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51Nod 1244 莫比乌斯函数之和

Description

求\(\sum_{i=a}^b\mu(i),1\leqslant l\leqslant r\leqslant 10^{10}\)

Solution

杜教筛..贴代码..

Code

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 2000500;const ll p = 1000000007;int pr[N],cp,b[N],mu[N],smu[N];void pre(int n) {	mu[1]=1;	for(int i=2;i<=n;i++) {		if(!b[i]) pr[++cp]=i,mu[i]=-1;		for(int j=1;j<=cp && i*pr[j]<=n;j++) {			b[i*pr[j]]=1;			if(i%pr[j]) mu[i*pr[j]]=-mu[i];			else break;		}	}for(int i=1;i<=n;i++) smu[i]=(smu[i-1]+mu[i]);//	for(int i=1;i<=10;i++) cout<<mu[i]<<" ";cout<<endl; }map<ll,ll> mp;ll S(ll n) {	if(n<=2000000) return smu[n];	if(mp.count(n)) return mp[n];	ll fn=1;	for(ll i=2,j;i<=n;i=j+1) {		j=n/(n/i);//		cout<<n<<" "<<i<<" "<<j<<endl;		fn=(fn-S(n/i)*(j-i+1));	}return mp[n]=fn;;}int main() {	pre(2000000);	ll l,r;	scanf("%lld%lld",&l,&r);//	cout<<S(r)<<" "<<smu[r]<<endl;//	cout<<S(l-1)<<" "<<smu[l-1]<<endl;	printf("%lld\n",S(r)-S(l-1));	return 0;}

  

51Nod 1244 莫比乌斯函数之和