SDOI2018 旧试题（莫比乌斯反演+三元环计数）

题目链接

题目大意

求：$$\sum _{i=1}^A\sum _{j=1}^B\sum _{k=1}^{C}d(ijk)$$
其中$d(i)$为$i$的因数个数。$A,B,C\le 10^5$

题解

居然真的会有这么美妙的结论qwq……
$$d(ijk)=\sum _{a|i}\sum _{b|j}\sum _{c|k}[\gcd (a,b)=1][\gcd (a,c)=1][\gcd (b,c)=1]$$
感性理解一下就是，每个质因数分开考虑。假设它在$i,j,k$中的指数分别为$x,y,z$，那么它总共会在上面的算式中出现恰好$x+y+z+1$次。由于所有质数贡献独立，因此上面的算式就等于因数个数。
暴力带入原式并使用莫比乌斯反演，可以得到：
$$\begin{gathered} \sum _{i=1}^A\sum _{j=1}^B\sum _{k=1}^C\sum _{a|i}\sum _{b|j}\sum _{c|k}\sum _{x|a,x|b}\mu (x)\sum _{y|a,y|c}\mu (y)\sum _{z|b,z|c}\mu (z) \\ =\sum _{i=1}^A\sum _{j=1}^B\sum _{k=1}^C\sum _{x|i,x|j}\sum _{y|i,y|k}\sum _{z|j,z|k}\mu (x)\mu (y)\mu (z)d\left(\frac{i}{lcm(x,y)}\right)d\left(\frac{j}{lcm(x,z)}\right)d\left(\frac{k}{lcm(y,z)}\right) \\ =\sum _{x=1}^A\sum _{y=1}^A\sum _{z=1}^B\mu (x)\mu (y)\mu (z)D\left(\frac{A}{lcm(x,y)}\right)D\left(\frac{B}{lcm(x,z)}\right)D\left(\frac{C}{lcm(y,z)}\right) \end{gathered}$$
其中$D(i)={\sum }_{j=1}^{i}d(i)$。
考虑$x,y,z$全都不相等的情况。我们可以建无向图，对于任意的$x<y,lcm(x,y)\le C$，连边$(x,y)$。然后找图中的三元环，并任意轮换3!次加入答案。
如果$x,y,z$中恰好有一对相等，则可以任意轮换3次计入答案。
如果全部相等，直接循环枚举即可。
三元环计数在本题中似乎暴力就能过（大雾）。

#include 
namespace IOStream {
	const int MAXR = 1 << 23;
	char _READ_[MAXR], _PRINT_[MAXR];
	int _READ_POS_, _PRINT_POS_, _READ_LEN_;
	inline char readc() {
	#ifndef ONLINE_JUDGE
		return getchar();
	#endif
		if (!_READ_POS_) _READ_LEN_ = fread(_READ_, 1, MAXR, stdin);
		char c = _READ_[_READ_POS_++];
		if (_READ_POS_ == MAXR) _READ_POS_ = 0;
		if (_READ_POS_ > _READ_LEN_) return 0;
		return c;
	}
	template<typename T> inline void read(T &x) {
		x = 0; register int flag = 1, c;
		while (((c = readc()) < '0' || c > '9') && c != '-');
		if (c == '-') flag = -1; else x = c - '0';
		while ((c = readc()) >= '0' && c <= '9') x = x * 10 + c - '0';
		x *= flag;
	}
	template<typename T1, typename ...T2> inline void read(T1 &a, T2 &...x) {
		read(a), read(x...);
	}
	inline int reads(char *s) {
		register int len = 0, c;
		while (isspace(c = readc()) || !c);
		s[len++] = c;
		while (!isspace(c = readc()) && c) s[len++] = c;
		s[len] = 0;
		return len;
	}
	inline void ioflush() {
		fwrite(_PRINT_, 1, _PRINT_POS_, stdout), _PRINT_POS_ = 0;
		fflush(stdout);
	}
	inline void printc(char c) {
		_PRINT_[_PRINT_POS_++] = c;
		if (_PRINT_POS_ == MAXR) ioflush();
	}
	inline void prints(char *s) {
		for (int i = 0; s[i]; i++) printc(s[i]);
	}
	template<typename T> inline void print(T x, char c = '\n') {
		if (x < 0) printc('-'), x = -x;
		if (x) {
			static char sta[20];
			register int tp = 0;
			for (; x; x /= 10) sta[tp++] = x % 10 + '0';
			while (tp > 0) printc(sta[--tp]);
		} else printc('0');
		printc(c);
	}
	template<typename T1, typename ...T2> inline void print(T1 x, T2... y) {
		print(x, ' '), print(y...);
	}
}
using namespace IOStream;
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
#define cls(a) memset(a, 0, sizeof(a))

const int MAXN = 100005, MOD = 1000000007;
int mu[MAXN], prime[MAXN], vis[MAXN], A, B, C = 100000, T;
ll d[MAXN], da[MAXN], db[MAXN], dc[MAXN];
vector<P> g[MAXN];
int main() {
	mu[1] = 1;
	for (int i = 2, cnt = 0; i <= C; i++) {
		if (!vis[i]) mu[prime[++cnt] = i] = -1;
		for (int j = 1; j <= cnt; j++) {
			int p = prime[j], q = i * p;
			if (q > C) break;
			vis[q] = 1;
			if (i % p == 0) break;
			mu[q] = -mu[i];
		}
	}
	for (int i = 1; i <= C; i++) {
		for (int j = i; j <= C; j += i) ++d[j];
		d[i] += d[i - 1];
	}
	for (scanf("%d", &T); T--;) {
		scanf("%d%d%d", &A, &B, &C);
		cls(vis);
		if (A > B) swap(A, B);
		if (A > C) swap(A, C);
		if (B > C) swap(B, C);
		for (int i = 1; i <= C; i++) {
			da[i] = d[A / i];
			db[i] = d[B / i];
			dc[i] = d[C / i];
			g[i].clear();
		}
		ll ans = 0;
		for (int i = 1; i <= C; i++) {
			for (int j = 1; j * i <= C; j++) {
				for (int k = j + 1; (ll)i * j * k <= C; k++) {
					int a = i * j, b = i * k, c = i * j * k;
					if (!mu[a] || !mu[b] || __gcd(j, k) > 1) continue;
					ans += mu[a] * mu[a] * mu[b] * (
						da[c] * db[c] * dc[a] +
						da[c] * db[a] * dc[c] +
						da[a] * db[c] * dc[c] );
					ans += mu[a] * mu[b] * mu[b] * (
						da[b] * db[c] * dc[c] +
						da[c] * db[b] * dc[c] +
						da[c] * db[c] * dc[b] );
					g[a].push_back(P(b, c));
				}
			}
			ans += mu[i] * mu[i] * mu[i] * da[i] * db[i] * dc[i];
		}
		for (int i = 1; i <= C; i++) if (mu[i]) {
			for (P j : g[i]) vis[j.first] = i, prime[j.first] = j.second;
			for (P j : g[i]) if (j.first > i && mu[j.first]) {
				int a = j.second;
				for (P k : g[j.first]) if (k.first > j.first && vis[k.first] == i && mu[k.first]) {
					int b = k.second, c = prime[k.first];
					ans += mu[i] * mu[j.first] * mu[k.first] * (
						da[a] * db[b] * dc[c] +
						da[a] * db[c] * dc[b] +
						da[b] * db[a] * dc[c] +
						da[b] * db[c] * dc[a] +
						da[c] * db[a] * dc[b] +
						da[c] * db[b] * dc[a] );
				}
			}
		}
		printf("%lld\n", (ans % MOD + MOD) % MOD);
	}
	return 0;
}