[bzoj 4176] Lucas的数论 (杜教筛 + 莫比乌斯反演)

题面

设$d(x)$为$x$的约数个数，给定$N$，求 $sum^{N}_{i=1}sum^{N}_{j=1} d(ij)$
$N<=10^9$

题目分析

有这样一个结论
$d(ij)=sum_{x|i}sum_{y|j}[(x,y)==1]$这道题就是下面这道题的数据增强版,那么这个结论的证明就不再赘述,请自行查看下面的(蒟蒻)博客 传送门:[SDOI2015][bzoj 3994][Luogu P3327] 约数个数和

$large Ans=sum_{k=1}^Nmu(k)left(sum_{x=1}^{⌊frac{N}{k}⌋}⌊frac{N}{kx}⌋ ight)^2$
由于数据范围的增强,我们不能预处理完整个$10^9$,于是就外层整除分块优化

• 内层杜教筛来算$mu$的前缀和,时间复杂度为$Theta (N^{frac 23})$
• 后面平方的底数实际上等于$left[1,⌊frac{N}{k}⌋ ight]$的约数个数和的前缀和,可以直接$Theta(sqrt {⌊frac{N}{k}⌋})$算,预处理出前$N^{frac23}$的约数个数和的前缀和后,总时间复杂度就如杜教筛一样为$Theta(N^frac 23)$

总时间复杂度为$Theta (N^{frac 23})$

AC code

#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
typedef long long LL;
const int N = 1e6 + 1;
const int mod = 1e9 + 7;
int Cnt, Prime[N], mu[N], d[N], a[N]; //a[i]存的是i的最小质因数的次数+1
bool IsnotPrime[N];
void init()
{
	mu[1] = d[1] = a[1] = 1;
	for(int i = 2; i < N; ++i)
	{
		if(!IsnotPrime[i])
			Prime[++Cnt] = i, mu[i] = -1, a[i] = d[i] = 2;
		for(int j = 1; j <= Cnt && i * Prime[j] < N; ++j)
		{
			IsnotPrime[i * Prime[j]] = 1;
			if(i % Prime[j] == 0)
			{
				mu[i * Prime[j]] = 0;
				d[i * Prime[j]] = d[i] / a[i] * (a[i * Prime[j]] = a[i] + 1);
				break;
			}
			mu[i * Prime[j]] = -mu[i];
			d[i * Prime[j]] = d[i] * (a[i * Prime[j]] = 2);
		}
	}
	for(int i = 1; i < N; ++i)
		(d[i] += d[i-1]) %= mod, (mu[i] += mu[i-1]) %= mod;
}

inline int sum_d(int n) //约数个数和的前缀和,也就是后面个平方的底数
{
	if(n < N) return d[n];
	int ret = 0;
	for(int i = 1, j; i <= n; i=j+1)
	{
		j = n/(n/i);
		ret = (ret + (LL)(n/i) * (j-i+1) % mod) % mod;
	}
	return ret;
}

map<int, int>s;
inline int sum_mu(int n)
{
	if(n < N) return mu[n];
	if(s.count(n)) return s[n];
	int ret = 1;
	for(int i = 2, j; i <= n; i=j+1)
	{
		j = n/(n/i);
		ret = (ret - (LL)sum_mu(n/i)*(j-i+1)%mod) % mod;
	}
	return s[n]=ret;
}

int solve(int n)
{
	int ret = 0, last = 0, tmp, tmp2;
	for(int i = 1, j; i <= n; i=j+1)
	{
		j = n/(n/i);
		tmp = sum_mu(j), tmp2 = sum_d(n/i), tmp2 = (LL)tmp2 * tmp2 % mod;
		//tmp2存后面那个平方的值
		ret = (ret + (LL)((tmp-last) % mod) * tmp2 % mod) % mod;
		last = tmp;//这利用了一个小优化,本来是sum_mu(j)-sum_mu(i-1),
				   //我们把sum_mu(i-1)的值存下来,就少计算一次,last存上一次答案
				   //然而我后来看发现这优化并没有什么卵用,本来就记忆化了...
	}
	return ret;
}

int main ()
{
	init(); int n;
	scanf("%d", &n);
	printf("%d
", (solve(n)+mod)%mod);
}

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少有的一A

二刷:bzoj rank 7

CODE

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1000005;
const int mod = 1e9 + 7;
int prime[MAXN/10], cnt, mu[MAXN], d[MAXN], a[MAXN];
bool vis[MAXN];
inline void Pre_Work(int n) {
	mu[1] = d[1] = a[1] = 1;
	for(int i = 2; i <= n; ++i) {
		if(!vis[i])
			prime[++cnt] = i, mu[i] = -1, d[i] = a[i] = 2;
		for(int j = 1; j <= cnt && i*prime[j] <= n; ++j) {
			vis[i*prime[j]] = 1;
			if(i % prime[j] == 0) {
				mu[i*prime[j]] = 0;
				d[i*prime[j]] = d[i] / a[i] * (a[i*prime[j]] = a[i]+1);
				break;
			}
			mu[i*prime[j]] = -mu[i];
			d[i*prime[j]] = d[i] * (a[i*prime[j]] = 2);
		}
	}
	for(int i = 2; i <= n; ++i)
		mu[i] += mu[i-1], (d[i] += d[i-1]) %= mod;
}
map<int, int>MU;
inline int sum_mu(int n) {
	if(n < MAXN) return mu[n];
	if(MU.count(n)) return MU[n];
	int re = 1;
	for(int i = 2, j; i <= n; i = j+1) {
		j = n/(n/i);
		re = (re - 1ll * (j-i+1) * sum_mu(n/i) % mod) % mod;
	}
	return MU[n]=re;
}
map<int, int>D;
inline int sum_d(int n) {
	if(n < MAXN) return d[n];
	if(D.count(n)) return D[n];
	int re = 0;
	for(int i = 1, j; i <= n; i = j+1) {
		j = n/(n/i);
		re = (re + 1ll * (j-i+1) * (n/i) % mod) % mod;
	}
	return D[n]=re;
}
inline int sqr(int x) { return 1ll*x*x%mod; }
inline int solve(int n) {
	int re = 0;
	for(int i = 1, j; i <= n; i = j+1) {
		j = n/(n/i);
		re = (re + 1ll * (sum_mu(j)-sum_mu(i-1)) % mod * sqr(sum_d(n/i)) % mod) % mod;
	}
	return re;
}
int main() {
	int n;
	scanf("%d", &n);
	Pre_Work(min(n, MAXN-1));
	printf("%d
", (solve(n) + mod) % mod);
}