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Statement:
Given two integer $$$n, k, p$$$, $$$(1 \leq k \leq n < p)$$$.
Count the number of array $$$a[]$$$ of size $$$k$$$ that satisfied
- $$$1 \leq a_1 < a_2 < \dots < a_k \leq n$$$
- $$$a_i \times a_j$$$ is perfect square $$$\forall 1 \leq i < j \leq n$$$
Since the number can be big, output it under modulo $$$p$$$.
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Solve for k = 1
The answer just simply be $$$n$$$
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Solve for k = 2
We need to count the number of pair $$$(a, b)$$$ that $$$1 \leq a < b \leq n$$$ and $$$a \times b$$$ is perfect square
Every positive integer $$$x$$$ can be represent uniquely as $$$x = u \times p^2$$$ for some positive integer $$$u, p$$$ and $$$u$$$ as small as possible ($$$u$$$ is squarefree number)
Let represent $$$x = u \times p^2$$$ and $$$y = v \times q^2$$$ (still, minimum $$$u$$$, $$$v$$$ ofcourse)
We can easily proove that $$$x \times y$$$ is a perfect square if and if only $$$u = v$$$
So for a fixed squarefree number $$$u$$$. You just need to count the number of ways to choose $$$p^2$$$
Therefore the answer just simply be
Implementation 1int solve(int n)
{
memset(is_squarefree, true, sizeof(is_squarefree[0]) * (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 += 1LL * (j - 1) * (j - 2) / 2;
}
res %= MOD;
return res;
}
Implementation 2int solve(int n)
{
memset(is_squarefree, true, sizeof(is_squarefree[0]) * (n + 1));
long long res = 0;
for (int i = 1; i <= n; ++i) if (is_squarefree[i])
{
for (int j = 1; i * j * j <= n; ++j)
{
is_squarefree[i * j * j] = false;
res += j - 1;
}
}
res %= MOD;
return res;
}
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Solve for general k
Using the same logic above, we can easily solve the problem with combinatorics
O(n) solutionconst int LIM = 1e7 + 17;
const int MOD = 1e9 + 7;
int fact[LIM + 1]; /// factorial: fact[n] = n!
int invs[LIM + 1]; /// inverse modular: invs[n] = n^(-1)
int tcaf[LIM + 1]; /// inverse factorial: tcaf[n] = (n!)^(-1)
void precal(int n = 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;
}
int solve(int n)
{
memset(is_squarefree, true, sizeof(is_squarefree[0]) * (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(j - 1, k);
}
res %= MOD;
return res;
}
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A better solution for k = 2
Let $$$f(n)$$$ is the number of $$$(a, b)$$$ that $$$1 \leq a \leq b \leq n$$$ and $$$(a, b, n)$$$ is a three-term geometric progression
Let $$$F(N) = \overset{n}{\underset{x=1}{\Large \Sigma}} f(x)$$$