51NOD 1222 最小公倍数计数 [莫比乌斯反演 杜教筛]

1222 最小公倍数计数

题意：求有多少数对\((a,b):a<b\)满足\(lcm(a,b) \in [1, n]\)

\(n \le 10^{11}\)

---

卡内存！

枚举\(gcd, \frac{a}{gcd}, \frac{b}{gcd}\)，然后\(\mu\)代入，就是

\[\sum_{d=1}^{\sqrt{n}}\mu(d) \sum_i \sum_j \sum_k [ijk \le \frac{n}{d^2}]
\]

问题就是怎么求后面的式子了

一开始我是

\[f(n) = \sum_i \sum_j \sum_k [ijk \le n] = \sum_{i=1}^n g(i) \\
g(n) = \sum_{i=1}^n \sigma(i)
\]

杜教筛\(g\)，方法是卷上\(\mu\)，当然还要杜教筛\(\sum\mu\)，但不影响复杂度，还是\(O(n^{\frac{2}{3}})\)

本机6.5s，改小预处理的大小后当然T了

然后又用分块的方法算\(g\)，预处理前\(O(n^{2/3})\)的\(\sigma\)剩下的分块\(O({\sqrt{n}})\)计算，复杂度也是\(O(n^{\frac{2}{3}})\)

本机4.6s，改小预处理大小又T了...

最后还是用了tangjz的方法，统计\(abc \le n\)的数对个数，规定\(a\le b \le c\)，然后a枚举\(n^{\frac{1}{3}}\)，b枚举\(\sqrt{\frac{n}{a}}\)。最后乘上\(3!\)再减去\(a= b <c,\ a<b = c,\ a=b=c\)

这一部分的复杂度\(T'(n)=O(\sum_{i=1}^{n^\frac{1}{3}}{\sqrt{\frac{n}{i}}})=O(n^\frac{2}{3})\)

总体复杂度\(T(n)=O(\sum_{d=1}^{\sqrt{n}}{T'(\frac{n}{d^2})})=?O(n^\frac{2}{3})​\)

求积分好像是\(\sqrt{n}\)啊，我也不知道这是怎么算的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long ll;
const int N=320000;
int U=316230;
inline ll read(){
	char c=getchar(); ll x=0,f=1;
	while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
	while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
	return x*f;
}
bool notp[N]; int p[N], mu[N];
void sieve(int n) {
	mu[1]=1;
	for(int i=2; i<=n; i++) {
		if(!notp[i]) p[++p[0]] = i, mu[i] = -1;
		for(int j=1; j <= p[0] && i*p[j] <= n; j++) {
			notp[ i*p[j] ] = 1;
			if(i % p[j] == 0) {mu[ i*p[j] ] = 0; break;}
			mu[ i*p[j] ] = -mu[i];
		}
	}
}
inline ll cal(ll n) {
	ll ans=0;
	for(ll i=1; i*i*i <= n; i++) {
		for(ll j=i; i*j*j <= n; j++) {
			ans += (n/i/j - j + 1) * 6;
			if(i == j) ans -= (n/i/j - j) * 3;
			if(i != j) ans -= 3;
		}
		ans -= 5;
	}
	return ans;
}
ll solve(ll n) {
	ll ans=0;
	int m = sqrt(n);
	for(int i=1; i<=m; i++) if(mu[i]) ans += mu[i]>0 ? cal(n/i/i) : -cal(n/i/i);
	return (ans + n) / 2;
}
int main() {
	freopen("in", "r", stdin);
	sieve(U);
	ll l=read(), r=read();
	printf("%lld", solve(r) - solve(l-1));
	printf("\n\n%lf\n",(double)clock()/CLOCKS_PER_SEC);
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <ctime>
using namespace std;
typedef long long ll;
const int N=22000005;
int U=22000000;
inline ll read(){
	char c=getchar(); ll x=0,f=1;
	while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
	while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
	return x*f;
}
bool notp[N]; int p[N/10], mu[N], lp[N]; ll si[N];
void sieve(int n) {
	mu[1]=1; si[1]=1;
	for(int i=2; i<=n; i++) {
		if(!notp[i]) p[++p[0]] = i, mu[i] = -1, si[i] = lp[i] = 2;
		for(int j=1; j <= p[0] && i*p[j] <= n; j++) {
			int t = i*p[j];
			notp[t] = 1;
			if(i%p[j] == 0) {
				mu[t] = 0;
				lp[t] = lp[i] + 1;
				si[t] = si[i] / lp[i] * lp[t];
				break;
			}
			mu[t] = -mu[i];
			lp[t] = 2;
			si[t] = si[i] * 2;
		}
		mu[i] += mu[i-1];
		si[i] += si[i-1];
	}
}
namespace ha {
	const int p = 1001001;
	struct ha {
		struct meow{int ne; ll val, r;} e[10000];
		int cnt, h[p];
		ha() {cnt=0; memset(h, 0, sizeof(h));}
		inline void insert(ll x, ll val) {
			int u = x%p;
			for(int i=h[u];i;i=e[i].ne) if(e[i].r == x) return;
			e[++cnt] = (meow){h[u], val, x}; h[u] = cnt;
		}
		inline ll quer(ll x) {
			int u = x%p;
			for(int i=h[u];i;i=e[i].ne) if(e[i].r == x) return e[i].val;
			return -1;
		}
	} hs, hu, me;
} using ha::hs; using ha::hu; using ha::me;
ll dj_u(ll n) { //return 1;
	if(n <= U) return mu[n];
	if(hu.quer(n) != -1) return hu.quer(n);
	ll ans = 1, r;
	for(ll i=2; i<=n; i=r+1) {
		r = n/(n/i);
		ans -= (r-i+1) * dj_u(n/i);
	}
	hu.insert(n, ans);
	return ans;
}
ll dj_s(ll n) {
	if(n <= U) return si[n];
	if(hs.quer(n) != -1) return hs.quer(n);
	dj_u(n);
	ll ans = n, r, now, last = dj_u(1);
	for(ll i=2; i<=n; i=r+1, last=now) {
		r = n/(n/i); now = dj_u(r);
		ans -= (now - last) * dj_s(n/i);
	}
	hs.insert(n, ans);
	return ans;
}
ll cal(ll n) {
	ll ans=0, r;
	for(ll i=1; i<=n; i=r+1) {
		r = n/(n/i);
		ans += (r-i+1) * dj_s(n/i);
	}
	return ans;
}
ll solve(ll n) {
	ll ans=0;
	int m = sqrt(n);
	for(ll i=1; i<=m; i++) if(mu[i] - mu[i-1]) ans += (mu[i] - mu[i-1]) * cal(n / (i*i));
	return (ans + n) / 2;
}
ll l, r;
int main() {
	freopen("in", "r", stdin);
	sieve(U);
	l=read(); r=read();
	//printf("%lld", dj_u(r));
	printf("%lld", solve(r) - solve(l-1));
	printf("\n\n%lf\n",(double)clock()/CLOCKS_PER_SEC);
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long ll;
const int N=22000005;
int U=22000000;
inline ll read(){
	char c=getchar(); ll x=0,f=1;
	while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
	while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
	return x*f;
}
bool notp[N]; int p[N/10], mu[N], lp[N]; ll si[N];
void sieve(int n) {
	mu[1]=1; si[1]=1;
	for(int i=2; i<=n; i++) {
		if(!notp[i]) p[++p[0]] = i, mu[i] = -1, si[i] = lp[i] = 2;
		for(int j=1; j <= p[0] && i*p[j] <= n; j++) {
			int t = i*p[j];
			notp[t] = 1;
			if(i%p[j] == 0) {
				mu[t] = 0;
				lp[t] = lp[i] + 1;
				si[t] = si[i] / lp[i] * lp[t];
				break;
			}
			mu[t] = -mu[i];
			lp[t] = 2;
			si[t] = si[i] * 2;
		}
		//mu[i] += mu[i-1];
		si[i] += si[i-1];
	}
}
inline ll g(ll n) { if(n <= U) return si[n];
	ll ans=0, r;
	for(ll i=1; i<=n; i=r+1) {
		r = n/(n/i);
		ans += (r-i+1) * (n/i);
	}
	return ans;
}
ll cal(ll n) {
	ll ans=0, r;
	for(ll i=1; i<=n; i=r+1) {
		r = n/(n/i);
		ans += (r-i+1) * g(n/i);
	}
	return ans;
}
ll solve(ll n) {
	ll ans = 0;
	int m = sqrt(n);
	for(ll i=1; i<=m; i++)
		if(mu[i]) ans += mu[i] * cal(n / (i*i));
	return (ans + n) / 2;
}
ll l, r;
int main() {
	freopen("in", "r", stdin);
	sieve(U);
	l=read(); r=read();
	printf("%lld", solve(r) - solve(l-1));
	printf("\n\n%lf\n",(double)clock()/CLOCKS_PER_SEC);
}