]

定义：\(S(n)=\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor\)

void init() {
	mu[1] = 1;
	for (int i = 2; i <= 50000; i++) {
		if (!np[i]) { p[++tot] = i; mu[i] = -1; }
		for (int j = 1; j <= tot && p[j] * i <= 50000; j++) {
			np[p[j] * i] = 1;
			if (!(i % p[j])) break;
			mu[p[j] * i] = -mu[i];
		}
	}
	for (int i = 2; i <= 50000; i++) mu[i] += mu[i - 1];
	for (int i = 1; i <= 50000; i++)
		for (int l = 1, r; l <= i; l = r + 1) {
			r = i / (i / l);
			S[i] += (r - l + 1) * (i / l);
		}
	return ;
}
 main() {
	init();
	int T;
	scanf("%lld", &T);
	while (T--) {
		int n, m;
		scanf("%lld%lld", &n, &m);
		if (n > m) swap(n, m);
		int ans = 0;
		for (int l = 1, r; l <= n; l = r + 1) {
			r = min(m / (m / l), n / (n / l));
			ans += (mu[r] - mu[l - 1]) * S[n / l] * S[m / l];
		}
		printf("%lld\n", ans);
	}
	return 0;
}

[SHOI2015]超能粒子炮·改

求\(\sum_{i=0}^{k}(^n_i)\)

根据\(Lucas\)定理

可以通过对模数相同的放在一块计算

再递归求解即可。

int Lucas(LL n, LL m) {
	if (!n || !m) return 1;
	return Lucas(n / P, m / P) * C[n % P][m % P] % P;
}
int f(LL n, LL k) {
	if (n <= 3000 && k <= 3000) return F[n][k];
	return (f(n % P, P - 1) * f(n / P, k / P - 1) % P +
	f(n % P, k % P) * Lucas(n / P, k / P) % P) % P;
}
void init() {
	for (int i = 0; i <= 3000; i++) C[i][0] = 1;
	for (int i = 1; i <= 3000; i++)
		for (int j = 1; j <= i; j++)
			C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
	for (int i = 0; i <= 3000; i++) {
		F[i][0] = C[i][0];
		for (int j = 1; j <= 3000; j++)
			F[i][j] = (F[i][j - 1] + C[i][j]) % P;
	}
	return ;
}

CF1097F Alex and a TV Show

因为只关心奇偶性，我们考虑使用\(bitset\)维护因子集合\(g(x)\)

操作2是异或

操作3是并

因为\(g=1\oplus f\)

\(f(x)\)是\(x\)在集合出现的次数

反演一下就是,\(f=\mu\oplus g\)

void prework(int n) {
	Mu.set();
	for (int i = 2; i * i <= n; i++)
		for (int j = 1; j * i * i <= n; j++)
			Mu[i * i * j] = 0;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j * i <= n; j++)
			mu[i][j * i] = Mu[j], p[j * i][i] = 1;
}
int main() {
	prework(7000);
	int n, Q;
	scanf("%d%d", &n, &Q);
	while (Q--) {
		int op, x, y, z;
		scanf("%d%d%d", &op, &x, &y);
		if (op == 1) S[x] = p[y];
		else if (op == 2) {
			scanf("%d", &z);
			S[x] = S[y] ^ S[z];
		}
		else if (op == 3) {
			scanf("%d", &z);
			S[x] = S[y] & S[z];
		}
		else printf("%d", (mu[y] & S[x]).count() & 1);
	}
	return 0;
}

Luogu5176 公约数

首先，\(gcd(ij,jk,ik)=\frac{gcd(i,j)gcd(j,k)gcd(i,k)}{gcd(i,j,k)}\)

那么，

\(Ans=\sum_i\sum_j\sum_k(i,j)^2(j,k)^2(i,k)^2\)

\(=\sum_i\sum_j\sum_k(i,j)^2(j,k)^2+\sum_i\sum_j\sum_k^2(j,k)^2(i,k)^2+\sum_i\sum_j\sum_k(i,j)^2(i,k)^2\)

那么定义\(F(n,m)=\sum_i\sum_j(i,j)^2\)，则\(Ans=F(n,m)*p+F(n,p)*m+F(m,p)*n\)

现在问题就是求\(F(n,m)\)

\(F(n,m)=\sum_i\sum_j(i,j)^2\)

\(=\sum_{d=1}^nd^2\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}[gcd(i,j)=1]\)

\(=\sum_{d=1}^nd^2\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}\sum_{k|i,k|j}\mu(k)\)

\(=\sum_{d=1}^nd^2\sum_{k=1}^{\lfloor\frac{n}{d}\rfloor}\mu(k)\lfloor\frac{n}{kd}\rfloor\lfloor\frac{m}{kd}\rfloor\)

到这一步已经可以做到\(O(n ^ \frac{3}{4})\)

考虑优化，

设\(T=kd\)

原式\(=\sum_{T=1}^n\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\sum_{d|T}d^2\mu(\frac{T}{d})\)

可以发现后面就是\(id^2*\mu\)

显然是个积性函数，考虑线性筛

\(f(p^0)=1\)

\(f(p)=p^2-1\)

\(f(p^k)=(p^2-1)p^{2(k-1)}=p^2f(p^{k-1})\)