用于快速求积性函数前缀和。
公式：若 $S$ 是 $f$ 的前缀和函数，则 $\displaystyle g(1)S(n) = \sum _ {i = 1} ^ n (f * g) (i) - \sum _ {i = 2} ^ n g(i) S (\left \lfloor \frac n i \right \rfloor)$ ，其中 $g$ 为任意构造的积性函数以方便计算。
$$
\mu * 1 = \epsilon
$$
$$
\varphi * 1 = id
$$
$$
\mu * id = \varphi
$$
$$
1 * 1 = d
$$
$$
id * 1 = \sigma
$$
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| #include <cstdio>
#include <map>
using namespace std;
typedef long long LL;
const int N = 2e6 + 10;
bool vis[N];
int cnt, p[N];
map<int, LL> Sphi, Smu;
LL phi[N], mu[N], sphi[N], smu[N];
void init()
{
vis[1] = true;
phi[1] = 1;
mu[1] = 1;
for (int i = 2; i < N; i++)
{
if (!vis[i])
{
p[++cnt] = i;
phi[i] = i - 1;
mu[i] = -1;
}
for (int j = 1; j <= cnt && i * p[j] < N; j++)
{
vis[i * p[j]] = true;
if (i % p[j] == 0)
{
mu[i * p[j]] = 0;
phi[i * p[j]] = phi[i] * p[j];
break;
}
phi[i * p[j]] = phi[i] * phi[p[j]];
mu[i * p[j]] = mu[i] * mu[p[j]];
}
}
for (int i = 1; i < N; i++)
{
sphi[i] = sphi[i - 1] + phi[i];
smu[i] = smu[i - 1] + mu[i];
}
}
LL calmu(LL n)
{
if (n < N)
return smu[n];
if (Smu.count(n))
return Smu[n];
LL res = 1;
for (LL l = 2, r; l <= n; l = r + 1)
{
r = n / (n / l);
res -= (r - l + 1) * calmu(n / l);
}
return Smu[n] = res;
}
LL calphi(LL n)
{
if (n < N)
return sphi[n];
if (Sphi.count(n))
return Sphi[n];
LL res = n * (n + 1) / 2;
for (LL l = 2, r; l <= n; l = r + 1)
{
r = n / (n / l);
res -= (r - l + 1) * calphi(n / l);
}
return Sphi[n] = res;
}
int main()
{
init();
int T;
scanf("%d", &T);
for (int n; T; T--)
{
scanf("%d", &n);
printf("%lld %lld\n", calphi(n), calmu(n));
}
return 0;
}
|