BZOJ4652 [Noi2016]循环之美 【数论 + 莫比乌斯反演 + 杜教筛】

题目链接

BZOJ

题解

orz

此题太优美了

我们令\(\frac{x}{y}\)为最简分数，则\(x \perp y\)即，\(gcd(x,y) = 1\)

先不管\(k\)进制，我们知道\(10\)进制下如果\(\frac{x}{y}\)是纯循环的，只要\(2 \perp y\)且\(5 \perp y\)

可以猜想在\(k\)进制下同样成立

证明：

若\(\frac{x}{y}\)为纯循环小数，设其循环节长度为\(l\)，那么一定满足

\[\{ \frac{xk^{l}}{y} \} = \{ \frac{x}{y}\}
\]

其中\(\{x\}\)指小数部分

可以写成

\[xk^{l}- \lfloor \frac{xk^{l}}{y} \rfloor * y= x - \lfloor \frac{x}{y} \rfloor * y
\]

可以看出就是同余的形式

\[xk^{l} \equiv x \pmod y
\]

由于\(x \perp y\)

\[k^{l} \equiv 1 \pmod y
\]

要使存在\(l\)，使得式子成立，那么一定\(k \perp y\)

所以我们有

\[\begin{aligned}
ans &= \sum\limits_{i = 1}^{n} \sum\limits_{j = 1}^{m} [i \perp j] [j \perp k] \\
&= \sum\limits_{j = 1}^{m} [j \perp k] \sum\limits_{i = 1}^{n} [(i,j) == 1] \\
&= \sum\limits_{j = 1}^{m} [j \perp k] \sum\limits_{i = 1}^{n} \epsilon((i,j)) \\
&= \sum\limits_{j = 1}^{m} [j \perp k] \sum\limits_{i = 1}^{n} \sum\limits_{d|(i,j)} \mu(d) \\
&= \sum\limits_{d = 1}^{n} \mu(d) \sum\limits_{d|i}^{n} \sum\limits_{d|j}^{m} [j \perp k] \\
&= \sum\limits_{d = 1}^{n} [d \perp k] \mu(d) \lfloor \frac{n}{d} \rfloor \sum\limits_{j}^{\lfloor \frac{m}{d} \rfloor} [j \perp k] \\
\end{aligned}
\]

我们令

\[f(n) = \sum\limits_{i}^{n} [i \perp k]
\]

根据\(gcd(i + k,k) = gcd(i,k)\)，我们有

\[f(n) = \lfloor \frac{n}{k} \rfloor f(k) + f(n \mod k)
\]

所以我们只需要\(O(klogk)\)暴力计算\(f(1....k)\)就可以\(O(1)\)计算后面的式子了

现在考虑前面的

\[\sum\limits_{d = 1}^{n} [d \perp k] \mu(d)
\]

为了使能整除分块，我们必须算出其前缀和

令

\[g(n,k) = \sum\limits_{i = 1}^{n} [i \perp k] \mu(d)
\]

用类似上面同样的方法：

\[\begin{aligned}
g(n,k) &= \sum\limits_{i = 1}^{n} [i \perp k] \mu(i) \\
&= \sum\limits_{i = 1}^{n} \mu(i) \sum\limits_{d|i,d|k} \mu(d) \\
&= \sum\limits_{d | k} \mu(d) \sum\limits_{d|i} \mu(i) \\
&= \sum\limits_{d | k} \mu(d) \sum\limits_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \mu(id) \\
&= \sum\limits_{d | k} \mu(d) \sum\limits_{i = 1}^{\lfloor \frac{n}{d} \rfloor} [i \perp d]\mu(i) * \mu(d) \\
&= \sum\limits_{d | k} \mu(d)^2 \sum\limits_{i = 1}^{\lfloor \frac{n}{d} \rfloor} [i \perp d]\mu(i) \\
&= \sum\limits_{d | k} \mu(d)^2 g(\lfloor \frac{n}{d} \rfloor,d) \\
\end{aligned}
\]

就可以递归求解

当\(n = 0\)时，\(g(0,k) = 0\)

当\(k = 1\)时，\(g(n,1) = \sum\limits_{i = 1}^{n} \mu(i)\)，上杜教筛即可

因为\(\lfloor \frac{n}{d} \rfloor\)只有\(\sqrt{n}\)种取值，\(d\)只能取\(k\)的因子，记其数量为\(p\)

那么求\(g(n,k)\)总的复杂度为\(O(p\sqrt{n} + n^{\frac{2}{3}})\)

于是乎我们就解决这道题了

总复杂度\(O(p\sqrt{n} + n^{\frac{2}{3}} + \sqrt{n} + klogk)\)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
#define mp(a,b) make_pair<LL,LL>(a,b)
using namespace std;
const int maxn = 1000005,maxm = 100005,INF = 1000000000;
map<LL,LL> _mu;
map<LL,LL>::iterator it;
map<pair<LL,LL>,LL> G;
map<pair<LL,LL>,LL>::iterator IT;
int N;
int p[maxn],pi,isn[maxn];
LL Smu[maxn],f[maxn],mu[maxn];
LL n,m,K;
int gcd(int a,int b){return b ? gcd(b,a % b) : a;}
void init(){
	N = 1000000;
	mu[1] = 1;
	for (int i = 2; i <= N; i++){
		if (!isn[i]) p[++pi] = i,mu[i] = -1;
		for (int j = 1; j <= pi && i * p[j] <= N; j++){
			isn[i * p[j]] = true;
			if (i % p[j] == 0){
				mu[i * p[j]] = 0;
				break;
			}
			mu[i * p[j]] = -mu[i];
		}
	}
	for (int i = 1; i <= N; i++) Smu[i] = Smu[i - 1] + mu[i];
	for (int i = 1; i <= K; i++) f[i] = f[i - 1] + (gcd(i,K) == 1);
}
LL S(LL n){
	if (n <= N) return Smu[n];
	if ((it = _mu.find(n)) != _mu.end()) return it->second;
	LL ans = 1;
	for (LL i = 2,nxt; i <= n; i = nxt + 1){
		nxt = n / (n / i);
		ans -= (nxt - i + 1) * S(n / i);
	}
	return _mu[n] = ans;
}
LL F(LL n){
	return (n / K) * f[K] + f[n % K];
}
LL g(LL n,LL k){
	if ((IT = G.find(mp(n,k))) != G.end())
		return IT->second;
	if (n == 0) return 0;
	if (k == 1) return S(n);
	LL ans = 0;
	for (LL i = 1; i * i <= k; i++){
		if (k % i == 0){
			if (mu[i]) ans += g(n / i,i);
			if (i * i != k && mu[k / i])
				ans += g(n / (k / i),k / i);
		}
	}
	return G[mp(n,k)] = ans;
}
int main(){
	cin >> n >> m >> K;
	init();
	LL ans = 0,now,last = 0;
	for (LL i = 1,nxt; i <= min(n,m); i = nxt + 1){
		nxt = min(n / (n / i),m / (m / i));
		now = g(nxt,K);
		ans += 1ll * (now - last) * (n / i) * F(m / i);
		last = now;
	}
	cout << ans << endl;
	return 0;
}