BZOJ 4176: Lucas的数论 [杜教筛]

4176: Lucas的数论

题意：求(sum_{i=1}^n sum_{j=1}^n sigma_0(ij)) (n le 10^9)

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代入(sigma_0(nm)=sum_{imid n}sum_{jmid m}[(i,j)=1])

反演得到

[sum_{d=1}^n mu(d) (g(frac{n}{d}))^2 \ g(n) = sum_{i=1}^n sigma_0(i) ]

杜教筛(mu sigma_0)的前缀和

当然和前面的题一样，(sigma_0)也可以用预处理+分块

复杂度(O(n^{frac{2}{3}}))

把上题TLE的杜教筛代码抄上就行了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long ll;
const int N=1664512, mo=1e9+7;
int U=1664510;
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;
}

inline void mod(int &x) {if(x>=mo) x-=mo; else if(x<0) x+=mo;}
inline void mod(ll &x) {if(x>=mo) x-=mo; else if(x<0) x+=mo;}
bool notp[N]; int p[N/10], mu[N], lp[N], 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, val, r;} e[10000];
		int cnt, h[p];
		ha() {cnt=0; memset(h, 0, sizeof(h));}
		inline void insert(int x, int 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 int quer(int 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; 
} using ha::hs; using ha::hu;

int dj_u(int n) {
	if(n <= U) return mu[n];
	if(hu.quer(n) != -1) return hu.quer(n);
	int ans = 1, r;
	for(int i=2; i<=n; i=r+1) {
		r = n/(n/i);
		mod(ans -= (r-i+1) * dj_u(n/i) %mo);
	}
	hu.insert(n, ans);
	return ans;
}

int dj_s(int n) {
	if(n <= U) return si[n];
	if(hs.quer(n) != -1) return hs.quer(n);
	int ans = n, r, now, last = dj_u(1);
	for(int i=2; i<=n; i=r+1, last=now) {
		r = n/(n/i); now = dj_u(r);
		mod(ans -= (ll) (now - last) * dj_s(n/i) %mo);
	}
	hs.insert(n, ans);
	return ans;
}

int solve(int n) { 
	dj_u(n); dj_s(n);
	int ans=0, r, last=0, now;
	for(int i=1; i<=n; i=r+1, last=now) {
		r = n/(n/i); now = dj_u(r); ll t = dj_s(n/i); //printf("hi [%d, %d]  %d %d  %lld
", i, r, now, last, t);
		mod(ans += (ll) (now - last) * t %mo * t %mo);
	}
	return ans;
}

int main() {
	freopen("in", "r", stdin);
	int n=read(); //U = pow(n, 2.0/3);
	sieve(U);
	printf("%d", solve(n));
}