Editing Openai/6943ca25-5f58-8006-882b-41249eb1ca7a (section)

=== Assistant: The checker is right: the problem expects YES / NO, but the code prints First / Second. ===
The checker is right: the problem expects <code>YES</code> / <code>NO</code>, but the code prints <code>First</code> / <code>Second</code>.

Your logic matches the sample; you just need to change the two output literals:
* winning for the starting player → YES
* losing for the starting player → NO

Here is the corrected code (only output strings changed):

<syntaxhighlight lang="cpp">#include <bits/stdc++.h>
using namespace std;

static inline int v2_int(int x){
    int c = 0;
    while ((x & 1) == 0) { ++c; x >>= 1; }
    return c;
}

struct SplitMix64Hash {
    size_t operator()(uint64_t x) const noexcept {
        x += 0x9e3779b97f4a7c15ULL;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9ULL;
        x = (x ^ (x >> 27)) * 0x94d049bb133111ebULL;
        return (size_t)(x ^ (x >> 31));
    }
};

struct Query { int n, m, idx; };

struct OddV2Solver {
    int m, g, half, maxN;
    vector<int> cold;        // cold positions >0 and <= maxN
    vector<uint8_t> b;       // b[i] = 1 if c_{i+1}-c_i = g+1, with b[0]=0
    unordered_map<uint64_t, uint8_t, SplitMix64Hash> memo;

    OddV2Solver(int m_, int maxN_) : m(m_), g(m_ + 1), half((m_ + 1) / 2), maxN(maxN_) {}

    inline uint64_t key(int i, int d) const {
        // d <= m <= 1e6 < 2^20
        return (uint64_t(i) << 20) | uint64_t(d);
    }

    bool singleton(int i, int d){
        uint64_t k = key(i, d);
        auto it = memo.find(k);
        if (it != memo.end()) return it->second;

        bool res;
        if (i == 0) {
            // After cold position 0: only n=m is singleton (n < m are immediate-win states)
            res = (d == m);
        } else {
            // Previous gap g+1 creates a special mark at distance (half-1).
            if (b[i-1] == 1 && d == half - 1) {
                res = false;
            }
            // Internal mark: singleton at d/2 marks d (only possible when d even).
            else if ((d & 1) == 0 && singleton(i, d >> 1)) {
                res = false;
            } else {
                // Mark from previous cold's singleton.
                int t = g + int(b[i-1]) + d;
                if ((t & 1) == 0) {
                    int dprev = t >> 1;
                    if (dprev <= m && singleton(i-1, dprev)) res = false;
                    else res = true;
                } else {
                    res = true;
                }
            }
        }

        memo.emplace(k, (uint8_t)res);
        return res;
    }

    void build(){
        cold.clear();
        b.clear();
        memo.clear();

        int approxK = maxN / g + 8;
        cold.reserve(approxK);
        b.reserve(approxK + 1);
        memo.reserve((size_t)approxK * 32);

        b.push_back(0); // b[0] corresponds to c0->c1, and c1 = g

        int c = g;
        int i = 1; // c_i index (c0=0, c1=g)
        while (c <= maxN) {
            cold.push_back(c);
            bool bi = singleton(i, half);     // is midpoint singleton?
            b.push_back((uint8_t)(bi ? 1 : 0));
            c += g + (bi ? 1 : 0);
            ++i;
        }
    }

    bool isCold(int n) const {
        return binary_search(cold.begin(), cold.end(), n);
    }
};

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

    int t;
    cin >> t;
    vector<Query> qs(t);
    for (int i = 0; i < t; i++) {
        int n, m;
        cin >> n >> m;
        qs[i] = {n, m, i};
    }

    sort(qs.begin(), qs.end(), [](const Query& a, const Query& b){
        return a.m < b.m;
    });

    vector<string> ans(t);

    int L = 0;
    while (L < t) {
        int R = L;
        int m = qs[L].m;
        int maxN = 0;
        while (R < t && qs[R].m == m) {
            maxN = max(maxN, qs[R].n);
            ++R;
        }

        // Trivial: if n <= m, first can finish immediately.
        if (maxN <= m) {
            for (int i = L; i < R; i++) ans[qs[i].idx] = "YES";
            L = R;
            continue;
        }

        int g = m + 1;
        int e = v2_int(g);

        // Easy case: v2(g) even => cold positions are multiples of g
        if ((e & 1) == 0) {
            for (int i = L; i < R; i++) {
                ans[qs[i].idx] = (qs[i].n % g == 0 ? "NO" : "YES");
            }
            L = R;
            continue;
        }

        // Hard case: v2(g) odd
        OddV2Solver solver(m, maxN);
        solver.build();
        for (int i = L; i < R; i++) {
            ans[qs[i].idx] = (solver.isCold(qs[i].n) ? "NO" : "YES");
        }

        L = R;
    }

    for (int i = 0; i < t; i++) {
        cout << ans[i] << '\n';
    }
    return 0;
}

</syntaxhighlight>

This will match the sample format exactly (YES/NO).