SP19985 GCDEX2 - GCD Extreme (hard) 题解

不用莫反……

$$ \begin{aligned} &\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n\gcd(i,j)\\ =&\sum\limits_{i=1}^n\sum\limits_{j=1}^{i-1}\gcd(i,j)\\ =&\sum\limits_{d=1}^nd\sum\limits_{i=1}^n\sum\limits_{j=1}^{i-1}[\gcd(i,j)=d]\\ =&\sum\limits_{d=1}^nd\sum\limits_{i=1}^{\lfloor\frac nd\rfloor}\sum\limits_{j=1}^{i-1}[\gcd(i,j)=1]\\ =&\sum\limits_{d=1}^nd\left(-1+\sum\limits_{i=1}^{\lfloor\frac nd\rfloor}\varphi(i)\right) \end{aligned} $$

数论分块套杜教筛即可。复杂度 $O(n^{\frac 23})$。

#include <cstdio>
#include <unordered_map>
using namespace std;
unordered_map<unsigned long long, unsigned long long> F;
unsigned long long n, f[10000050];
int T, p[10000050];
bool v[10000050];
unsigned long long P(unsigned long long x)
{
    if (x <= 1e7)
        return f[x];
    if (F[x])
        return F[x];
    F[x] = x & 1 ? x * (x + 1 >> 1) : (x >> 1) * (x + 1);
    for (unsigned long long l = 2, r; l <= x; l = r + 1)
        r = x / (x / l), F[x] -= (r - l + 1) * P(x / l);
    return F[x];
}
unsigned long long Q(unsigned long long l, unsigned long long r)
{
    if (l + r & 1)
        return (r - l + 1 >> 1) * (l + r);
    else
        return (l + r >> 1) * (r - l + 1);
}
int main()
{
    f[1] = 1;
    for (int i = 2; i <= 1e7; ++i)
    {
        if (!v[i])
            p[++p[0]] = i, f[i] = i - 1;
        for (int j = 1; i * p[j] <= 1e7; ++j)
        {
            v[i * p[j]] = 1;
            if (i % p[j])
                f[i * p[j]] = f[i] * (p[j] - 1);
            else
            {
                f[i * p[j]] = f[i] * p[j];
                break;
            }
        }
    }
    for (int i = 1; i <= 1e7; ++i)
        f[i] += f[i - 1];
    scanf("%d", &T);
    while (T--)
    {
        scanf("%llu", &n);
        unsigned long long q = 0;
        for (unsigned long long l = 1, r; l <= n; l = r + 1)
            r = n / (n / l), q += Q(l, r) * (P(n / l) - 1);
        printf("%llu\n", q);
    }
    return 0;
}