题目大意：求 $$\sum\limits_{i=1}^{a\sum\limits_{j=1}}b[gcd(i,j)=c]$$

题解：学会了狄利克雷卷积。

将艾弗森表达式转化成卷积的形式，在调换求和顺序，消去不必要的和式。最后利用除法分块+预处理的莫比乌斯函数前缀和在 \(O(\sqrt n)\) 时间内单次回答询问。

代码如下

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int maxn = 5e4 + 10;

int mu[maxn], sum[maxn];

vector<int> primes;

bool vis[maxn];

void RunLinearSieve() {

  mu[1] = 1, vis[1] = 1;

  int n = 5e4;

  for (int i = 2; i <= n; i++) {

    if (!vis[i]) {

      primes.push_back(i);

      mu[i] = -1;

    }

    for (int j = 0; i * primes[j] <= n; j++) {

      vis[i * primes[j]] = 1;

      if (i % primes[j] == 0) {

        mu[i * primes[j]] = 0;

        break;

      } else {

        mu[i * primes[j]] = -mu[i];

      }

    }

  }

  for (int i = 1; i <= n; i++) {

    sum[i] = sum[i - 1] + mu[i];

  }

}

int main() {

  ios::sync_with_stdio(false);

  cin.tie(0), cout.tie(0);

  int T;

  cin >> T;

  RunLinearSieve();

  while (T--) {

    LL a, b, c;

    cin >> a >> b >> c;

    a /= c, b /= c;

    LL range = min(a, b);

    LL ans = 0;

    for (int i = 1; i <= range; i++) {

      int j = min(a / (a / i), b / (b / i));

      ans += (LL)(sum[j] - sum[i - 1]) * (a / i) * (b / i);

      i = j;

    }

    cout << ans << endl;

  }

  return 0;

}