P2257 YY的GCD
对于每个质数 $n$ ，有
$$
\begin {aligned}
f(n) = & \sum_{i = 1} ^ a \sum_{j = 1} ^ b [ gcd(i, j) = n ] \\
= & \sum_{i = 1} ^ {\frac a n} \sum_{j = 1} ^ {\frac b n} \sum_{d | gcd(i, j)} \mu(d) \\
= & \sum_{i = 1} ^ {\frac a n} \sum_{j = 1} ^ {\frac b n} \sum_{d | gcd(i, j)} \mu(d) \\
= & \sum_{d = 1} ^ {min({\frac a n}, {\frac b n})} \mu (d) \left \lfloor \frac a {dn} \right \rfloor \left \lfloor \frac b {dn} \right \rfloor \\
\end {aligned}
$$
令
$$
T = dn
$$
则
$$
f(n) = \sum_{T = 1} ^ {min(a, b)} \mu (\frac T n) \left \lfloor \frac a T \right \rfloor \left \lfloor \frac b T \right \rfloor \\
$$
其中前半部分可以预处理，对于每个质数的倍数累加进去。
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 cnt;
LL low[N], primes[N], mu[N], f[N], s[N];
void init()
{
mu[1] = 1;
low[1] = 1;
for (int i = 2; i < N; i++)
{
if (!low[i])
{
primes[++cnt] = i;
mu[i] = -1;
low[i] = i;
}
for (int j = 1; j <= cnt && i * primes[j] < N; j++)
{
if (i % primes[j] == 0)
{
if (low[i] == primes[j])
mu[i * primes[j]] = 0;
else
mu[i * primes[j]] = mu[i / low[i]] * mu[primes[j] * low[i]];
low[i * primes[j]] = low[i] * primes[j];
break;
}
mu[i * primes[j]] = mu[i] * mu[primes[j]];
low[i * primes[j]] = primes[j];
}
}
for (int i = 1; i <= cnt; i++)
for (int j = 1; j * primes[i] < N; j++)
f[j * primes[i]] += mu[j];
for (int i = 1; i < N; i++)
s[i] = s[i - 1] + f[i];
}
int main()
{
init();
int T;
scanf("%d", &T);
for (int n, m; T; T--)
{
scanf("%d%d", &n, &m);
if (n > m)
swap(n, m);
LL res = 0;
for (int l = 1, r; l <= n; l = r + 1)
{
r = min(n / (n / l), m / (m / l));
res += (s[r] - s[l - 1]) * (LL)(n / l) * (LL)(m / l);
}
printf("%lld\n", res);
}
return 0;
}
|