P4450 双亲数
$$
\begin{aligned}
f(n)
= & \sum_{i = 1} ^ a \sum_{j = 1} ^ b [ gcd(i, j) = n ] \\
= & \sum_{i = 1} ^ {\frac a n} \sum_{j = 1} ^ {\frac b n} \sum_{d | gcd(i, j)} \mu(d) \\
= & \sum_{i = 1} ^ {\frac a n} \sum_{j = 1} ^ {\frac b n} \sum_{d | gcd(i, j)} \mu(d) \\
= & \sum_{d = 1} ^ {min({\frac a n}, {\frac b n})} \mu (d) \left \lfloor \frac a {dn} \right \rfloor \left \lfloor \frac b {dn} \right \rfloor
\end{aligned}
$$
查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
| #include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e6 + 10;
int n, m, k;
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%d%d", &n, &m, &k);
n /= k, m /= k;
LL res = 0;
for (int l = 1, r; l <= min(n, m); 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;
}
|