complet this func: int Solution::solve(vector<vect...

Here’s a clean, fast way to do it.

Idea

• Values are small (A[i] ≤ 100) ⇒ any XOR fits in 7 bits (0..127).
• Walk the array left→right and keep, for every possible last value v (1..100) and last popcount pc (0..7), the set of XORs reachable by a valid subsequence that ends at such a state.
• Transitions allowed: from any state with last_value < a and last_pc ≤ pc(a) to the new state (a, pc(a)). Start a new subsequence with the single element a as well.
• To aggregate “all states with last_value < a” quickly, use a Fenwick tree (BIT) over the value axis for each pc. Each node stores a bitset<128> (the reachable XORs).
• Each step costs O( (pc(a)+1) * log 100 * 128 ) bit operations → ~90M simple boolean ops worst-case, which is fine in C++.

C++ implementation

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

struct Fenwick {
    static const int MAXX = 128;
    int n;
    // tree[p][i] = bitset of XORs for popcount p, Fenwick node i
    vector<array<bitset<MAXX>, 8>> tree;
    Fenwick(int n_=100) : n(n_), tree(n_+1) {}

    void update(int p, int idx, const bitset<MAXX>& add) {
        for (int i = idx; i <= n; i += i & -i) tree[i][p] |= add;
    }
    bitset<MAXX> query(int p, int idx) const {
        bitset<MAXX> res;
        for (int i = idx; i > 0; i -= i & -i) res |= tree[i][p];
        return res;
    }
};

static inline int popc(int x) { return __builtin_popcount((unsigned)x); }

// permute bits: result[y ^ k] = b[y]
static inline bitset<128> xor_permute(const bitset<128>& b, int k) {
    bitset<128> r;
    for (int i = 0; i < 128; ++i) if (b[i]) r[i ^ k] = 1;
    return r;
}

// ----- Asked function -----
int Solution::solve(vector<int> &A) {
    const int MAXV = 100;
    Fenwick fw(MAXV);                     // per-popcount Fenwick over values
    bitset<128> all;                      // all reachable XORs overall

    for (int a : A) {
        int pc = popc(a);

        // Gather unions of all states with last_value < a and last_pc <= pc
        bitset<128> unionPrev;
        for (int p = 0; p <= pc; ++p) {
            unionPrev |= fw.query(p, a-1);
        }

        // Extend those with current 'a' and also start a new subsequence {a}
        bitset<128> newSet = xor_permute(unionPrev, a);
        newSet[a] = 1;                    // start with single element a

        // Push to structures: state (a, pc)
        fw.update(pc, a, newSet);
        all |= newSet;
    }

    // Count distinct XORs attainable
    return (int)all.count();
}

Minimal main (reads input separately)

text
/*
Input:
n
a1 a2 ... an
Output:
single integer = number of distinct X values
*/
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n; 
    if (!(cin >> n)) return 0;
    vector<int> A(n);
    for (int i = 0; i < n; ++i) cin >> A[i];
    Solution s;
    cout << s.solve(A) << "\n";
    return 0;
}

This satisfies both conditions:

1. values strictly increase along the subsequence, and
2. popcounts are non-decreasing,
   and returns the count of distinct XORs X over all valid subsequences.