[bzoj 4176] Lucas的数论 (杜教筛 + 莫比乌斯反演)

题面

设为的约数个数，给定，求

题目分析

有这样一个结论
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​​

由于数据范围的增强,我们不能预处理完整个,于是就外层整除分块优化

• 内层杜教筛来算的前缀和,时间复杂度为
• 后面平方的底数实际上等于的约数个数和的前缀和,可以直接算,预处理出前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

总时间复杂度为

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\n", (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\n", (solve(n) + mod) % mod);
}