Overview

Given \(N\) gems, select 2 or more to maximize the GCD (Greatest Common Divisor), then find the minimum number \(K\) of gems needed to achieve that GCD, and output \(N - K\). The conclusion is that \(K = 2\) always holds, so the answer is \(N - 2\).

Analysis

Key Insight 1: How to find the maximum GCD \(G\)

For a value \(g\) to be achievable as the GCD of a set of 2 or more gems, there must be at least 2 multiples of \(g\) in the array.

Conversely, if there are 2 or more multiples of \(g\), selecting any 2 of them guarantees GCD ≥ \(g\).

Therefore, \(G\) can be found as the largest \(g\) such that “there exist at least 2 multiples of \(g\) in the array.”

Key Insight 2: \(K = 2\) always holds

Since \(G\) is the maximum value described above, there are at least 2 multiples of \(G\) in the array. If we pick any 2 of them, letting \(d\) be their GCD:

• Since both are multiples of \(G\), we have \(d \geq G\)
• If \(d > G\), then there would also be at least 2 multiples of \(d\) in the array, contradicting the maximality of \(G\)

Therefore \(d = G\) always holds, and we can achieve GCD \(= G\) by selecting just 2 elements.

Conclusion: \(K = 2\), and the answer is always \(N - 2\)

Why a naive approach would be problematic

If we naively scan the array for each \(g\) to count multiples, it would be \(O(N \cdot \max(A))\) which risks TLE. We use a sieve-like technique for efficiency.

Algorithm

1. Compute the frequency count \(\text{cnt}[v]\) for each value in array \(A\).
2. Using a sieve-like approach, for each \(g = 1, 2, \ldots, \max(A)\), sum up the frequencies at \(g, 2g, 3g, \ldots\) to obtain \(\text{divisor\_count}[g]\) = “the number of multiples of \(g\).”
3. Iterate \(g\) from largest to smallest, and the first \(g\) with \(\text{divisor\_count}[g] \geq 2\) is \(G\).
4. From the analysis above, \(K = 2\), so output \(N - 2\).

Concrete example: \(A = [6, 10, 15]\) (\(N = 3\))

• \(\text{divisor\_count}[5] = 2\) (10 and 15), \(\text{divisor\_count}[6] = 1\), \(\text{divisor\_count}[10] = 1\), …
• The largest \(g\) with count ≥ 2 is \(g = 5\). \(G = 5\).
• \(\gcd(10, 15) = 5 = G\). \(K = 2\). Answer: \(3 - 2 = 1\).

Complexity

• Time complexity: \(O(M \log M)\) (\(M = \max(A_i)\), harmonic series \(\sum_{g=1}^{M} \lfloor M/g \rfloor = O(M \log M)\))
• Space complexity: \(O(M)\)

Implementation Notes

• The sieve computation iterates from \(g = 1\) to \(\max(A)\), scanning multiples of \(g\) for each \(g\). The total number of loop iterations is the harmonic series \(O(M \log M)\), which is sufficiently fast.

• The maximality of \(G\) automatically guarantees \(K = 2\), so no search for the minimum subset is needed.

• The essence of the problem lies in realizing that “\(K = 2\) always holds.”

  Source Code

import sys

def solve():
    input_data = sys.stdin.buffer.read().split()
    N = int(input_data[0])
    A = [int(input_data[i + 1]) for i in range(N)]
    
    MAX_VAL = max(A) + 1
    
    # Count occurrences of each value
    cnt = [0] * MAX_VAL
    for a in A:
        cnt[a] += 1
    
    # For each possible GCD value g (from large to small),
    # count how many elements in A are divisible by g.
    # If count >= 2, then g is achievable as a GCD of some subset.
    # But we need the actual GCD of those elements to be exactly g, not a multiple of g.
    # 
    # Wait, actually if we pick elements divisible by g, their GCD is a multiple of g.
    # We need to find the maximum g such that there exist >= 2 elements whose GCD is exactly g.
    #
    # Better approach: For each g from large to small, count how many A_i are divisible by g.
    # If that count >= 2, then we can pick a subset of those elements. The GCD of all elements
    # divisible by g is some multiple of g, say g' >= g. But g' would have been considered
    # before g (since we go from large to small). So if g' > g, then g' already has count >= 2
    # and we would have found g' first.
    #
    # Actually, we need to be more careful. Let me think again.
    #
    # We want the maximum G such that there exist >= 2 elements whose GCD equals G.
    # 
    # For each g from MAX_VAL-1 down to 1:
    #   Collect all A_i divisible by g. Let there be c(g) such elements.
    #   If c(g) >= 2, compute the GCD of all those elements divided by g.
    #   If that GCD is 1, then the GCD of those elements is exactly g.
    #   But we want subset GCD = g, and we want minimum subset size.
    #   Actually, if c(g) >= 2, we can always find a subset with GCD exactly g IF
    #   there's no larger g' that all of them share. But some subset of them might have GCD = g.
    #
    # Simpler correct approach:
    # For each g from large to small, count c(g) = number of A_i divisible by g.
    # If c(g) >= 2, then the GCD of ANY subset of size >= 2 from those elements is a multiple of g.
    # The GCD of ALL c(g) elements is the GCD of (A_i/g for A_i divisible by g) times g.
    # If that overall GCD equals g, then picking all of them gives GCD = g, but we want min subset.
    # 
    # Key insight: G = max g such that c(g) >= 2.
    # Because if c(g) >= 2, pick any 2 elements divisible by g. Their GCD is >= g.
    # But for any g' > g, c(g') < 2, so no pair has GCD > g. Hence G = g.
    # And K = minimum subset size with GCD = g among elements divisible by g.
    # Since we want GCD exactly g, we need GCD of chosen subset = g.
    # The minimum such subset: we need to find minimum number of elements (among those divisible by g)
    # whose GCD (after dividing by g) is 1.
    
    # Find G: largest g with c(g) >= 2
    # Compute c(g) for all g using sieve-like approach
    
    divisor_count = [0] * MAX_VAL
    for g in range(1, MAX_VAL):
        for multiple in range(g, MAX_VAL, g):
            divisor_count[g] += cnt[multiple]
    
    G = 1
    for g in range(MAX_VAL - 1, 0, -1):
        if divisor_count[g] >= 2:
            G = g
            break
    
    # Now find K: minimum subset size among elements divisible by G with GCD = G
    # Collect elements divisible by G, divide by G, find min subset with GCD 1
    elements = []
    for a in A:
        if a % G == 0:
            elements.append(a // G)
    
    # We need minimum subset of `elements` with GCD = 1.
    # Since overall GCD of all elements must be 1 (otherwise G wouldn't be the max),
    # and we want minimum subset size.
    # Greedy: start with first element, keep taking GCD with next elements, stop when GCD = 1.
    # But greedy order matters. With N up to 10^6, trying all pairs: if any pair has GCD 1, answer is 2.
    # Otherwise answer is at most small. But finding exact minimum is hard in general.
    # However, K=2 if any pair among them has GCD=1 (i.e., coprime after dividing by G).
    # Since G is maximum, there's no g > G with 2+ multiples, so K=2 always.
    # Because if all pairs had GCD > 1 after dividing by G, then... actually that's possible.
    # But K is always 2 because G is the max g with >=2 multiples. Pick any two such elements: their GCD is exactly G.
    # Wait, their GCD could be a multiple of G. But then that multiple would have count >= 2, contradicting maximality of G.
    
    K = 2
    print(N - K)

solve()

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