【Luogu P2257】 YY的GCD
题目链接：
题目大意：
快速求：
\[\sum_{p\in Prime}\sum_{i=1}^{n}\sum_{j=1}^{m}\left[\operatorname{gcd}(i,j)==p\right]
\]
正文：
按照莫比乌斯反演的常规套路，将式子化简成我们能接受的时限：
\[\begin{aligned}\sum_{p\in Prime}\sum_{i=1}^{n}\sum_{j=1}^{m}\left[\operatorname{gcd}(i,j)==p\right] &=\sum_{p\in Prime}\sum_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{p}\rfloor}\left[\operatorname{gcd}(i,j)==1\right]\\&=\sum_{p\in Prime}\sum_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{p}\rfloor}\sum_{d|\operatorname{gcd}(i,j)}\mu(d)\\&=\sum_{p\in Prime} \sum_{d=1}\mu(d)\sum_{d|i}^{\lfloor\frac{n}{p}\rfloor}\sum_{k|j}^{\lfloor\frac{m}{p}\rfloor}1\\&=\sum_{p\in Prime}\sum_{d=1}\mu(d)\left\lfloor\frac{n}{pd}\right\rfloor\left\lfloor\frac{m}{pd}\right\rfloor\end{aligned}
\]
设 \(k=pd\) 枚举 \(k\)：
\[\begin{aligned}\sum_{p\in Prime}\sum_{d=1}\mu(d)\left\lfloor\frac{n}{pd}\right\rfloor\left\lfloor\frac{m}{pd}\right\rfloor&=\sum_{k=1}\left\lfloor\frac{n}{k}\right\rfloor\left\lfloor\frac{m}{k}\right\rfloor\sum_{p|k,p\in Prime}\mu(\frac{k}{p})\end{aligned}
\]
预处理 \(\mu(\frac{k}{p})\)，就能 \(O(t\sqrt{n})\)。
代码：
稍微卡卡常就能过了。
inline void prework()
{
miu[1] = 1;
for (int i = 2; i <= N - 10; i++)
{
if(!vis[i]) {pri[++cnt] = i, miu[i] = -1;}
for (int j = 1; j <= cnt && pri[j] * i <= N - 10; j++)
{
vis[pri[j] * i] = 1;
if (i % pri[j] == 0)
{
miu[i * pri[j]] = 0;
break;
}
else
miu[i * pri[j]] = -miu[i];
}
}
for (int i = 1; i <= cnt; i++)
for (int j = 1; j * pri[i] <= N - 10; j++)
sum[j * pri[i]] += miu[j];
for (int i = 1; i <= N - 10; i++)
sum[i] += sum[i - 1];
}
int main()
{
prework();
for (read(t); t--; )
{
ans = 0LL;
read(n);read(m);
if(n > m)
{
ll c = n; n = m; m = c;
}
for (register int l = 1, r; l <= n; l = r + 1)
{
r = min (n / (n / l), m / (m / l));
ans += (sum[r] - sum[l - 1]) * (n / l) * (m / l);
}
print(ans);
}
return 0;
}
