A

按照题意模拟即可
code

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    cout << a[k] << '\n';
    return 0;
}

B

直接双指针即可
假设我们现在的区间是[i, j]
那么可以证明不可能存在一个小于i的数k，能与大于j的数u匹配
所以可以双指针

#include <bits/stdc++.h>

using namespace std;

const int N = 3e5 + 10;

int n, k, a[N], c[N];

int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> k;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
        c[i] = c[i - 1] + (a[i] == 0);
    int j = 0, res = 0, t1 = n + 1, t2 = n + 1; // 为了处理边界情况而赋值
    for (int i = 1; i <= n; i++) {
        while (j + 1 <= n && c[j + 1] - c[i - 1] <= k)
            j++;
        if (res < j - i + 1) {
            res = j - i + 1;
            t1 = i, t2 = j;
        }
    }
    cout << res << '\n';
    for (int i = 1; i <= n; i++)
        if (i >= t1 && i <= t2)
            cout << 1 << ' ';
        else
            cout << a[i] << ' ';
    return 0;
}

C题
对于每次操作，我们假设新添加的值是x和y
那么我们树的直径只有三种可能，原来的，$a->x$, $b->x$。这里选择$a->y$,$b->y$也是可以的。
之后就简单了，上代码
code

#include <bits/stdc++.h>

using namespace std;

const int N = 1000010, K = 20;

int n, m, f[N][K], d[N];

int lca(int a, int b) {
    if (d[a] < d[b])
        swap(a, b);
    for (int k = K - 1; k >= 0; k--)
        if (d[f[a][k]] >= d[b])
            a = f[a][k];
    if (a == b)
        return a;
    for (int k = K - 1; k >= 0; k--)
        if (f[a][k] != f[b][k])
            a = f[a][k], b = f[b][k];
    return f[a][0];
}

int get_dist(int a, int b) {
    int LCA = lca(a, b);
    return d[a] + d[b] - d[LCA] * 2;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> m;
    for (int i = 2; i <= 4; i++)
        f[i][0] = 1, d[i] = 1;
    n = 4;
    int A = 2, B = 3, res = 2;
    while (m--) {
        int u;
        cin >> u;
        int x = n + 1, y = n + 2;
        n += 2;
        d[x] = d[y] = d[u] + 1, f[x][0] = f[y][0] = u;
        for (int k = 1; k < K; k++)
            f[x][k] = f[f[x][k - 1]][k - 1], f[y][k] = f[f[y][k - 1]][k - 1];
        int da = get_dist(A, x), db = get_dist(B, x);
        if (da > db && da > res)
            res = da, B = x;
        else if (da <= db && db > res)
            res = db, A = x;
        cout << res << '\n';
    }
    return 0;
}