我的思路就是模2k之后用树状数组维护一下每个 $a_i$ 所点亮的区间。

wrong answer 2003rd numbers differ - expected: '877248279', found: '877248280'

感觉没道理会出这个错啊

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr int MAXN = 200007;
template <typename T>
class TreeArray {
   private:
    vector<T> a, b;
    int lowbit(int x) { return x & -x; }
    void update(int pos, T val) {
        for (int i = pos; i < a.size(); i += lowbit(i)) {
            a[i] += val;
            b[i] += pos * val;
        }
    }

   public:
    TreeArray(int n) : a(n + 1), b(n + 1) {}
    void updateRange(int l, int r, T val) {
        update(l, val);
        update(r + 1, -val);
    }
    T prefixSum(int pos) {
        T as = 0, bs = 0;
        for (int i = pos; i > 0; i -= lowbit(i)) {
            as += a[i];
            bs += b[i];
        }
        return (pos + 1) * as - bs;
    }
    T rangeSum(int l, int r) { return prefixSum(r) - prefixSum(l - 1); }
    void buildTree(vector<T>& arr) {
        for (int i = 1; i < arr.size(); i++) {
            update(i, arr[i] - arr[i - 1]);
        }
    }
};

int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int T;
    cin >> T;
    while (T--) {
        int n, k;
        cin >> n >> k;
        TreeArray<ll> t(2 * k + 1);
        vector<int> a(n + 1), c(n + 1);
        for (int i = 1; i <= n; i++) {
            cin >> a[i];
            c[i] = a[i] % (2 * k) + 1;
            if (c[i] + k - 1 <= 2 * k) {
                t.updateRange(c[i], c[i] + k - 1, 1);
            } else {
                t.updateRange(c[i], 2 * k, 1);
                t.updateRange(1, c[i] - k - 1, 1);
            }
        }
        ll base = *max_element(a.begin() + 1, a.end());
        ll ans = -1;
        for (int i = 1; i <= 2 * k; i++) {
            if (t.rangeSum(i, i) == n) {
                ll rel = base % (2 * k) + 1;
                if (rel <= i) {
                    ans = i - rel + base;
                } else {
                    ans = 2 * k - rel + i + base;
                }
                break;
            }
        }
        cout << ans << "\n";
    }
    return 0;
}