A
先考虑用填2然后用1填补，若可填补则成功

# include <bits/stdc++.h>
using namespace std;
#define int long long 
signed main()
{
int t;
cin>>t;
while(t--)
{
    int a,b;
    cin>>a>>b;
    int x=0;
    bool st=0;
    if(b%2==0)
    x=0;
    else x=2;
    if(x==2&&a-2>=0&&(a-2)%2==0)
    st=1;
    if(x==0&&a%2==0)
    st=1;
    if(st)
    cout<<"YES";
    else  cout<<"NO";
    cout<<endl;
}

}

B
题意模拟即可

# include <bits/stdc++.h>
using namespace std;
#define int long long
signed main()
{
    int t;
    cin>>t;
    //cout<<(5%3==0)||(5%3==1);
    while(t--)
    {
    int n;
    string s=" ",s1;
    cin>>n>>s1;
    s+=s1;
    bool st=0;
    int x=sqrt(n);
    if(x*x!=n)
    {
    cout<<"NO"<<endl;
    continue;
    }
    for(int i=1;i<=n;i++)
    {
    if((i>=1&&i<=x))
    {
        if(s[i]=='0')
        st=1;
    }
    if((i>=n-x+1&&i<=n))
    {
        if(s[i]=='0')
        st=1;    
    }
    else if(i>x&&i<n-x+1)
    {
    if(i%x==0||i%x==1)
    {
        //cout<<i<<" ";
        if(s[i]=='0')
        {
        st=1;
        //cout<<i<<endl;
        }
    }
    else 
    {
        //cout<<i<<" ";
        if(s[i]=='1')
        st=1;
    }    
    } 
    }    
    if(st)
    cout<<"NO";
    else cout<<"YES";
    cout<<endl;
    }
return 0;
}

C
暴力模拟即可

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

D
模拟样例会发现形成环 同一个环上的点可以相互到达 dfs环上的点
存在重复计算 所以递归到底层时不断返回记录答案即可

#include <bits/stdc++.h>
using namespace std;
#define int long long 
const int N=200010;
int a[N],ans[N];
int n;
string s;
bool vis[N];
int cnt;
int dfs(int p)
{
    if(vis[p])
    {
        return cnt;
    }
    vis[p]=1;
    if(s[p]=='0')
        cnt++;
    ans[p]=dfs(a[p]);
    return cnt;
}

signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    cin>>n;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    s="";
    cin>>s;
    s=" "+s;
    memset(ans,-1,sizeof(ans));
    for(int i=1;i<=n;i++)
    {
        if(ans[i]==-1)
        {
            cnt=0;
            memset(vis,0,sizeof(vis));
            ans[i] = dfs(i);
        }
    }
    for(int i=1;i<=n;i++)
        cout<<ans[i]<<" ";
    cout<<endl;
    }
}

E
分类讨论
对于偶数只执行操作二，奇数执行一次操作一。用前缀和的思想，记 odd[i][j] 为前 i个数，奇数位字符为 j的个数，even[i][j] 为偶数位字符为 j的个数，再枚举删去哪个点，最终序列的奇数位是这个点前面的奇数位和后面的偶数位，再分别枚举奇数位和偶数位的最终字符，这样就能得到正确答案。
偶数直接计算，奇数因为要删去一位所以对于奇数位考虑前面奇数后面的偶数，偶数位同理。

# include <bits/stdc++.h>
using namespace std;
#define int long long 
const int N = 2e5 + 5;
int odd[N][26], even[N][26];
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    int n;
    string s;
    cin>>n>>s;
    s=' '+s;
    int ans=0x3f3f3f3f;
    for(int i = 1; i <= n; ++i)
    {
        for(int j = 0; j <= 25; ++j) odd[i][j] = even[i][j] = 0;
        if(i%2==1) odd[i][s[i] - 'a'] = 1;
        else even[i][s[i] - 'a'] = 1;
        for(int j = 0; j <= 25; ++j)
        odd[i][j] += odd[i - 1][j], even[i][j] += even[i - 1][j];
    }
    if(n%2==1)
    {
    for(int i = 1; i <= n; ++i)
    {
        for(int j = 0; j <= 25; ++j)
            for(int k = 0; k <= 25; ++k)
            {
            int x = odd[i - 1][j] + even[n][j] - even[i][j];
            int y = even[i - 1][k] + odd[n][k] - odd[i][k];
            ans = min(ans, n - 1 - x - y);
            }
    }
    cout<<ans+1;
    }
    else 
    {
    for(int i = 0; i <= 25; ++i)
        for(int j = 0; j <= 25; ++j)
            ans = min(ans, n - (odd[n][i] + even[n][j]));
    cout<<ans;
    }
    cout<<endl;
    }
}