C：

n个元素，n个人轮流取数，可以取数组头或尾元素，你在第m轮取，并且你可以控制k个人取什么元素，问你能取得最大元素是多少(前面的人不一定都取最大的)

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
const int maxn = 4e3;
const int inf = 0x3f3f3f3f;
int a[maxn];
 
inline const int read()
{
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + ch - '0'; ch = getchar(); }
    return x * f;
}
 
inline void write(int x)
{
    if (x < 0) { putchar('-'); x = -x; }
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("input.txt", "r", stdin);
#endif
    int t = read();
    while (t--)
    {
        int n = read(), m = read(), k = read(), ans = 0;
        k = min(k, m - 1);
        for (int i = 1; i <= n; i++) a[i] = read();
        for (int i = 0; i <= k; i++) // 枚举前k个人取左端数目,也就是强制k-i个放在右边
        {
            int mn = inf;
            for (int j = i + 1; j <= i + m - k; j++) // 枚举后（m-k-1）个人取完后的左端点，不受控制的人从0个到m-k个排在左边
                mn = min(mn, max(a[j], a[j + n - m]));///这里就是可取的两端了
            ans = max(ans, mn);
        }
        write(ans); putchar(10);
    }
    return 0;
}

　　D