[51Nod 1237] 最大公约数之和 (杜教筛+莫比乌斯反演)

题目描述

求题目分析

乍一看十分像裸莫比乌斯反演,然而$n$的范围让人望而却步
于是先变化一下式子
$Ans=\sum_{i=1}^n\sum_{j=1}^n(i,j)$
枚举$T=(i,j)$
$=\sum_{T=1}^n\sum_{i=1}^{\lfloor\frac nT\rfloor}\sum_{j=1}^{\lfloor\frac nT\rfloor}[(i,j)==1]\\=\sum_{T=1}^n\sum_{i=1}^{\lfloor\frac nT\rfloor}\sum_{j=1}^{\lfloor\frac nT\rfloor}\sum_{d|i,d|j}\mu(d)\\=\sum_{T=1}^nT\sum_{d=1}^{\lfloor\frac nT\rfloor}\mu(d){\lfloor\frac n{Td}\rfloor}^2$
令k=Td
$=\sum_{k=1}^n{\lfloor\frac n{k}\rfloor}^2\varphi(k)$
则此时可以整除分块优化,每次算出${\lfloor\frac n{k}\rfloor}$相等的上下界$i,j$后用莫比乌斯反演计算$(S\varphi(j)-S\varphi(i-1))$
由于计算$\varphi$的前缀和时记忆化处理过,所以在杜教筛外面再套了一个整除分块优化不会影响时间复杂度,复杂度仍是$\Theta(n^{\frac23})$

AC Code

#include <cstdio>
#include <map>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
const int MAXN = 5e6+1;
const int inv2 = 500000004;
map<LL, LL> S; LL s[MAXN];
int Prime[MAXN], Cnt, phi[MAXN];
bool IsnotPrime[MAXN];

void init()
{
  phi[1] = 1;
  for(int i = 2; i < MAXN; ++i)
  {
    if(!IsnotPrime[i]) Prime[++Cnt] = i, phi[i] = i-1;
    for(int j = 1; j <= Cnt && i * Prime[j] < MAXN; ++j)
    {
      IsnotPrime[i * Prime[j]] = 1;
      if(i % Prime[j] == 0)
      {
        phi[i * Prime[j]] = phi[i] * Prime[j];
        break;
      }
      phi[i * Prime[j]] = phi[i] * phi[Prime[j]];
    }
  }
  for(int i = 1; i < MAXN; ++i) s[i] = (s[i-1] + phi[i]) % mod;
}

inline LL sum(LL n)
{
  if(n < MAXN) return s[n];
  if(S.count(n)) return S[n];
  LL ret = (n%mod) * ((n+1)%mod) % mod * inv2 % mod;
  for(LL i = 2, j; i <= n; i=j+1)
  {
    j = n/(n/i);
    ret = (ret - sum(n/i) * ((j-i+1)%mod) % mod) % mod;
  }
  return S[n] = ret;
}

inline LL solve(LL n)
{
  LL ret = 0;
  for(LL i = 1, j; i <= n; i=j+1)
  {
    j = n/(n/i);
    ret = (ret + ((n/i)%mod) * ((n/i)%mod) % mod * ((sum(j)-sum(i-1))%mod) % mod) % mod;
  }
  return ret;
}
int main ()
{
  init(); LL n;
  scanf("%lld", &n);
  printf("%lld\n", (solve(n)+mod)%mod);
}