补一个 $Q\le 10^6$ 数据的代码。顺便还附带了打表的程序片段。

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 1000000;
int n, T, Q;
// 打表看看
namespace TEST
{
    vector<int> a, b;
    void out()
    {
        int tim = 1;
        for (int i = 1; i <= n; i++)
            a.push_back(i);
        cout << tim++ << ": ";
        for (int x : a)
            cout << x << " ";
        cout << "\n";
        while (T--)
        {
            b.clear();
            for (int i = 0; i < n; i += 2)
                b.push_back(a[i]);
            for (int i = 1; i < n; i += 2)
                b.push_back(a[i]);
            a = b;
            cout << tim++ << ": ";
            for (int x : a)
                cout << x << " ";
            cout << "\n";
        }
    }
}
int nxt[MAXN + 5];
bool vis[MAXN + 5];
vector<int> e[MAXN + 5];
int ans[MAXN + 5];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> T >> Q;
    // TEST::out();
    int base = n / 2 + (n & 1); // 奇数数量
    for (int i = 1; i <= n; i++)
    {
        if (i % 2 == 1)
            nxt[i] = 1 + i / 2;
        else
            nxt[i] = base + i / 2;
    }
    // 每个点入度出度都是 1，所以必然是形成了几个环，现在把每个环都找到存下来
    // 存成 vector 之后方便一口气做一个 T 轮的轮换
    for (int i = 1; i <= n; i++)
    {
        int now = i;
        while (!vis[now])
        {
            vis[now] = true;
            e[i].push_back(now);
            now = nxt[now];
        }
    }
    // getAns
    for (int i = 1; i <= n; i++)
    {
        if (e[i].empty())
            continue;
        int siz = e[i].size();
        for (int j = 0; j < e[i].size(); j++)
            ans[e[i][j]] = e[i][(j + T) % siz];
    }
    // 输出 Q 次询问的答案
    while (Q--)
    {
        int b;
        cin >> b;
        cout << ans[b] << " ";
    }
    return 0;
}