D1. Infinite Sequence (Easy Version) 【Codeforces Round 1007 (Div. 2)】

D1. Infinite Sequence (Easy Version)

思路

为了方便起见，假设$n$为奇数（如果不是，则将$n$递增并单独处理边缘情况）。首先预先计算前$2n$项$a_1,a_2,\dots ,a_{2n}$。对于索引不超过$2n$的查询，直接返回预先计算的值。

当查询的索引满足$2m>n$时，观察到以下关系：

$$a_{2m}=a_1\oplus a_2\oplus \dots \oplus a_m=a_{2m+1}.$$

定义$p=a_1\oplus a_2\oplus \dots \oplus a_n$。当$m>n$时，可以将XOR和分解为：

$$\begin{array}{rl}{a}_{2m}& =a_1\oplus a_2\oplus \dots \oplus a_m\\ & =p\oplus (a_{n+1}\oplus a_{n+2})\oplus (a_{n+3}\oplus a_{n+4})\oplus \dots \oplus a_m.\end{array}$$

由于$n$为奇数，配对项$(a_{n+1}\oplus a_{n+2}),(a_{n+3}\oplus a_{n+4}),\dots$会相互抵消。因此公式简化为：

$$a_{2m}=a_{2m+1}=\{\begin{array}{ll}p& \text{若 }m\text{ 为奇数},\\ p\oplus a_m& \text{若 }m\text{ 为偶数}.\end{array}$$

最终，通过递归地将$m$不断减半直到$m\le 2n$，并在每一步根据奇偶性应用上述规则，即可计算$a_m$。

总时间复杂度为：$O(n+\log (m))$。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
#define int long long
#define pb push_back
#define pii pair<int, int>
#define FU(i, a, b) for (int i = (a); i <= (b); ++i)
#define FD(i, a, b) for (int i = (a); i >= (b); --i)
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

void solve() {
    int n, l, r, p;
    cin >> n >> l >> r;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    vector<int> pref(n + 1);
    for (int i = 1; i <= n; i++) {
        pref[i] = pref[i - 1] + a[i];
    }
    if (n % 2 == 0) {
        n++;
        int cur = pref[n / 2] & 1;
        a.push_back(cur);
        pref.push_back(pref.back() + cur);
    }
    for (int i = n + 1; i <= n * 2; i++) {
        a.push_back(pref[i / 2] & 1);
        pref.push_back(pref[i - 1] + a[i]);
    }
    p = pref[n] & 1;
    int ret = 0;
    while (true) {
        if (l <= n * 2) {
            ret ^= a[l];
            break;
        }
        ret ^= p;
        if ((l / 2 - n) % 2 == 0) {
            break;
        }
        l /= 2;
    }
    cout << ret << endl;
}

signed main() {
    cin.tie(0)->ios::sync_with_stdio(0);
    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    }
    return 0;
}