Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

SP7001 VLATTICE - Visible Lattice Points

LuoGu: SP7001 VLATTICE - Visible Lattice Points

SPOJ: VLATTICE - Visible Lattice Points

将贡献拆为所有坐标均不为 $0$ 的,一个坐标为 $0$ 的,两个坐标为 $0$ 的。

贡献分别为
$$
\sum _ {i = 1} ^ n \mu(i) S(\lfloor \frac n i \rfloor) ^ 3
$$

$$
3\sum _ {i = 1} ^ n \mu(i) S(\lfloor \frac n i \rfloor) ^ 2
$$

$$
3
$$

数论分块。

查看代码
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#include <cstdio>
using namespace std;
typedef long long LL;
const int N = 1e6 + 10;
int n;
bool vis[N];
int cnt, primes[N], mu[N], smu[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[i * primes[j]] = true;
if (i % primes[j] == 0)
{
mu[i * primes[j]] = 0;
break;
}
mu[i * primes[j]] = -mu[i];
}
}
for (int i = 1; i < N; i++)
smu[i] = smu[i - 1] + mu[i];
}
int main()
{
init();
int T;
scanf("%d", &T);
while (T--)
{
scanf("%d", &n);
LL res = 3;
for (int l = 1, r; l <= n; l = r + 1)
{
r = n / (n / l);
res += (LL)(smu[r] - smu[l - 1]) * (n / l) * (n / l) * (n / l + 3);
}
printf("%lld\n", res);
}
return 0;
}