51nod1238. 最小公倍数之和 V3（莫比乌斯反演）

Posted 2021-01-27 imaginec

题目链接

https://www.51nod.com/Challenge/Problem.html#!#problemId=1238

题解

本来想做个杜教筛板子题结果用另一种方法过了......

所谓的“另一种方法”用到的技巧还是挺不错的，因此这里简单介绍一下。

首先还是基本的推式子：

[egin{aligned}sum_{i = 1}^n sum_{j = 1}^n { m lcm}(i, j) &= sum_{i = 1}^n sum_{j = 1}^n frac{ij}{{ m gcd}(i, j)} \ &= sum_{d = 1}^{n} sum_{i = 1}^n sum_{j = 1}^n[{ m gcd}(i, j) = d]frac{ij}{d} \ &= sum_{d = 1}^n dsum_{i = 1}^{leftlfloorfrac{n}{d} ight floor}sum_{j = 1}^{leftlfloorfrac{n}{d} ight floor}[{ m gcd}(i, j) = 1]ijend{aligned}]

设 (f(x) = sum_limits{i = 1}^x sum_limits{j = 1}^x [{ m gcd}(i, j) = 1]ij)，那么答案即为 (sum_limits{d = 1}^n d imes f(leftlfloorfrac{n}{d} ight floor))。显然答案可以数论分块求，因此我们的任务就是求 (f) 函数。

首先考虑当 (x) 较小时，我们能否直接预处理出 (f(x))。考虑差分：当 (x > 1) 时，(f(x)) 较 (f(x - 1)) 而言，多了的部分为：[left(sum_limits{i = 1}^xsum_limits{j = 1}^x [{ m gcd}(i, j) = 1]ij ight) - left(sum_limits{i = 1}^{x - 1}sum_limits{j = 1}^{x - 1} [{ m gcd}(i, j) = 1]ij ight) = 2x sum_{i = 1}^x [{ m gcd}(i, x) = 1]i]

而由于小于等于 (x(x > 1)) 且与 (x) 互质的数的和为 (frac{varphi(x)x}{2})（证明提示：当 (n geq 2) 时，若 ({ m gcd}(d, n) = 1) 必然有 ({ m gcd}(n - d, n) = 1)，与 (n) 互质的 (d) 共有 (varphi(n)) 个），因此我们就得到了 (f(x)) 的递推式：(f(x) = f(x - 1) + 2x imes frac{varphi(x)x}{2})，即 (f(x) = f(x - 1) + varphi(x)x^2)。

不过这只能处理 (x) 较小的情况。当 (x) 较大时，我们仍然得另谋他路。在 (f(x) = sum_limits{i = 1}^x sum_limits{j = 1}^x [{ m gcd}(i, j) = 1]ij) 当中，对 (f(x)) 有贡献的 (i, j) 满足 (i) 与 (j) 是互质的，我们考虑补集转化，用总和减去不互质的 (i, j) 的贡献：

[egin{aligned} f(x) &= sum_{i = 1}^xsum_{j = 1}^x ij - sum_{d = 2}^x sum_{i = 1}^x sum_{j = 1}^x [{ m gcd}(i, j) = d]ij \ &= left(frac{x(x+1)}{2} ight)^2 - sum_{d = 2}^x d^2 sum_{i = 1}^{leftlfloorfrac{x}{d} ight floor} sum_{j = 1}^{leftlfloorfrac{x}{d} ight floor} [{ m gcd}(i, j) = 1]ij \ &=left(frac{x(x+1)}{2} ight)^2 - sum_{d = 2}^x d^2 f(leftlfloorfrac{x}{d} ight floor) end{aligned}]

这样，我们就能够递归地去求解当 (x) 较大时 (f(x)) 的值了。不难发现，该求解方法的时间复杂度和杜教筛是一样的，为 (O(n^{frac{2}{3}}))，且非常好写。

代码

#include<bits/stdc++.h>

using namespace std;

const int mod = 1000000007, inv2 = 500000004, inv6 = 166666668, up = 10000001;

int main() {
  function<int (int, int)> mul = [&] (int x, int y) {
    return (long long) x * y % mod;
  };
  function<void (int&, int)> add = [&] (int& x, int y) {
    x += y;
    if (x >= mod) {
      x -= mod;
    }
  };
  function<void (int&, int)> sub = [&] (int& x, int y) {
    x -= y;
    if (x < 0) {
      x += mod;
    }
  };
  vector<bool> is_prime(up, true);
  vector<int> phi(up), primes;
  phi[1] = 1;
  for (int i = 2; i < up; ++i) {
    if (is_prime[i]) {
      primes.push_back(i);
      phi[i] = i - 1;
    }
    for (auto v : primes) {
      int d = v * i;
      if (d >= up) {
        break;
      }
      is_prime[d] = false;
      if (i % v == 0) {
        phi[d] = mul(phi[i], v);
        break;
      } else {
        phi[d] = mul(phi[i], phi[v]);
      }
    }
  }
  for (int i = 2; i < up; ++i) {
    phi[i] = mul(mul(phi[i], i), i);
    add(phi[i], phi[i - 1]);
  }
  function<int (long long)> sum_pow2 = [&] (long long n) {
    n %= mod;
    return mul(mul(mul(n, n + 1), (n * 2 + 1)), inv6);
  };
  map<long long, int> value;
  function<int (long long)> f = [&] (long long n) {
    if (value.count(n)) {
      return value[n];
    } else {
      int result = 0;
      if (n < up) {
        result = phi[n];
      } else {
        int x = n % mod;
        x = mul(mul(x, x + 1), inv2);
        result = mul(x, x);
        for (long long i = 2, last; i <= n; i = last + 1) {
          last = n / (n / i);
          sub(result, mul((sum_pow2(last) - sum_pow2(i - 1) + mod) % mod, f(n / i)));
        }
      }
      return value[n] = result;
    }
  };
  long long n;
  scanf("%lld", &n);
  int answer = 0;
  for (long long i = 1, last; i <= n; i = last + 1) {
    last = n / (n / i);
    add(answer, mul(mul(mul((i + last) % mod, (last - i + 1) % mod), inv2), f(n / i)));
  }
  printf("%d
", answer);
  return 0;
}