loj#6076「2017 山东一轮集训 Day6」三元组 莫比乌斯反演 + 三元环计数

题目大意：

给定\(a, b, c\)，求\(\sum \limits_{i = 1}^a \sum \limits_{j = 1}^b \sum \limits_{k = 1}^c [(i, j) = 1][(j, k) = 1][(i, k) = 1]\)

$a, b, c \leq 5*10^4 $

---

首先莫比乌斯反演

$Ans = \sum \limits_{i = 1}^a \sum \limits_{j = 1}^b \sum \limits_{k = 1}^c [(i, j) = 1][(j, k) = 1][(i, k) = 1] $

\(= \sum \limits_{i} \sum \limits_{j} \sum \limits_{k} \sum \limits_{x |i \;x|j} \mu(x) \sum \limits_{y|j\;y|k} \mu(y) \sum \limits_{z |i\;z|k} \mu(z)\)

\(= \sum \limits_{x} \sum \limits_{y} \sum \limits_{z} \mu(x) \mu(y) \mu(z) \frac{A}{lcm(x, y)} \frac{B}{lcm(x, z)} \frac{C}{lcm(y, z)}\)

那么考虑计算这个式子

注意到其实有效的三元组\((x, y, z)\)是十分稀少的

我们考虑用一种高效的办法来找到所有的三元组

三元环计数是一个十分便利的算法

如果\(\mu(u), \mu(v) \neq 0, lca(u, v) \leq C\)，那么我们连边\((u, v)\)

怎么连边呢？

我们先枚举\(lca(u, v)\)，然后枚举\(u\)，之后再枚举\(gcd(u, v)\)判断即可

对于有两个数相同的情况和三个数相同的情况进行特判即可

复杂度不会算，反正跑的挺快的

ps：我怎么感觉dfs也能过呢？

---

#include <bits/stdc++.h>
using namespace std;

#define mp make_pair
#define pii pair <int, int>

#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

const int sid = 5e4 + 5;
const int cid = 2e6 + 5;
const int mod = 1e9 + 7;

inline void inc(int &a, int b) { a += b; if(a >= mod) a -= mod; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }

int a, b, c, id, ans, tot;
int mu[sid], pr[sid], nop[sid];
int eu[cid], ev[cid], ew[cid], d[sid], vis[sid], vv[sid];
vector <pii> go[sid];
vector <int> fac[sid];

inline void Init() {
	mu[1] = 1;
	for (int i = 2; i <= 50000; i ++) {
		if (!nop[i]) { pr[++ tot] = i; mu[i] = mod - 1; }
		for (int j = 1; j <= tot; j ++) {
			int p = i * pr[j];
			if(p > 50000) break; nop[p] = 1;
			if(i % pr[j] == 0) break; if(mu[i]) mu[p] = mod - mu[i];
		}
	}

	for (ri i = 1; i <= tot; i ++)
	for (ri j = pr[i]; j <= 50000; j += pr[i])
		fac[j].push_back(pr[i]);
}

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return 1ll * a * b / gcd(a, b);}

inline void calc() {
	if(c < a) swap(a, c);
	if(c < b) swap(b, c);
	if(b < a) swap(a, b);

	for (ri x = 1; x <= a; x ++) // x = y = z
	if(mu[x]) inc(ans, mul(mu[x], 1ll * (a / x) * (b / x) * (c / x) % mod));

	for (ri L = 1; L <= c; L ++) if(mu[L]) {
		int v = fac[L].size();
		for (ri S = 0; S <= (1 << v) - 1; S ++) {
			int x = 1;
			rep(i, 0, v - 1) if(S & (1 << i)) x *= fac[L][i];
			if(x > b) continue;
			for (ri T = S & (S - 1); ; T = (T - 1) & S) {
				int D = 1;
				rep(j, 0, v - 1) if(T & (1 << j)) D *= fac[L][j];
				int y = 1ll * L * D / x;
				if(x > y && y <= a) {
					d[x] ++; d[y] ++;
					eu[++ id] = x; ev[id] = y; ew[id] = L;
					inc(ans, mul(mu[y], 1ll * (a / L) * (b / L) * (c / x) % mod));
					inc(ans, mul(mu[y], 1ll * (a / x) * (b / L) * (c / L) % mod));
					inc(ans, mul(mu[y], 1ll * (a / L) * (b / x) * (c / L) % mod));
					inc(ans, mul(mu[x], 1ll * (a / L) * (b / L) * (c / y) % mod));
					inc(ans, mul(mu[x], 1ll * (a / y) * (b / L) * (c / L) % mod));
					inc(ans, mul(mu[x], 1ll * (a / L) * (b / y) * (c / L) % mod));
				}
				if(!T) break;
			}
		}
	}

	for (ri i = 1; i <= id; i ++) {
		int u = eu[i], v = ev[i];
		if(d[u] > d[v]) swap(u, v);
		go[u].push_back(mp(v, ew[i]));
	}

	for (ri x = 1; x <= b; x ++) {
		for (auto Y : go[x]) vis[Y.first] = x, vv[Y.first] = Y.second;
		for (auto Y : go[x]) for (auto Z : go[Y.first]) if(vis[Z.first] == x) {
			static int res = 0, cer = 0;
			int y = Y.first, z = Z.first, xy = Y.second, yz = Z.second, xz = vv[z];
			res = 0; cer = mul(mu[x], mul(mu[y], mu[z]));
			inc(res, 1ll * (a / xy) * ((b / xz) * (c / yz) + (b / yz) * (c / xz)) % mod);
			inc(res, 1ll * (b / xy) * ((a / xz) * (c / yz) + (a / yz) * (c / xz)) % mod);
			inc(res, 1ll * (c / xy) * ((a / xz) * (b / yz) + (a / yz) * (b / xz)) % mod);
			inc(ans, mul(cer, res));
		}
	}	

	cout << ans << endl;
}

int main() {
	cin >> a >> b >> c;
	Init();	calc();
	return 0;
}