give solution in cpp #include <bits/stdc++.h> usi...

cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const int M = 500000;
const int INF = 1000000000;

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

    // 1) Precompute odd divisors for all n <= M
    static vector<int> divs[M+1];
    for(int d = 1; d <= M; d += 2) {
        for(int x = d; x <= M; x += d) {
            divs[x].push_back(d);
        }
    }

    // 2) dpNodes[f] = minimum size of a subtree (excluding the global root)
    //    whose A-value = f.  Only odd f.
    //    D[g] = minimal sum of dpNodes[d_i] over any factorization g = ∏ d_i (all d_i odd > 1)
    static int dpNodes[M+1], D[M+1];
    fill(dpNodes, dpNodes+M+1, INF);
    fill(D,       D+M+1,       INF);

    dpNodes[1] = 0;   // no nodes needed for A=1
    D[1]       = 0;   // empty product for g=1

    // Build dpNodes in increasing odd f
    for(int f = 3; f <= M; f += 2) {
        int g = f - 2;  // product of child A-values

        // Compute D[g]: best way to factor g into odd >1 pieces
        if (g > 1) {
            int best = INF;
            for(int d : divs[g]) {
                if (d == 1) continue;
                best = min(best, dpNodes[d] + D[g/d]);
            }
            D[g] = best;
        }

        // Option 1: path shape cost = (f-1)/2
        dpNodes[f] = (f - 1) / 2;
        // Option 2: use an internal node of A=f => cost = 1 + D[g]
        dpNodes[f] = min(dpNodes[f], 1 + D[g]);
    }

    // 3) dpRoot[m] = minimum total vertices in a full tree with exactly m colorings,
    //    where root is green (so m = ∏ A[child]).
    static int dpRoot[M+1];
    fill(dpRoot, dpRoot+M+1, INF);

    dpRoot[1] = 1;  // only the root

    for(int m = 3; m <= M; m += 2) {
        // Single-child: root -> subtree with A = m
        if (dpNodes[m] < INF) {
            dpRoot[m] = 1 + dpNodes[m];
        }
        // Multi-child splits: m = f * (m/f)
        for(int f : divs[m]) {
            if (f == 1 || f == m) continue;
            dpRoot[m] = min(dpRoot[m],
                            dpRoot[m/f] + dpNodes[f]);
        }
    }

    // 4) Answer queries
    int T;
    cin >> T;
    while (T--) {
        int m;
        cin >> m;
        if (m == 1) {
            cout << 1 << "\n";
        } else if ((m & 1) == 0) {
            cout << -1 << "\n";
        } else {
            int ans = dpRoot[m];
            cout << (ans >= INF ? -1 : ans) << "\n";
        }
    }
    return 0;
}

Explanation of Key Steps

1. Divisor Precomputation
   We only care about odd divisors (even divisors don’t contribute to the product for A-values), so we sieve them in $O(M \log M)$.
2. dpNodes[f] Calculation
   • A “path” shape uses $(f-1)/2$ nodes to achieve $f$ colorings.
   • An “internal” node with children factorizing $g = f-2$ costs $1 + D[g]$, where
     $D[g]$ is the best sum of subtree sizes over any factorization of $g$.
3. dpRoot[m] Calculation
   We consider making the root green (so the total colorings is the product of its children’s A-values). We either attach one subtree with $A=m$, or split $m$ into factors and combine optimal solutions.
4. Query Handling
   For each $m$, output the computed minimum size or $-1$ if impossible.