A

#include <bits/stdc++.h>
using namespace std;
#define int long long 
int n;

signed  main()
{
        int T;
    cin>>T;
    while(T--)
    {
    cin>>n;
    int t=0;
    while(n>3)
    {
    n=n/4;
    t++;
    }
    int ans=1;
    for(int i=1;i<=t;i++)
    ans*=2;
    cout<<ans<<endl;
    }
    return 0;
}

B
数位和为9 3倍数得可以整除9 3
个位为0或5得可整除5
至于7可拆解运算 https://www.cnblogs.com/matinalcosmos/p/18634230

#include <bits/stdc++.h>
using namespace std;
int n,d;
int a[6]={0,1,3,5,7,9};
vector<int>v;
int main()
{
        //cout<<8*7*6*5*4*3*2*5;
        int T;
    cin>>T;
    while(T--)
    {
    v.clear();
    cin>>n>>d;
    bool st=0;
    int ans=1;
    for(int i=2;i<=min(int(9),n);i++)
    ans*=i;
    ans*=d;
    //cout<<ans<<endl;
    if(ans%3==0)v.push_back(3);
    if(ans%9==0)v.push_back(9);
    if(d==0||d==5)
    v.push_back(5);
    v.push_back(1);
    if(d%7==0||n>=3)
    v.push_back(7);
    sort(v.begin(),v.end());
    for(int i=0;i<v.size();i++)
    cout<<v[i]<<" ";
    cout<<endl;
    }
    return 0;
}

C
我们可以找到不为1和-1的数分开考虑 左侧右侧得最大最小值即为最大最小子段和dp即可 注意段长可以取0 对于加上x后数值取值范围（不可以用子段和，无法确定子段是否与x直接相连）则需要计算出x左侧的最大后缀和右侧得最大前缀和

#include<bits/extc++.h>
#define int long long
using namespace std;
const int maxn = 2e5 + 5;
int n,val,pos,sum;
int max1,min1,max2,min2;
int a[maxn],dp1[maxn],dp2[maxn];
set<int>s;
void solve()
{
    s.clear();
    cin>>n;
    pos=val=0;
    for (int i=1;i<=n;i++)
    {
        cin>>a[i];
        if (a[i]!=1&&a[i]!=-1)
        {
            val=a[i];
            pos=i;
        }
    }
   max1=-1e18;
   min1=1e18;
   for(int i=1;i<=n;i++)
   dp1[i]=dp2[i]=0;
   for(int i=1;i<=pos-1;i++)
   {
   dp1[i]=max(a[i],dp1[i-1]+a[i]);
   max1=max(max1,dp1[i]);
   dp2[i]=min(dp2[i-1]+a[i],a[i]);
   min1=min(min1,dp2[i]);
   }
   max1=max(max1,0ll);
   min1=min(min1,0ll);
   //cout<<max1<<" "<<min1<<endl;
   max2=-1e18;
   min2=1e18;
   for(int i=pos+1;i<=n;i++)
   {
   dp1[i]=max(a[i],dp1[i-1]+a[i]);
   max2=max(max2,dp1[i]);
   dp2[i]=min(dp2[i-1]+a[i],a[i]);
   min2=min(min2,dp2[i]);
   }
   max2=max(max2,0ll);
   min2=min(min2,0ll);
   //cout<<max2<<" "<<min2<<endl;
   for(int i=min(min1,min2);i<=max(max1,max2);i++)
   s.insert(i);
   int sum=0;
   int mx1=-1e18,mn1=1e18;
   for(int i=pos-1;i>=1;i--)
   {
   sum+=a[i];
   mx1=max(mx1,sum);
   mn1=min(mn1,sum);
   }
   mx1=max(mx1,0ll);
   mn1=min(mn1,0ll);
   //cout<<mx1<<" "<<mn1<<endl;
   int mx2=-1e18,mn2=1e18;
   sum=0;
   for(int i=pos+1;i<=n;i++)
   {
   sum+=a[i];
   mx2=max(mx2,sum);
   mn2=min(mn2,sum);
   }
   mx2=max(mx2,0ll);
   mn2=min(mn2,0ll);
   //cout<<mx2<<" "<<mn2<<endl;
   for(int i=val+mn1+mn2;i<=val+mx1+mx2;i++)
   s.insert(i);
   cout<<s.size()<<endl;
   for(auto x:s)
   cout<<x<<" ";
   cout<<endl;

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