A

# include <bits/stdc++.h>
using namespace std;
#define int long long 
int n,k;
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n>>k;
        bool st=0;
        for(int i=1;i<=n;i++)
        {
        int x;
        cin>>x;
        if(x==k)
        {
        st=1;    
        }
        }
        if(st)
        cout<<"YES";
        else cout<<"NO";
        cout<<endl;
    }
}

B
奇数一定无法整除偶数

# include <bits/stdc++.h>
using namespace std;
#define int long long
int n;
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    cin>>n;
    int l=1;
    for(int i=1;i<=n;i++)
    {
    cout<<l<<" ";
    l+=2;
    }
    cout<<endl;
    }
    return 0;
}

C
贪心的考虑 所凑出的数位开头前k个到结尾后k数之间

# include <bits/stdc++.h>
using namespace std;
#define int long long 
int n,k,s;
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int mn=0,mx=0;
        cin>>n>>k>>s;
        mn=(1+k)*k/2;
        mx=(n+n-k+1)*k/2;
        if(s>=mn&&s<=mx)
        cout<<"YES";
        else cout<<"NO";
        cout<<endl;

    }
    return 0;
}

E
根据与的操作性质都为一时才为一，否则为零.a[i][j]表示前i个数第j位为1的个数。区间具有单调性当1越多时数越大，反之越小。二分即可

# include <bits/stdc++.h>
using namespace std;
#define int long long
using namespace std;
const int N = 2e5+10;
int t;
int a[N][40];
int l, k;
bool cheek(int l, int r, int k)
{
    int s=0;
    for (int i = 32; i >= 1; i--)
    {
       if(a[r][i]-a[l-1][i]==r-l+1)//全1才是1
        {
            s+=1<<(i-1);
            if(s>=k)
            return true;//满足直接退出即可
        }

    }
    return false;
}


signed main()
{
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
            int tmp;
            cin >> tmp;
            for (int j = 1; tmp; j++)
            {
                a[i][j] = tmp % 2;
                tmp >>= 1;
            }
        }

        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= 32; j++)
                a[i][j] += a[i - 1][j];
        }
        int q;
        cin >> q;
        while (q--)
        {
            cin >> l >> k;
            int ll=l-1, rr = n;
            while (ll < rr)
            {
                int mid = (ll + rr + 1) >> 1;
                if (cheek(l, mid, k))ll = mid;
                else rr = mid - 1;
            }

            if (ll<l||ll>n)cout << -1 << " ";
            else cout << ll << " ";
        }
        cout << endl;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= 32; j++)a[i][j] = 0;
    }

    return 0;
}