SP7001 VLATTICE - Visible Lattice Points

题意

求： \[ 3+3\sum_{i=1}^n\sum_{j=1}^n[\gcd(i,j)==1]+\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n[\gcd(i,j,k)==1] \]

分析

没啥分析的，直接考虑化简

对于三元的部分有： \[ \begin{aligned} \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n[\gcd(i,j,k)==1]&=\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n\sum_{d|\gcd(i,j,k)}\mu(d)\\ &=\sum_{i=1}^n\mu(i)\left\lfloor\frac ni\right\rfloor^3 \end{aligned} \] 对于二元的部分有： \[ \begin{aligned} 3\sum_{i=1}^n\sum_{j=1}^n[\gcd(i,j)==1]&=3\sum_{i=1}^n\sum_{j=1}^n\sum_{d|\gcd(i,j)}\mu(d)\\ &=3\sum_{i=1}^n\mu(i)\left\lfloor\frac ni\right\rfloor^2 \end{aligned} \] 所以可以稍微整理一下： \[ \begin{aligned} ans&=3+3\sum_{i=1}^n\sum_{j=1}^n[\gcd(i,j)==1]+\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n[\gcd(i,j,k)==1]\\ &=3+3\sum_{i=1}^n\mu(i)\left\lfloor\frac ni\right\rfloor^2+\sum_{i=1}^n\mu(i)\left\lfloor\frac ni\right\rfloor^3\\ &=3+\sum_{i=1}^n\mu(i)\left\lfloor\frac ni\right\rfloor\times(3\left\lfloor\frac ni\right\rfloor+\left\lfloor\frac ni\right\rfloor^2) \end{aligned} \] 然后数论分块一下就可以了

Code

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int __read()
{
    int x(0), t(1);
    char o (getchar());
    while (o < '0' || o > '9') {
        if (o == '-') t = -1;
        o = getchar();
    }
    for (; o >= '0' && o <= '9'; o = getchar()) {
        x = (x << 1) + (x << 3) + (o ^ 48);
    }
    return x * t;
}

int T, n;
int mu[maxn], pr[maxn];
bool vis[maxn];

inline void init()
{
    mu[1] = 1;
    for (int i = 2; i < maxn; ++i) {
        if (!vis[i]) pr[++pr[0]] = i, mu[i] = - 1;
        for (int j = 1; j <= pr[0] && i * pr[j] < maxn; ++j) {
            vis[i * pr[j]] = 1;
            if (i % pr[j]) mu[i * pr[j]] = -mu[i];
            else break;
        }
    }
    for (int i = 1; i < maxn; ++i) mu[i] += mu[i - 1];
}

int main()
{
    init();
    T = __read();
    while (T--) {
        n = __read();
        ll ans(3);
        for (int l = 1, r; l <= n; l = r + 1) {
            r = n / (n / l);
            ans += 1ll * (mu[r] - mu[l - 1]) * (n / l) * (3ll * (n / l) + 1ll * (n / l) * (n / l));
        }
        printf ("%lld\n", ans);
    }
}