我有罪，休息了好久，这次来补补题
签到，找到规律就可以
void solve(){
int x;
cin >> x;
if(x%3 == 1){
cout << "YES\n";
}else{
cout << "NO\n";
}
}
找规律，发现如果他如果前缀和组成了完全平方数，与后面的进行交换就能防止
倒序输出bfs序即可
如果n是偶数，则n+1为奇数
$a{n+1}\bigoplus a{n + 2} = 0$可以得出更多的规律，由此可以推出计算的公式
进行递归计算就可
bool judge(int x){
return int(sqrt(x))*int(sqrt(x)) == x;
}
int sum[N];
int a[N];
void solve(){
int n;
cin >> n;
for(int i = 1;i <= n;i++){
sum[i] = 0;
}
for(int i = 1;i <= n;i++){
a[i] = i;
}
int ans = n*(n+1)/2;
if(judge(ans)){
cout << -1 << '\n';
return;
}
for(int i = 1;i <= n;i++){
sum[i] = sum[i-1] + a[i];
}
for(int i = 1;i <= n;i++){
if(judge(sum[i])){
std::swap(a[i],a[i+1]);
}
}
for(int i = 1;i <= n;i++){
cout << a[i] <<' ';
}
cout << '\n';
}
void solve(){
cin >> n >> st >> en;
for(int i = 1;i <= n;i++){
e[i].nxt = 0;
head[i] = 0;
vis[i] = 0;
e[i].to = 0;
}
for(int i = 1;i < n;i++){
int u,v;
cin >> u >> v;
add(u,v);
add(v,u);
}
vector<int> ans;
queue<int> q;
q.push(en);
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = 1;
ans.push_back(u);
for(int i = head[u];i;i = e[i].nxt){
int v = e[i].to;
if(!vis[v]){
q.push(v);
}
}
}
std::reverse(ans.begin(),ans.end());
for(int i = 0;i < ans.size();i++){
cout << ans[i] << " ";
}
cout << '\n';
}
int cal(int x){
if ( x <= n ) return a[x];
else if ( x / 2 <= n ) return pre[x / 2];
else if ( (x / 2 - n) % 2 ){
return pre[n] ^ cal(x / 2);
}else{
return pre[n];
}
}
void solve(){
cin >> n >> l >> r;
for(int i = 1; i <= n; i++){
cin >> a[i];
}
for(int i = 1;i <= n;i++){
pre[i] = pre[i-1] ^ a[i];
}
if(!(n&1)){
n++;
a[n] = pre[n/2];
pre[n] = pre[n - 1]^a[n];
}
cout << cal(r) << '\n';
}
signed main(){
std::ios::sync_with_stdio(false);
cin.tie(0);
int T = 1;
cin >> T;
while(T--) solve();
return 0;
}