[WOJ3575] Lucas的数论[莫比乌斯反演+杜教筛]

传送门

反演一波

枚举k,l

发现n/t 总共有根号n个取值, 每算一次前缀和是根号n的

可以证明时间复杂度是n^3/4, mu用杜教筛

#include<bits/stdc++.h>
#define N 2000050
#define Mod 1000000007
#define LL long long
#define M 50050
using namespace std;
int prim[N],mu[N],isp[N],tot;
int n; LL val[N];
map<int,int> Mu;
void prework(){
	mu[1] = 1;
	for(int i=2;i<=N-50;i++){
		if(!isp[i]) prim[++tot] = i, mu[i] = -1;
		for(int j=1;j<=tot;j++){
			if(i*prim[j] > N - 50) break;
			isp[i*prim[j]] = 1;
			if(i%prim[j]==0) break;
			mu[i*prim[j]] = -mu[i]; 
		}
	}
	for(int i=2;i<=N-50;i++) mu[i] += mu[i-1];
	for(int i=1;i<=M-50;i++){
		for(int l=1,r;l<=i;l=r+1){
			int v = i/l; r = i/v;
			val[i] += (LL)(r-l+1) * (LL)v % Mod;
			val[i] %= Mod; 
 		}
	}
}
LL getval(int x){
	if(x<=M-50) return val[x];
	LL ans = 0;
	for(int l=1,r;l<=x;l=r+1){
		int v = x/l; r = x/v;
		ans += (LL)(r-l+1) * (LL)v % Mod;
		ans %= Mod; 
	} return ans;
}
int getf(int x){
	if(x<=N) return mu[x];
	if(Mu[x]) return Mu[x];
	int ans = 1;
	for(int l=2,r;l<=x;l=r+1){
		int v = x/l; r = x/v;
		ans -= (r-l+1) * getf(v);
	} return Mu[x] = ans;
}
int main(){
	prework();
	scanf("%d",&n); LL ans = 0;
	for(int l=1,r;l<=n;l=r+1){
		int v = n/l; r = n/v;
		LL tmp = getval(v); tmp = (tmp * tmp) % Mod;
		ans += (LL)(getf(r) - getf(l-1)) * tmp % Mod;
		ans = (ans % Mod + Mod) % Mod; 
	} printf("%lld",ans); return 0;
}