P2568 GCD
对于每一个素数 $p$ ，对答案的贡献为：
$$
\begin{aligned}
&\sum _ {i = 1} ^ n \sum _ {j = 1} ^ n [gcd(i, j) = p]\\
=& \sum _ {i = 1} ^ n \sum _ {j = 1} ^ n \sum _ {d | gcd(\left \lfloor \frac i p\right \rfloor, \left \lfloor \frac j p\right \rfloor)} \mu (d)\\
=& \sum _ {i = 1} ^ \left \lfloor \frac n p\right \rfloor \sum _ {j = 1} ^ \left \lfloor \frac n p\right \rfloor \sum _ {d | i \wedge d | j} \mu (d)\\
=& \sum _ {d = 1} ^ \left \lfloor \frac n p\right \rfloor \mu (d) \sum _ {i = 1} ^ \left \lfloor \frac n p\right \rfloor [d |i] \sum _ {i = 1} ^ \left \lfloor \frac n p\right \rfloor [d |j]\\
=& \sum _ {d = 1} ^ \left \lfloor \frac n p\right \rfloor \mu (d) \left \lfloor \frac n {pd} \right \rfloor ^ 2\\
\end{aligned}
$$
筛出所有质数，同时筛出 $\mu$ 并预处理前缀和，询问时可以数论分块。
查看代码
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| #include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e7 + 10;
int n;
bool vis[N];
int cnt, primes[N], mu[N];
LL s[N];
void init()
{
mu[1] = 1;
for (int i = 2; i < N; i++)
{
if (!vis[i])
{
primes[++cnt] = i;
mu[i] = -1;
}
for (int j = 1; j <= cnt && i * primes[j] < N; j++)
{
vis[primes[j] * i] = true;
if (i % primes[j] == 0)
{
mu[primes[j] * i] = 0;
break;
}
mu[primes[j] * i] = -mu[i];
}
}
for (int i = 1; i < N; i++)
s[i] = s[i - 1] + mu[i];
}
int main()
{
init();
scanf("%d", &n);
LL res = 0;
for (int i = 1; primes[i] <= n; i++)
for (int l = 1, r; l <= n / primes[i]; l = r + 1)
{
r = n / (n / l);
res += (s[r] - s[l - 1]) * (LL)(n / primes[i] / l) * (LL)(n / primes[i] / l);
}
printf("%lld", res);
return 0;
}
|