D. Xor of 3

>>> 0^0^0
0
>>> 0^0^1
1
>>> 0^1^1
0
>>> 1^1^1
1

1. 整个数列异或和为0时，无解
2. 奇数乱搞
3. 偶数长度时，将问题拆解成两个奇数长度且偶数xor和

#include <bits/stdc++.h>
#define sc(x) scanf("%lld", &(x))
#define pr(x) printf("%lld\n", (x))
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 2e5 + 7;
const int mod = 1e9 + 7;
int n, t, a[N];

void out(int l, int r, vector<int>& a) {
    for (int i = r - 2; i >= l; i -= 2) a.push_back(i);
    for (int i = l; i < r; i += 2) a.push_back(i);
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> t;
    while (t--) {
        cin >> n;
        vector<int> ans;
        for (int i = 1; i <= n; ++i) cin >> a[i], a[i] ^= a[i - 1];
        if (!a[n]) {
            if (n & 1)
                out(1, n, ans);
            else
                for (int i = 1; i <= n; i += 2)
                    if (a[i] == 0) {
                        out(1, i, ans);
                        out(i + 1, n, ans);
                        break;
                    }
        }
        if (ans.size()) {
            cout << "YES\n" << ans.size() << '\n';
            for (int x : ans) cout << x << ' ';
            cout << '\n';
        } else
            cout << "NO\n";
    }
}