B

//这个题，傻逼了，明显就是构造一个二进制数，而且二进制数每位是二倍关系，
//联想高中的斐波那契数列的关系，显然有2^n-1+2^n=2^(n+1)-2^(n-1);
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

void solve() {
    ll n;
    cin >> n;

    vector<int> arr(32, 0);


    for (int i = 0; i < 30; i++) {

        if ((n >> i) &(ll)1) {
            if (arr[i] == 1) {  
                arr[i] = 0;      
                arr[i + 1] = 1;  
            } else if (i > 0 && arr[i - 1] == 1) { 
                arr[i + 1] = 1; 
                arr[i - 1] = -1; 
            } else {
                arr[i] = 1; 
            }
        }
    }


    cout << 31 << "\n";
    for (int i = 0; i < 31; i++) {
        cout << arr[i] << " ";
    }
    cout << "\n";
}

int main() {
    int t;
    cin >> t; 
    while (t--) {
        solve();
    }
    return 0;
}

B

//这个题，建议打表看性质，bitset是很好用的容器
//正确性说明：由于连续，所以，+1必然是从最低为开始；
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll qmi(ll m, ll k)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k&1) res = res * t;
        t = t * t;
        k >>= 1;
    }
    return res;
}
int lowbit(ll x) {
    return x & -x;
}
void solve() {
    ll a,b;
    cin>>a>>b;

    ll o=a^b;
    cout<<(2,lowbit(o))<<"\n";


}

int main() {
    int t;
    cin >> t; 
    while (t--) {
        solve();
    }
    return 0;
}

C

//构造使得 连加s/ki<s化简得到 连加1/ki<1;
//检查k，其实检查最大公倍数>硬币数的连加！！！！！
//ps.我是乱搞出来的
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gggcd(ll a,ll b){
    return a*b/(__gcd(a,b));
}
void solve() {
    ll n;
    cin>>n;
    vector<ll> vc(n+1);
    ll num=1;
    for(int i=1;i<=n;i++){
        cin>>vc[i];
        num=gggcd(num,vc[i]);
       //cout<<vc[i]<<" ";
    }
    vector<int> ans(n+1);
    ll sum=0;
    for(int i=1;i<=n;i++){
        ans[i]=num/vc[i];
        sum+=ans[i];
    }
    if(sum<num){
    for(int i=1;i<=n;i++){
        cout<<ans[i]<<" ";
    }}else{
        cout<<-1;
    }

    cout<<"\n";

}

int main() {
    int t;
    cin >> t; 
    while (t--) {
        solve();
    }
    return 0;
}

C

//规律不难看出
//但是实现有点屎
//要看出两个东西，1对角交换，（n-i）*2的增量，相邻也是，为零加速，优先用对角，不够就直接i和i+k
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    ll T;
    cin >> T;

    while (T--) {
        int n; 
        ll k; 
        cin >> n >> k;

        ll max_s = 0;
        ll index=n;
        for (int i = 1; i <= n; i++) {
            max_s+=abs(i-index--);
        }

        if (k&1 || k > max_s) {
            cout << "No\n";
        } else {
            cout << "Yes\n";
            vector<int> p(n);
            iota(p.begin(), p.end(), 0);  // 初始化排列为 [0, 1, 2, ..., n-1]

            k /= 2; 
            for (int i = 0; k > 0; i++) {
                if (k >= n - 1 - 2 * i) {
                    swap(p[i], p[n - 1 - i]);
                    k -= n - 1 - 2 * i; //收益
                } else {
                    swap(p[i], p[i + k]);//剩余交换
                    k = 0;
                }
            }

            for (int i = 0; i < n; i++) {
                cout << p[i] + 1 << " ";
            }
            cout << "\n";
        }
    }

    return 0;
}