題目連結

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

題解

本來想做個杜教篩板子題結果用另一種方法過了......

所謂的“另一種方法”用到的技巧還是挺不錯的，因此這裡簡單介紹一下。

首先還是基本的推式子：

\[\begin{aligned}\sum_{i = 1}^n \sum_{j = 1}^n {\rm lcm}(i, j) &= \sum_{i = 1}^n \sum_{j = 1}^n \frac{ij}{{\rm gcd}(i, j)} \\ &= \sum_{d = 1}^{n} \sum_{i = 1}^n \sum_{j = 1}^n[{\rm gcd}(i, j) = d]\frac{ij}{d} \\ &= \sum_{d = 1}^n d\sum_{i = 1}^{\left\lfloor\frac{n}{d}\right\rfloor}\sum_{j = 1}^{\left\lfloor\frac{n}{d}\right\rfloor}[{\rm gcd}(i, j) = 1]ij\end{aligned}\]

設 \(f(x) = \sum_\limits{i = 1}^x \sum_\limits{j = 1}^x [{\rm gcd}(i, j) = 1]ij\)，那麼答案即為 \(\sum_\limits{d = 1}^n d\times f(\left\lfloor\frac{n}{d}\right\rfloor)\)。顯然答案可以數論分塊求，因此我們的任務就是求 \(f\) 函式。

首先考慮當 \(x\) 較小時，我們能否直接預處理出 \(f(x)\)。考慮差分：當 \(x > 1\) 時，\(f(x)\) 較 \(f(x - 1)\) 而言，多了的部分為：\[\left(\sum_\limits{i = 1}^x\sum_\limits{j = 1}^x [{\rm gcd}(i, j) = 1]ij\right) - \left(\sum_\limits{i = 1}^{x - 1}\sum_\limits{j = 1}^{x - 1} [{\rm gcd}(i, j) = 1]ij\right) = 2x \sum_{i = 1}^x [{\rm gcd}(i, x) = 1]i\]

而由於小於等於 \(x(x > 1)\) 且與 \(x\) 互質的數的和為 \(\frac{\varphi(x)x}{2}\)（證明提示：當 \(n \geq 2\) 時，若 \({\rm gcd}(d, n) = 1\) 必然有 \({\rm gcd}(n - d, n) = 1\)，與 \(n\) 互質的 \(d\) 共有 \(\varphi(n)\) 個），因此我們就得到了 \(f(x)\) 的遞推式：\(f(x) = f(x - 1) + 2x \times \frac{\varphi(x)x}{2}\)，即 \(f(x) = f(x - 1) + \varphi(x)x^2\)

不過這隻能處理 \(x\) 較小的情況。當 \(x\) 較大時，我們仍然得另謀他路。在 \(f(x) = \sum_\limits{i = 1}^x \sum_\limits{j = 1}^x [{\rm gcd}(i, j) = 1]ij\) 當中，對 \(f(x)\) 有貢獻的 \(i, j\) 滿足 \(i\) 與 \(j\) 是互質的，我們考慮補集轉化，用總和減去不互質的 \(i, j\) 的貢獻：

\[\begin{aligned} f(x) &= \sum_{i = 1}^x\sum_{j = 1}^x ij - \sum_{d = 2}^x \sum_{i = 1}^x \sum_{j = 1}^x [{\rm gcd}(i, j) = d]ij \\ &= \left(\frac{x(x+1)}{2}\right)^2 - \sum_{d = 2}^x d^2 \sum_{i = 1}^{\left\lfloor\frac{x}{d}\right\rfloor} \sum_{j = 1}^{\left\lfloor\frac{x}{d}\right\rfloor} [{\rm gcd}(i, j) = 1]ij \\ &=\left(\frac{x(x+1)}{2}\right)^2 - \sum_{d = 2}^x d^2 f(\left\lfloor\frac{x}{d}\right\rfloor) \end{aligned}\]

這樣，我們就能夠遞迴地去求解當 \(x\) 較大時 \(f(x)\) 的值了。不難發現，該求解方法的時間複雜度和杜教篩是一樣的，為 \(O(n^{\frac{2}{3}})\)，且非常好寫。

程式碼

#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\n", answer);
  return 0;
}