int solve(int n)
{
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (n + 1));
    long long res = 0;

    int t = sqrt(n);
    while (t * t < n) ++t;
    while (t * t > n) --t;
    for (int j = t; j > 1; --j)
    {
        for (int i = 1; i * j * j <= n; ++i)
        {
            if (!used[i * j * j])
            {
                used[i * j * j] = true;
                res += j - 1;
            }
        }
    }

    res %= MOD;
    return res;
}

Well, you see, since $$$(a, b, n)$$$ form a three-term geometric progression then you have $$$b^2 = an$$$.

It is not hard to prove that $$$F(N)$$$ will be our answer as we count over all possible squarefree $$$u$$$ for every fixed $$$p^2$$$.

We use $$$g(n)$$$ because its beautiful property, which make the calculation a lot easier.

$$$F(N) = \overset{n}{\underset{p=1}{\Large \Sigma}} f(p)$$$.

and

$$$f(n) = g(n) + 1$$$.

We have

$$$F(n)$$$

$$$= \overset{n}{\underset{p=2}{\Large \Sigma}} g(p) - n$$$

$$$= \overset{n}{\underset{p=2}{\Large \Sigma}} \underset{d^2 | p}{\Large \Sigma} \phi(d) - n$$$

Break the loop, make it simpler.

Reduce a for loop to constant.

$$$F(N)$$$

$$$= \overset{n}{\underset{p=1}{\Large \Sigma}} f(p)$$$

$$$= -n + \overset{n}{\underset{p=2}{\Large \Sigma}} g(p)$$$

$$$= -n + \overset{n}{\underset{p=2}{\Large \Sigma}} \underset{d^2 | p}{\Large \Sigma} \phi(d)$$$

$$$= -n + \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{p=1}{\Large \Sigma}} \underset{1 \leq u \times p^2 \leq n}{\Large \Sigma} \phi(p)$$$

$$$= -n + \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{p=1}{\Large \Sigma}} \phi(p) \times \left \lfloor \frac{n}{p^2} \right \rfloor$$$

$$$= \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{p=2}{\Large \Sigma}} \phi(p) \times \left \lfloor \frac{n}{p^2} \right \rfloor$$$

Well, you can sieve $$$\phi(p)$$$ for all $$$2 \leq p \leq \sqrt{n}$$$ in $$$O\left(\sqrt{n} \log \log \sqrt{n} \right)$$$ or improve it with linear sieve to $$$O(\sqrt{n})$$$.

Therefore you can solve the whole problem in $$$O(\sqrt{n})$$$.

#include <iostream>
#include <cstring>
#include <numeric>
#include <cmath>

using namespace std;

const int MOD = 1e9 + 7;
const int LIM = 1e7 + 17;
const int SQRT_LIM = ceil(sqrt(LIM) + 1) + 1;

int euler[SQRT_LIM];
void sieve_phi(int n)
{
    iota(euler, euler + n + 1, 0);
    for (int x = 2; x <= n; x++) if (euler[x] == x)
        for (int j = x; j <= n; j += x)
            euler[j] -= euler[j] / x;
}

int solve(int n)
{
    sieve_phi(ceil(sqrt(n) + 1) + 1);
    
    long long res = 0;
    for (int p = 2; p * p <= n; ++p)
        res += 1LL * euler[p] * (n / (p * p));

    res %= MOD;
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    
    int n;
    cin >> n;
    cout << solve(n);
    return 0;
}

#include <iostream>
#include <cstring>
#include <numeric>
#include <vector>
#include <cmath>

using namespace std;

const int MOD = 1e9 + 7;

vector<int> lpf;
vector<int> prime;
vector<int> euler;
void linear_sieve_phi(int n)
{
    lpf.assign(n + 1, 0);
    euler.assign(n + 1, 1);
    for (int x = 2; x <= n; ++x)
    {
        if (lpf[x] == 0)
        {
            prime.push_back(lpf[x] = x);
            euler[x] = x - 1;                    
        }
        for (int i = 0; i < prime.size() && x * prime[i] <= n; ++i)
        {
            lpf[x * prime[i]] = prime[i];
            if (x % prime[i] == 0) {
                euler[x * prime[i]] = euler[x] * prime[i];    
                break;
            }
            euler[x * prime[i]] = euler[x] * euler[prime[i]];
        }
    }
}

int solve(int n)
{
    linear_sieve_phi(ceil(sqrt(n) + 1) + 1);
    
    long long res = 0;
    for (int p = 2; p * p <= n; ++p)
        res += 1LL * euler[p] * (n / (p * p));

    res %= MOD;
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    
    int n;
    cin >> n;
    cout << solve(n);
    return 0;
}

$$$s_k(n) = C^{n-1}_{k-1}$$$

$$$f_k(n) = F_k(n) - F_k(n-1) = s_k(g_k(n)) = \underset{d^2 | n}{\Large \Sigma} (\mu \times s_k)(d) = \underset{d^2 | n}{\Large \Sigma} \underset{p | d}{\Large \Sigma} \mu(p) \times s_k\left(\frac{d}{p}\right)$$$

$$$F_k(n) = \overset{n}{\underset{p=1}{\Large \Sigma}} f_k(p)$$$

$$$= F_k(0) + \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{d = 1}{\Large \Sigma}}(\mu \times s_k)(d) \times \left \lfloor \frac{n}{d^2} \right \rfloor$$$

$$$= \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{d = 1}{\Large \Sigma}} \left \lfloor \frac{n}{d^2} \right \rfloor \times \left ( \underset{p | d}{\Large \Sigma} \mu(p) \times s_k\left(\frac{d}{p}\right) \right)$$$

const int LIM = 5e6 + 56;
const int SQRT_LIM = ceil(sqrt(LIM) + 1) + 1;
const int MOD = 1e9 + 7;

/// Precalculating factorials under prime modulo
int fact[SQRT_LIM + 10]; /// fact[n] = n!
int invs[SQRT_LIM + 10]; /// invs[n] = n^(-1)
int tcaf[SQRT_LIM + 10]; /// tcaf[n] = (n!)^(-1)
void precal_nck(int n = SQRT_LIM)
{
    fact[0] = fact[1] = 1;
    invs[0] = invs[1] = 1;
    tcaf[0] = tcaf[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
        fact[i] = (1LL * fact[i - 1] * i) % MOD;
        invs[i] = MOD - 1LL * (MOD / i) * invs[MOD % i] % MOD;
        tcaf[i] = (1LL * tcaf[i - 1] * invs[i]) % MOD;
    }
}

/// Calculating binomial coefficient queries
int nck(int n, int k)
{
    k = min(k, n - k);
    if (k < 0) return 0;

    long long res = fact[n];
    res *= tcaf[k];         res %= MOD;
    res *= tcaf[n - k];     res %= MOD;
    return res;
}

/// Linear Sieve
vector<int> prime;           /// prime list              = A000040
bool isPrime[SQRT_LIM + 10]; /// characteristic function = A010051
int lpf[SQRT_LIM + 10];      /// lowest prime factor     = A020639
int mu[SQRT_LIM + 10];       /// mobius                  = A008683
void linear_sieve(int n)
{
    if (n < 1) return ;
    /// Extension Sieve || You can add something more
    memset(lpf, 0, sizeof(lpf[0]) * (n + 1));
    fill_n(mu, n + 1, 1);
    /// Main Sieve || Without this, you barely able to achive linear complexity
    prime.clear();
    prime.reserve(n / log(n - 1));
    memset(isPrime, true, sizeof(isPrime[0]) * (n + 1));
    isPrime[0] = isPrime[1] = false;
    for (int x = 2; x <= n; ++x) /// For each number
    {
        if (isPrime[x]) /// Func[Prime]
        {
            mu[x] = -1;
            lpf[x] = x;
            prime.push_back(x);
        }
        for (int p : prime) /// Func[Prime * X] <- Func[Prime]
        {
            if (p > lpf[x] || x * p > n) break;
            isPrime[x * p] = 0;
            lpf[x * p] = p;
            mu[x * p] = (lpf[x] == p) ? 0 : -mu[x];
        }
    }
}

/// Divisor sieve
vector<int> divisors[SQRT_LIM];
void precal_div(int n) /// O(n log n)
{
    for (int u = n; u >= 1; --u)
    {
        divisors[u].clear();
        for (int v = u; v <= n; v += u)
            divisors[v].push_back(u);
    }
}

/// Solving for n, k
long long solve(int n, int k)
{
    /// We only care for d that 1 <= d <= sqrt(n)
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);
    precal_div(t);

    long long res = 0;
    for (int d = 1; d * d <= n; ++d) /// For each fixed p^2
    {
        long long sum = 0;
        for (int p : divisors[d]) /// For each (p | d)
            sum += mu[d / p] * nck(p - 1, k - 1);

        sum %= MOD;
        res += sum * (n / (d * d));
    }

    res %= MOD;
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    /// Assumming constant p = 10^9 + 7
    int n, k;
    cin >> n >> k;
    cout << solve(n, k);
    return 0;
}

vector<int> prime;           /// prime list              = A000040
bool isPrime[SQRT_LIM + 10]; /// characteristic function = A010051
int lpf[SQRT_LIM + 10];      /// lowest prime factor     = A020639
void linear_sieve(int n)
{
    if (n < 1) return ;
    prime.clear();
    prime.reserve(n / log(n - 1));
    memset(lpf, 0, sizeof(lpf[0]) * (n + 1));
    memset(isPrime, true, sizeof(isPrime[0]) * (n + 1));
    isPrime[0] = isPrime[1] = false;
    for (int x = 2; x <= n; ++x)
    {
        if (isPrime[x]) /// Func[Prime]
        {
            lpf[x] = x;
            prime.push_back(x);
        }
        for (int p : prime) /// Func[Prime * X] <- Func[Prime]
        {
            if (p > lpf[x] || x * p > n) break;
            isPrime[x * p] = 0;
            lpf[x * p] = p;
        }
    }
}

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        res[d] = nck(d - 1, k - 1);

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

The complexity depends on Möbius function, traverse over all divisor of each number and calculating binomial coefficients

We only care about $$$d$$$ that $$$1 \leq d \leq \sqrt{n}$$$

We can use factorial formula for binomial coefficients as $$$1 \leq k \leq n < p$$$

As we only use $$$d$$$ that $$$d^2 \leq n$$$, let $$$t = \sqrt{n}$$$ for convenient

We can use linear sieve for $$$\mu{d}$$$ in $$$O(t)$$$

We can use divisor sieve for finding all divisors $$$p$$$ for each $$$v$$$ in $$$O(t \log t)$$$

We can calculate the result in $$$O(t \log t)$$$

Therefore the overall complexity is $$$O(t \log t)$$$

The complexity depends on sieve, calculating result and calculating binomial coefficients

The sieve and binomial coefficients is obviously $$$O(t)$$$ complexity for $$$t = \sqrt{n}$$$

The calculating part is just simply related to the sum of inversion of primes

Therefore we achived $$$O(t \log t)$$$ complexity for $$$t = \sqrt{n}$$$

Literally what we do up to now just optimize the counting part. The counting, which is the core is as the same as original.

The counting part is only about the number of valid sequences bounded by $$$n$$$, or you can say, the number of possible $$$p^2$$$ you can choose when you fix squarefree number $$$u$$$.

And those stuffs above $$$\mu(n)$$$, $$$\phi(n)$$$, sieving or anything else are just optimizations.

$$$\binom{n + k - 1}{k}$$$

and when you fix one number as a similar idea:

$$$s_k(n) = \binom{n + k - 2}{k - 1}$$$

It is standard stars and bars problem.

Instead of thinking the number of way to select $$$a_i$$$.

You can think of the number of value $$$a_i$$$ you have used.

Let $$$x_v$$$ is the number of value $$$v$$$ you used, or the number of $$$a_i = v$$$ in this sequence.

Each sequence can be represented uniquely as a sequence of $$$x_1, x_2, \dots, x_n$$$.

Since we selected $$$k$$$ in total we have $$$x_1 + x_2 + \dots + x_n = k$$$.

Let $$$x_v$$$ is the number of value $$$v$$$ you used, or the number of $$$a_i = v$$$ in this sequence.

For each box $$$v = 1 \dots n$$$, draw $$$x_v$$$ stars ⭐.

You can make $$$k - 1$$$ cut between these box (not cut into the box), and split into $$$k$$$ intervals.

The sum of each interval is the value of selected element.

Since there are $$$n + k - 1$$$ boxes, and you use $$$k - 1$$$ cuts, there are the total of $$$\binom{n + k - 1}{k}$$$ different ways.

int fact[SQRT_LIM + 10];
int invs[SQRT_LIM + 10];
int tcaf[SQRT_LIM + 10];
void precal_nck(int n = SQRT_LIM)
{
    fact[0] = fact[1] = 1;
    invs[0] = invs[1] = 1;
    tcaf[0] = tcaf[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
        fact[i] = (1LL * fact[i - 1] * i) % MOD;
        invs[i] = MOD - 1LL * (MOD / i) * invs[MOD % i] % MOD;
        tcaf[i] = (1LL * tcaf[i - 1] * invs[i]) % MOD;
    }
}

int nck(int n, int k)
{
    k = min(k, n - k);
    if (k < 0) return 0;

    long long res = fact[n];
    res *= tcaf[k];         res %= MOD;
    res *= tcaf[n - k];     res %= MOD;
    return res;
}

bool is_squarefree[LIM];
int solve(int n, int k)
{
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (n + 1));
    precal_nck(2 * n + 1);

    long long res = 0;
    for (int i = 1, j; i <= n; ++i) if (is_squarefree[i]) 
    {
        for (j = 1; i * j * j <= n; ++j)
            is_squarefree[i * j * j] = false;

        res += nck(k + j - 2, k);
    }

    res %= MOD;
    return res;
}

const int LIM = 5e6 + 56;
const int SQRT_LIM = ceil(sqrt(LIM) + 1) + 1;
const int MOD = 1e9 + 7;

/// Precalculating factorials under prime modulo
int fact[SQRT_LIM + 10]; /// fact[n] = n!
int invs[SQRT_LIM + 10]; /// invs[n] = n^(-1)
int tcaf[SQRT_LIM + 10]; /// tcaf[n] = (n!)^(-1)
void precal_nck(int n = SQRT_LIM)
{
    fact[0] = fact[1] = 1;
    invs[0] = invs[1] = 1;
    tcaf[0] = tcaf[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
        fact[i] = (1LL * fact[i - 1] * i) % MOD;
        invs[i] = MOD - 1LL * (MOD / i) * invs[MOD % i] % MOD;
        tcaf[i] = (1LL * tcaf[i - 1] * invs[i]) % MOD;
    }
}

/// Calculating binomial coefficient queries
int nck(int n, int k)
{
    k = min(k, n - k);
    if (k < 0) return 0;

    long long res = fact[n];
    res *= tcaf[k];         res %= MOD;
    res *= tcaf[n - k];     res %= MOD;
    return res;
}

/// Linear Sieve
vector<int> prime;           /// prime list              = A000040
bool isPrime[SQRT_LIM + 10]; /// characteristic function = A010051
int lpf[SQRT_LIM + 10];      /// lowest prime factor     = A020639
int mu[SQRT_LIM + 10];       /// mobius                  = A008683
void linear_sieve(int n)
{
    if (n < 1) return ;
    /// Extension Sieve || You can add something more
    memset(lpf, 0, sizeof(lpf[0]) * (n + 1));
    fill_n(mu, n + 1, 1);
    /// Main Sieve || Without this, you barely able to achive linear complexity
    prime.clear();
    prime.reserve(n / log(n - 1));
    memset(isPrime, true, sizeof(isPrime[0]) * (n + 1));
    isPrime[0] = isPrime[1] = false;
    for (int x = 2; x <= n; ++x) /// For each number
    {
        if (isPrime[x]) /// Func[Prime]
        {
            mu[x] = -1;
            lpf[x] = x;
            prime.push_back(x);
        }
        for (int p : prime) /// Func[Prime * X] <- Func[Prime]
        {
            if (p > lpf[x] || x * p > n) break;
            isPrime[x * p] = 0;
            lpf[x * p] = p;
            mu[x * p] = (lpf[x] == p) ? 0 : -mu[x];
        }
    }
}

/// Divisor sieve
vector<int> divisors[SQRT_LIM];
void precal_div(int n) /// O(n log n)
{
    for (int u = n; u >= 1; --u)
    {
        divisors[u].clear();
        for (int v = u; v <= n; v += u)
            divisors[v].push_back(u);
    }
}

/// Solving for n, k
long long solve(int n, int k)
{
    /// We only care for d that 1 <= d <= sqrt(n)
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);
    precal_div(t);

    long long res = 0;
    for (int d = 1; d * d <= n; ++d) /// For each fixed p^2
    {
        long long sum = 0;
        for (int p : divisors[d]) /// For each (p | d)
            sum += mu[d / p] * nck(d + k - 2, k - 1);

        sum %= MOD;
        res += sum * (n / (d * d));
    }

    res %= MOD;
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    /// Assumming constant p = 10^9 + 7
    int n, k;
    cin >> n >> k;
    cout << solve(n, k);
    return 0;
}

int fact[SQRT_LIM + 10];
int invs[SQRT_LIM + 10];
int tcaf[SQRT_LIM + 10];
void precal_nck(int n = SQRT_LIM)
{
    fact[0] = fact[1] = 1;
    invs[0] = invs[1] = 1;
    tcaf[0] = tcaf[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
        fact[i] = (1LL * fact[i - 1] * i) % MOD;
        invs[i] = MOD - 1LL * (MOD / i) * invs[MOD % i] % MOD;
        tcaf[i] = (1LL * tcaf[i - 1] * invs[i]) % MOD;
    }
}

int nck(int n, int k)
{
    k = min(k, n - k);
    if (k < 0) return 0;

    long long res = fact[n];
    res *= tcaf[k];         res %= MOD;
    res *= tcaf[n - k];     res %= MOD;
    return res;
}

vector<int> prime;           /// prime list              = A000040
bool isPrime[SQRT_LIM + 10]; /// characteristic function = A010051
int lpf[SQRT_LIM + 10];      /// lowest prime factor     = A020639
void linear_sieve(int n)
{
    if (n < 1) return ;
    prime.clear();
    prime.reserve(n / log(n - 1));
    memset(lpf, 0, sizeof(lpf[0]) * (n + 1));
    memset(isPrime, true, sizeof(isPrime[0]) * (n + 1));
    isPrime[0] = isPrime[1] = false;
    for (int x = 2; x <= n; ++x)
    {
        if (isPrime[x]) /// Func[Prime]
        {
            lpf[x] = x;
            prime.push_back(x);
        }
        for (int p : prime) /// Func[Prime * X] <- Func[Prime]
        {
            if (p > lpf[x] || x * p > n) break;
            isPrime[x * p] = 0;
            lpf[x * p] = p;
        }
    }
}

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t + k);
    precal_nck(t + k);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        res[d] = nck(d + k - 2, k - 1);

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

The complexity is depended mainly on sieving part and binomial coefficient precalculation

Sieving can be done in linear

Binomial coefficient can also done been linear assumming $$$p$$$ is a sufficient large prime greater than $$$max(n, k)$$$

Like task B the algorithm is depended on calculating result and binomial coefficient precalculation

Yet we proved calculating result only take $$$O(\sqrt{n} \log \sqrt{n})$$$ [2] and $$$O(\sqrt{n} \log \log \sqrt{n})$$$ [3]

And we need to calculate binomial coefficient upto $$$\binom{d + k - 2}{k - 1}$$$ for $$$d = \left \lfloor \sqrt{n} \right \rfloor$$$ so we need $$$O(\sqrt{n} + k)$$$ precalculation time.

You optimize the problem using this

Then you can solve the problem in $$$O(\sqrt{n} \log \log \sqrt{n} + \sqrt{p} \log^q p)$$$ ($$$q$$$ is depended on how you implement it).

/// SPyofgame linear template for precalculating factorials under large prime modulo
int fact[SQRT_LIM + 10]; /// fact[n] = n!
int invs[SQRT_LIM + 10]; /// invs[n] = n^(-1)
int tcaf[SQRT_LIM + 10]; /// tcaf[n] = (n!)^(-1)
void precal_nck(int n = SQRT_LIM)
{
    fact[0] = fact[1] = 1;
    invs[0] = invs[1] = 1;
    tcaf[0] = tcaf[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
        fact[i] = (1LL * fact[i - 1] * i) % MOD;
        invs[i] = MOD - 1LL * (MOD / i) * invs[MOD % i] % MOD;
        tcaf[i] = (1LL * tcaf[i - 1] * invs[i]) % MOD;
    }
}

/// Calculating binomial coefficient queries
int nck(int n, int k)
{
    k = min(k, n - k);
    if (k < 0) return 0;

    long long res = fact[n];
    res *= tcaf[k];         res %= MOD;
    res *= tcaf[n - k];     res %= MOD;
    return res;
}

vector<int> factor;
int factorize(int n) /// Calculating phi(n) while factorizing (n) in O(sqrt n)
{
    factor.clear();
    int phi = n;

    if (!(n & 1))
    {
        n >>= __builtin_ctz(n);
        factor.push_back(2);
        phi -= phi / 2;
    }

    for (int x = 3; x * x <= n; x += 2)
    {
        if (n % x == 0)
        {
            do n /= x; while (n % x == 0);
            factor.push_back(x);
            phi -= phi / x;
        }
    }

    if (n > 1)
    {
        factor.push_back(n);
        phi -= phi / n;
    }

    return phi;
}

int f[LIM];    /// f[x] = nck(n, x)
int fact[LIM]; /// n! 
int tcaf[LIM]; /// n!^(-1)
int divp[LIM]; /// x but ignore all factors p[i]
int cntp[LIM][LOG_LIM]; /// cntp[x][i] = Number of time factor p[i] appear in 1..x
void precal(int MOD) /// Calculate f[x] for all x = 1 -> n in O(n log mod + sqrt mod)
{
    int PHIMOD = factorize(MOD);
    for (int x = 1; x <= n; ++x) /// For each part x in n!
    {
        int &t = divp[x] = x;
        for (int i = 0; i < factor.size(); ++i) /// Ignore all factor p[i] of p
        {
            cntp[x][i] = cntp[x - 1][i];
            for (; t % factor[i] == 0; t /= factor[i]) /// Count how many times p[i] appears in 1..n
                ++cntp[x][i];
        }
    }

    fact[0] = fact[1] = 1;
    tcaf[0] = tcaf[1] = 1;
    for (int x = 2; x <= n; ++x) /// Finding n! and n!^(-1)
    {
        fact[x] = (1LL * fact[x - 1] * divp[x]) % MOD;
        tcaf[x] = powMOD(fact[x], PHIMOD - 1, MOD);
    }

    memset(f, 0, sizeof(f[0]) * k);
    for (int x = k; x <= n; ++x)
    {
        /// Calculate nck % p normally
        f[x] = fact[x];
        mulMOD(f[x], tcaf[k], MOD);
        mulMOD(f[x], tcaf[x - k], MOD);
        for (int i = 0; i < factor.size(); ++i) /// Bringing those factors back
        {
            int p = cntp[x][i] - cntp[k][i] - cntp[x - k][i];
            f[x] = 1LL * f[x] * powMOD(factor[i], p, MOD) % MOD;
        }
    }
}

Each number is visited once and the transistion only cost $$$O(1)$$$

As you might notice $$$p$$$ is atleast prime, or atmost a squarefree number.

Since the complexity also depends on the factors of $$$p$$$, in the worst case, $$$p = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times \dots$$$ having most factors that is about $$$\log(p)$$$.

The total complexity also depended on factorizing $$$p$$$ and calculating $$$\phi{p}$$$.

Yet for normal trial division you can solve in $$$O(\sqrt{p})$$$ time.

You can also do all of this in $$$O(n \log p) + Õ(p^{1/4})$$$

The sequence contain positive integers only.

The sequence contain negative integers only.

The sequence contain a zero, and the rest are positive integers.

The sequence contain a zero, and the rest are negative integers.

There cant be both negative number and positive number at the same time

You can flip all the sign to positive, thus negative integers sequence are now the same as positive sequences.

Without zero, the problem has no difference from the original task.

With zero, any product with it is already a perfect square.

Hence zeros dont affect anything other than the remaining number you have not selected.

The sequence contain both negative and positive that satified the problem is not existed as their product is not a perfect square (we only care in integers not complex)

The sequence contain positive integers only:

• This is the original task as positive integers only means $$$1 \leq a_i \leq n$$$.

The sequence contain negative integers only:

• Flip all the sign doesnt change the product for each pairs.
• The problem now is the same as the sequence contain positive integers only.

The sequence contain a zero, and the rest are positive integers:

• The product of a zero with any number is still a perfect square.
• So the zero only reduce the number you have to select.
• Therefore this is same as the original task but now you have to select $$$k - 1$$$ numbers.

The sequence contain a zero, and the rest are negative integers:

• Just also flip the sign, the answer is the same as above.

The sequence contain zeros only.

The sequence contain positive integers only.

The sequence contain negative integers only.

The sequence contain $$$0 < t < k$$$ positive integers, and the rest $$$k - t$$$ are zeros.

The sequence contain $$$0 < t < k$$$ positive integers, and the rest $$$k - t$$$ are zeros.

The sequence contain zeros only.

The sequence contain $$$0 < t \leq k$$$ non-zero integers, and the rest $$$k - t$$$ are zeros.

You might wondering where is the $$$2 \times \dots$$$ come from.

Yet it is the same as unique number cases.

You have the answer for positive numbers are the same as negative number when you flip the sign.

So you have $$$1 \times \dots$$$ for positive, and $$$1 \times \dots$$$ for negative.

Therefore you have $$$2 \times \dots$$$ in total

You might wondering why we dont have a formula like this:

$$$task_E(n, k) = 1 + \overset{k}{\underset{t = 1}{\Large \Sigma}} C^t_k \times task_B(n, t)$$$.

But you know, since the sequence is ordered.

No matter which number you selected as zero there is one and only one way to sort it back.

You see, there is only one case for fully zeros sequence.

Isnt this sum is familiar to some standard math problem related to binomial coefficient ?

This formula is related to $$$\overset{k}{\underset{t = 0}{\Large \Sigma}} \binom{n + t}{t} = \binom{n + k + 1}{k}$$$

$$$s_k(n) = \overset{k}{\underset{t = 1}{\Large \Sigma}} \binom{d + t - 2}{t - 1} \times 2 = \binom{d + k - 1}{k - 1} \times 2$$$

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
    {
        res[d] += (k >= 1) * nck(d - 1, k - 1) * 2;
        res[d] += (k >= 2) * nck(d - 1, k - 2) * 2;  
    }

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

bool is_squarefree[LIM];
int brute(int n, int k)
{
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (n + 1));
    precal_nck(2 * n + 1);

    long long res = 0;
    for (int i = 1, j; i <= n; ++i) if (is_squarefree[i]) 
    {
        for (j = 1; i * j * j <= n; ++j)
            is_squarefree[i * j * j] = false;

        res += nck(k + j - 2, k);
    }

    res %= MOD;
    return res;
}


long long solve(int n, int k)
{
    long long res = 1;
    for (int t = 1; t <= k; ++t)
        res += brute(n, t) * 2;

    res %= MOD;
    return res;
}

long long res[SQRT_LIM + 10];
long long brute(int n, int k)
{
    /// We only care for d that 1 <= d <= sqrt(n)
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t + k);
    precal_div(t);

    long long res = 0;
    for (int d = 1; d * d <= n; ++d) /// For each fixed p^2
    {
        long long sum = 0;
        for (int p : divisors[d]) /// For each (p | d)
            sum += mu[d / p] * nck(d + k - 2, k - 1);

        sum %= MOD;
        res += sum * (n / (d * d));
    }

    res %= MOD;
    return res;
}

long long solve(int n, int k)
{
    long long res = 1;
    for (int t = 1; t <= k; ++t)
        res += brute(n, t) * 2;

    res %= MOD;
    return res;
}

long long res[SQRT_LIM + 10];
long long brute(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t + k);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        res[d] = nck(d + k - 2, k - 1);

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

long long solve(int n, int k)
{
    long long res = 1;
    for (int t = 1; t <= k; ++t)
        res += brute(n, t) * 2;

    res %= MOD;
    return res;
}

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t + k);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        for (int t = 1; t <= k; ++t)
            res[d] += nck(d + t - 2, t - 1) * 2;

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 1;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t + k);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        res[d] += nck(d + k - 1, k - 1) * 2;

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 1;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

Case $$$1$$$: $$$L \leq 0 \leq R$$$

Case $$$2$$$: $$$0 \leq L \leq R$$$

Case $$$3$$$: $$$L \leq R \leq 0$$$

Case $$$4$$$: $$$L > R$$$

It is $$$task_B(|L|, k) + task_B(|L|, k - 1) + task_B(|R|, k) + task_B(|R|, k - 1)$$$

This is what we are going to discuss about.

A simple formula is fox each squarefree $$$u$$$ in $$$1 \dots R$$$.

Count the number of $$$p^2$$$ in $$$[L, R]$$$.

Using combinatorics to count the number of way to select $$$p^2$$$ for each $$$u$$$.

The answer is sum over all result for $$$u$$$.

Same as case 2 when you flip the sign but also remember to swap $$$(L, R)$$$ too.

Just simply $$$0$$$.

Highly notice that fixed squarefree $$$u$$$ might also in $$$[1, L - 1]$$$ and not just in $$$[L, R]$$$.

Also remember, just count the $$$p^2$$$ in $$$[L, R]$$$.

Ofcourse we can just optimize it by using squarefree sieve in $$$O(R - L + 1)$$$ time.

Yet again, let use linear sieve and linear precalculation for binomial coefficient too.

We just need $$$O(R - L + 1)$$$ space.

But this lead us to another problem, how can we check if $$$[u]$$$ is squarefree when we literally remove the ability to check for any $$$u < L$$$ ???

A simple approach is just to redefine the squarefree sieve.

You see, when we go to the first $$$u$$$ that not marked in the sieve, we mark all form of $$$u \times p^2$$$ except $$$u$$$.

Why dont we change a little bit, let say, we check for the first $$$u \times c^2$$$ for some $$$c > 0$$$. And then mark all form of $$$u \times p^2$$$ except $$$u \times c^2$$$.

Yes, we just move the $$$u$$$ to the a number $$$u \times c^2$$$ that suitable for it and marked only once for each unique $$$u$$$.

How do we find $$$c$$$ ?

We can take $$$c$$$ as the smallest number that $$$u \times c^2 \geq L$$$.

So by this definition, $$$c = \left \lceil \sqrt{\frac{L}{u}} \right \rceil$$$.

• Yes

• Yes

• Yes

• Yes

Factorizing numbers.

Well, not sure what "bad" you are talking about.

But if it is about the complexity then each number $$$x$$$ in $$$[L, R]$$$ can be factorized in $$$O(\sqrt{x})$$$.

Yet for $$$L \approx 1$$$ this will about to be $$$O(R \sqrt{R} \log{R})$$$ if you implement badly.

Though it is so, let just ignore the time complexity for now.

Yes, you can use pollard rho to factorize in expected time $$$Õ(x^{1/4})$$$.

Also remember to combine it with miller rabin for prime cases.

For more detail, you can read this https://en.wikipedia.org/wiki/Pollard%27s_rho_algorithm

Yes, you can sieve for first $$$\sqrt{r}$$$ number for primes.

And for each number in $$$[L, R]$$$ you just need $$$min(\sqrt{x}, \sqrt{r} / \log {r})$$$ time to factorize it.

But in practice this is significant fast.

Though I am not quite sure what is the actual complexity of doing so.

There are rare amount of papers related about this specific topics and even worse that most of it are not well researched.

You can just iterative for each form of $$$u \times p^2$$$ that lies in $$$[L, R]$$$.

Also remember that each number should be visited once.

Not sure what you mean though.

When you factorize $$$x$$$, you can get the form $$$x = u \times p^2$$$.

Just remember dont take the whole $$$p^2$$$ but take $$$p$$$ and iterate using $$$p^2$$$, each step increase $$$p$$$ by one.

But for each prime factor $$$f$$$ in $$$x$$$ that appear more than once, we reduce $$$x$$$ by $$$f^2$$$ and increase $$$p$$$ by $$$f$$$ where $$$p$$$ is initially $$$1$$$.

• Yes

• Yes

• Yes

• Yes

• Yes

• Yes

• Yes

Sieve all prime up to $$$\sqrt{R}$$$.

Initialize $$$u_x$$$ — the squarefree part of each number $$$x$$$ in $$$[L, R]$$$ is $$$x$$$ itself.

Also initialize $$$p_x$$$ — the squarefree factor part of each number $$$x$$$ in $$$[L, R]$$$ is simply $$$1$$$.

Now for each prime $$$p$$$ we sieved, you factorize each number $$$x$$$, reduce $$$u_x$$$ by $$$p^2$$$ while increase $$$p_x$$$ by $$$p$$$.

Now for each number $$$x$$$ in $$$[L, R]$$$, you can do squarefree sieve like normal, with the steps of $$$p_x$$$ initially with $$$u_x$$$.

bool is_squarefree[LIM + 10];
int solve(int l, int r, int k)
{
    if (l > r) return 0;
    if (r < 0) return solve(-r, -l, k);
    if (l <= 0 && 0 <= r)
    {
        long long res = 0LL + taskB(abs(l), k) + taskB(abs(l), k - 1) + taskB(abs(r), k) + taskB(abs(r), k - 1);
        while (res >= MOD) res -= MOD;
        return res;
    }

    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (r + l - 1));
    precal_nck(r - l + 1);

    int tot = r - l + 1;
    long long res = 0;
    for (int i = 1; i <= r; ++i)
    {
        int j = sqrt(l / i);
        while (i * j * j > l) --j;
        while (i * j * j < l) ++j;
        if (is_squarefree[i * j * j - l])
        {
            int cnt = 0;
            for (; i * j * j <= r; ++j)
            {
                ++cnt;
                --tot;
                is_squarefree[i * j * j - l] = false;
            }
            res += nck(cnt, k);
            if (tot == 0) break;
        }
    }

    res %= MOD;
    return res;
}

long long res[SQRT_LIM + 10];
long long taskB(int n, int k)
{
    if (k == 0) return 0;
    if (k == 1) return n;

    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        res[d] = nck(d - 1, k - 1);

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

bool is_squarefree[LIM + 10];
int solve(int l, int r, int k)
{
    if (l > r) return 0;
    if (r < 0) return solve(-r, -l, k);
    if (l <= 0 && 0 <= r)
    {
        long long res = 0LL + taskB(abs(l), k) + taskB(abs(l), k - 1) + taskB(abs(r), k) + taskB(abs(r), k - 1);
        while (res >= MOD) res -= MOD;
        return res;
    }
    
    int t = ceil(sqrt(r + 1) + 1) + 1;
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (r + l - 1));
    precal_nck(r - l + 1);
    linear_sieve(t);

    long long res = 0;
    for (int x = l; x <= r; ++x) if (is_squarefree[x - l])
    {
        int u = x;
        int v = 1;
        for (int p : prime)
        {
            int q = p * p;
            if (q > u) break;
            while (u % q == 0)
            {
                u /= q;
                v *= p;
            }
        }

        int cnt = 0;
        for (; u * v * v <= r; ++v)
        {
            is_squarefree[u * v * v - l] = false;
            ++cnt;
        }

        res += nck(cnt, k);
    }

    res %= MOD;
    return res;
}

long long res[SQRT_LIM + 10];
long long taskB(int n, int k)
{
    if (k == 0) return 0;
    if (k == 1) return n;

    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        res[d] = nck(d - 1, k - 1);

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

#include <algorithm>
#define all(x) (x).begin(), (x).end()
bool is_squarefree[LIM + 10];
int squarefree[LIM + 10];
int sqf_factor[LIM + 10];
int solve(int l, int r, int k)
{
    if (l > r) return 0;
    if (r < 0) return solve(-r, -l, k);
    if (l <= 0 && 0 <= r)
    {
        long long res = 0LL + taskB(abs(l), k) + taskB(abs(l), k - 1) + taskB(abs(r), k) + taskB(abs(r), k - 1);
        while (res >= MOD) res -= MOD;
        return res;
    }
    
    int t = ceil(sqrt(r + 1) + 1) + 1;
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (r + l - 1));
    precal_nck(r - l + 1);
    linear_sieve(t);
    
    for (int x = l; x <= r; ++x)
    {
        squarefree[x - l] = x;
        sqf_factor[x - l] = 1;
    }

    long long res = 0;
    for (int p : prime)
    {
        for (int q = p * p, x = max(q, (l + q - 1) / q * q); x <= r; x += q)
        {
            while (squarefree[x - l] % q == 0)
            {
                squarefree[x - l] /= q;
                sqf_factor[x - l] *= p;
            }
        }
    }
    
    for (int x = l; x <= r; ++x)
    {
        int u = squarefree[x - l];
        int p = sqf_factor[x - l];
        if (is_squarefree[x - l])
        {
            int cnt = 0;
            for (; u * p * p <= r; ++p)
            {
                is_squarefree[u * p * p - l] = false;
                ++cnt;
            }
    
            res += nck(cnt, k);   
        }
    }
    
    res %= MOD;
    return res;
}

Obviously $$$res = 0$$$.

Obviously $$$res = n$$$.

Obviously $$$res = \frac{n(n-1)}{2}$$$.

Becareful of overflow.

$$$a_1 < a_2 < \dots < a_k$$$.

What if $$$a_1 \times a_2 \times a_3$$$ is a perfect cube and also does $$$a_1 \times a_2 \times a_4$$$

Let $$$a_x = b_x \times c_x^3$$$ for $$$b_x$$$ cubefree and $$$c_x$$$ is some integer.

Let $$$b_{12} = b_1 \times b_2$$$

We have $$$b_{12} \times b_3$$$ and $$$b_{12} \times b_4$$$ are all cubes.

The fundamental theorem of arithmetic states that every positive integer can be represented in exactly one way apart from rearrangement as a product of prime with positive integer power hence we can express number as prime factorization form as:

$$$b_x = p_0^{f_0} \times p_1^{f_1} \times \dots$$$ for unique prime $$$p_i$$$ and positive integer $$$f_i$$$.

By definition cubefree $$$b_x$$$ we have $$$f_z \in {0, 1, 2}$$$ for all $$$x, z$$$.

[1] If $$$b_x \times b_y$$$ is a cube and you know $$$b_x$$$ then you can easily express $$$b_y$$$ uniquely (based on the theorem above)

[2] We also have $$$b_{12} \times b_3$$$ and $$$b_{12} \times b_4$$$ are all cubes.

From [1] and [2] you can prove that $$$b_3 = b_4$$$.

Without loss of generality we can conclude that for any different $$$4$$$ numbers you selected $$$a_x, a_y, a_z, a_t$$$.

Then you can easily prove that $$$b_x = b_y = b_z = b_t$$$ for $$$b_p$$$ is cubefree part of $$$a_p$$$

We have discussed that in extra task B we count things for squarefree.

So we just make a squarefree sieve, and for each fixed squarefree $$$u$$$ we count the number of ways to select $$$p^2$$$.

In order to make it faster, we just reverse the loop, using combinatorics to count $$$u$$$ for each fixed $$$p^2$$$.

For those $$$p > \sqrt{n}$$$ or you can say $$$p^2 > n$$$, the answer is obviously $$$0$$$.

So the complexity is depended on how you count $$$u$$$ for each $$$p^2$$$.

We are discussing that in extra task G we count things for cubefree.

So we just make a cubefree sieve, and for each fixed cubefree $$$u$$$ we count the number of ways to select $$$p^3$$$.

In order to make it faster, we just reverse the loop, using combinatorics to count $$$u$$$ for each fixed $$$p^3$$$.

For those $$$p > \sqrt[3]{n}$$$ or you can say $$$p^3 > n$$$, the answer is obviously $$$0$$$.

So the complexity is depended on how you count $$$u$$$ for each $$$p^3$$$.

Let $$$f_k^3(n)$$$ is the number of set $$$(a_1, a_2, \dots, a_k, n)$$$ that $$$1 \leq a_1 < a_2 < \dots < a_k \leq n$$$ and $$$(a_1, a_2, \dots, a_k, n)$$$ is a $$$(k+1)$$$-term geometric progression in 3 dimensional space.

Let $$$g_k^3(n)$$$ is the number of set $$$(a_1, a_2, \dots, a_k, n)$$$ that $$$1 \leq a_1 \leq a_2 \leq \dots \leq a_k \leq n$$$ and $$$(a_1, a_2, \dots, a_k, n)$$$ is a $$$(k+1)$$$-term geometric progression in 3 dimensional space.

Let $$$F_k^3(n)$$$ once again, is our answer.

Let $$$s_k^3(n)$$$ is the number of way to choose $$$p^3$$$ among those $$$k$$$ numbers when you fix cubefree $$$u$$$ (though we are doing in reverse).

What is the differences when you count cubefree instead of squarefree ?

$$$s_k^3(n) = C^{n-1}_{k-1}$$$.

Yes, just that simple.

It just count the number of way to select set of size $$$k$$$ among those possibility of choosing $$$p^3$$$.

The formula is not depended by squarefree, cubefree stuffs but only the parameters changes (the "$$$n$$$") ins the formula.

$$$F_k^3(p) = \overset{n}{\underset{p=1}{\Large \Sigma}} f_k^3(p)$$$

Just the same way they does, summarize those $$$f_k^3(p)$$$ means you make every possibilities to fix $$$p^3$$$ among those you can fix.

The calculation is not overlap each other, as for each $$$f_k^3(p)$$$, the sequence is unique to each other as its alway end at $$$p$$$.

$$$f_k^3(n) = F_k^3(n) - F_k^3(n-1) = s_k^3(g_k^3(n)) = \underset{d^3 | n}{\Large \Sigma} (\mu \times s_k^3)(d) = \underset{d^3 | n}{\Large \Sigma} \underset{p | d}{\Large \Sigma} \mu(p) \times s_k^3\left(\frac{d}{p}\right)$$$

Well, the formula isnt that much difference.

The only thing changed is that we try to fix $$$d^3$$$ instead of $$$d^2$$$.

The $$$\mu(n)$$$ function is not depended on squarefree or cubefree, so it shouldnt be in higher dimensional Möbius function.

Even if it is expected to somewhat related to calculate squarefree, it is not how we use in the formula as it just appear when you swap two for loop using dirichlet convolution.

$$$F_k^3(n) = \overset{n}{\underset{p=1}{\Large \Sigma}} f_k^3(p)$$$

$$$= F_k^3(0) + \overset{\left \lfloor \sqrt[3]{n} \right \rfloor}{\underset{d = 1}{\Large \Sigma}}(\mu \times s_k^3)(d) \times \left \lfloor \frac{n}{d^3} \right \rfloor$$$

$$$= \overset{\left \lfloor \sqrt[3]{n} \right \rfloor}{\underset{d = 1}{\Large \Sigma}} \left \lfloor \frac{n}{d^3} \right \rfloor \times \left ( \underset{p | d}{\Large \Sigma} \mu(p) \times s_k^3\left(\frac{d}{p}\right) \right)$$$

So now we got the formula, just apply it to the problem

You can also reduce $$$O(\log n)$$$ factor to $$$O(\log \log n)$$$ too.

Let $$$a_x = b_x \times c_x^3$$$ for $$$b_x$$$ cubefree and $$$c_x$$$ is some integer.

The fundamental theorem of arithmetic states that every positive integer can be represented in exactly one way apart from rearrangement as a product of prime with positive integer power hence we can express number as prime factorization form as:

$$$b_x = p_0^{f_0} \times p_1^{f_1} \times \dots$$$ for unique prime $$$p_i$$$ and positive integer $$$f_i$$$.

By definition cubefree $$$b_x$$$ we have $$$f_z \in {0, 1, 2}$$$ for all $$$x, z$$$.

If $$$b_x \times b_y$$$ is a cube and you know $$$b_x$$$ then you can easily express $$$b_y$$$ uniquely (based on the theorem above)

int cnt[LIM];
vector<int> appear[LIM];
int solveG(int n, int k)
{
    linear_sieve(n * n);
    for (int i = 1; i <= n; ++i)
        appear[i].clear();

    for (int i = 1; i <= n; ++i)
    {
        for (int j = i + 1; j <= n; ++j)
        {
            int u = 1;
            for (int x = i * j; x > 1; )
            {
                int a = lpf[x], b = a * a * a;
                for (; x % b == 0; x /= b);
                for (; x % a == 0; x /= a) u *= a;
            }
            appear[j + 1].push_back(u);
        }
    }

    long long res = 0;
    memset(cnt, 0, sizeof(cnt[0]) * (n * n + 1));
    for (int i = 1; i <= n; ++i)
    {
        for (int u : appear[i]) ++cnt[u];
        int v = 1;
        for (int x = i; x > 1; )
        {
            int a = lpf[x], b = a * a * a;
            for (; x % b == 0; x /= b);
            if (x % (a * a) == 0)
            {
                x /= a * a;
                v *= a;
            }
            else if (x % a == 0)
            {
                x /= a;
                v *= a * a;
            }
        }

        res += cnt[v];
    }    

    res %= MOD;
    return res;
}

bool is_cubefree[LIM + 10];
int solve(int n, int k)
{
    if (k == 0) return 0;
    if (k == 1) return n;
    if (k == 2) return (1LL * n * (n - 1) / 2) % MOD;
    if (k == 3) return solveG(n, k);

    memset(is_cubefree, true, sizeof(is_cubefree[0]) * (n + 1));
    precal_nck(n);

    long long res = 0;
    for (int i = 1, j; i <= n; ++i) if (is_cubefree[i]) 
    {
        for (j = 1; i * j * j * j<= n; ++j)
        {
            is_cubefree[i * j * j * j] = false;
        }

        res += nck(j - 1, k);
    }

    res %= MOD;
    return res;
}

vector<int> divisor[LIM];
int solveG(int n, int k)
{
    linear_sieve(n);
    for (int i = 1; i <= n; ++i)
        divisor[i].clear();
 
    for (int j = 1; j <= n; ++j)
    {
        int x = 1;
        for (int t = j; t > 1; )
        {
            if (x > n) break;
            int a = lpf[t];
            ll b = 1LL * a * a * a;
            while (t % b == 0)
            {
                t /= b;
                if (1LL * x * a > n) goto skip;
                x *= a;
            }
 
            if (t % a == 0)
            {
                if (1LL * x * a > n) goto skip;
                x *= a;
                do t /= a; while (t % a == 0);
            }
        }
 
        for (int i = x; i <= n; i += x)
            divisor[i].push_back(j);
 
        skip:{};
    }
 
    int res = 0;
    for (int i = 1; i <= n; ++i)
    {
        ll t = 1LL * i * i * i;
        for (int x = 0; x < divisor[i].size(); ++x)
        {
            for (int y = x + 1; y < divisor[i].size(); ++y)
            {
                int a = divisor[i][x];
                int b = divisor[i][y];
                int c = t / (a * b);
                res += (n >= c && c > b && 1LL * a * b * c == 1LL * i * i * i);
            }
        }
    }
 
    return res;
}
 
int res[LIM + 10];
int solve(int n, int k)
{
    if (k == 0) return 0;
    if (k == 1) return n;
    if (k == 2) return (1LL * n * (n - 1) / 2) % MOD;
    if (k == 3) return solveG(n, k);
 
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d * d <= n; ++d) 
        res[d] = nck(d - 1, k - 1);

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d * d <= n; ++d)
        ans += res[d] * (n / (d * d * d));

    ans %= MOD;
    return ans;
}

You can reduce the $$$O(\log n)$$$ part to $$$O(\log \log n)$$$.

If the problem is mixed with task J, then it might be able to improve furthur more, about $$$O(n \sqrt n)$$$ or even linear.

No papers in Russian, Chinese, English related to this task has been found.

Not a single useful sequence related to this sequence $$$(k = q = 3)$$$ has been found.

Yet this is very hard and specific problem related deep into number theory even with tricks related GEF, DFT, FFT, NTT as recommended.

The complexity is $$$O\left( \underset{p=1}{\Large \Sigma} \left \lfloor \frac{n}{f(p)} \right \rfloor \right)$$$ for $$$f(p) = min(x \times p | x \times p \text{ is cube and } x \in \mathbb{N}^*)$$$

Ignore the fact that having $$$O(\log x)$$$ of factorizing but can be optimized to ignore that the part

The complexity is $$$O\left( \underset{p=1}{\Large \Sigma} d(p)^2 \right)$$$ for $$$d(p)$$$ is the number of divisors of $$$p^3$$$ that in $$$[1, n]$$$.

Though it might be possible to improve this from $$$O(d(p)^2)$$$ to $$$O(d(p)^{1 + ε})$$$

https://terrytao.wordpress.com/2008/09/23/the-divisor-bound/

https://www2.math.upenn.edu/WilF/gfologyLinked2.pdf

https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function

https://en.wikipedia.org/wiki/Prime_zeta_function

It seems possible to be $$$O(\sqrt{n})$$$ complexity with heavy optimization and far deep into the world of number theory to prove.

Needed significant more time and work though.