A

# include <bits/stdc++.h>
using namespace std;
#define int long long 
typedef long long LL;
int n;
map<int,int>ma;
void solve()
{
    ma.clear();
    cin>>n;
    int sum=0;
    for(int i=1;i<=n;i++)
    {
    int x;
    cin>>x;
    ma[x]++;
    }
    for(auto [x,y]:ma)
    sum+=y/2;
    cout<<sum<<endl;
}
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    solve();
    }
}

B
题意为在序列中寻炸两数乘积为k-2 枚举两个数的位置，第二个数用二分查找

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e6+10;
int a[N];
signed main()
{
    int t;
    cin>>t;
    while(t--)
    {
    int k;
    cin>>k;
    for(int i=1;i<=k;i++)cin>>a[i];
    sort(a+1,a+k+1);
    int y=k-2;
    for(int i=1;i<=k;i++)
    {
        int l=1,r=k;
        while(l<r)
        {
            int mad=(l+r)/2;
            if(a[mad]*a[i]>=y)r=mad;
            else l=mad+1;
        }
        if(a[l]*a[i]==y)
        {
            cout<<a[i]<<" "<<a[l]<<'\n';
            break;
        }
    }
    }
}

C
考虑 奇奇奇偶偶偶 会存在一个交替和为奇数,9即为最小的复合整数 n<=8时特判即可

# include <bits/stdc++.h>
using namespace std;
#define int long long 
typedef long long LL;
int n;
void solve()
{
    cin>>n;
    if (n < 5)
    {
        cout << -1 << "\n";
        return;
    }
    else if (n == 7)
    {
        cout << "1 3 5 4 6 2 7\n";
        return;
    }
    else if (n == 6)
    {
        cout << "1 3 5 4 2 6\n";
        return;
    }
    else if (n == 5)
    {
        cout << "1 3 5 4 2\n";
        return;
    }
    else 
    {
    for(int i=1;i<=n;i+=2)
    {
           if(i!=7)
    cout<<i<<" ";
    }
    cout<<7<<" "<<2<<" ";
    for(int i=4;i<=n;i+=2)
    cout<<i<<" ";
    cout<<endl;
    }

}
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    solve();
    }
}

D
题目并未限定跳跃次数 我们可以进行任意次跳跃，只要可以跳过所有障碍物即可，贪心的考虑我们肯定要先选取较大的能量值，用优先队列维护能量值并判断每个障碍物是否能通过

#include <bits/stdc++.h>
#define int  long long
using namespace std;
const int N=2e5+10;
struct Node 
{
  int l, r;
}a[N], b[N];
int n,m,l;
int ans,cnt = 1;
priority_queue <int> q;
void solve()
{
    while (!q.empty())
    q.pop();
    ans = 0;
    cin >> n >> m >> l;
    for (int i = 1; i <= n; i ++) 
    cin >> a[i].l >> a[i].r;
    for (int i = 1; i <= m; i ++) 
    cin >> b[i].l >> b[i].r;
    cnt=1;
    int sum=1;
    for (int i = 1; i <= n; i ++)
    {
       while(cnt<=m&&b[cnt].l<a[i].l)
       {
       q.push(b[cnt].r);
       cnt++;
       }
       while(!q.empty()&&sum<=a[i].r-a[i].l+1)
       {
       sum+=q.top();
       q.pop();
       ans++;
       }
       if(sum<=a[i].r-a[i].l+1)
       {
       cout<<-1<<endl;
       return ;
       }
    }
    cout << ans << endl;
}
signed main ()
{
    int T;
    cin >> T;
    while (T --) solve();
    return 0;
}