• 相向双指针
• a[l] 和 a[r] 哪个离 x 远就删哪个

#include <bits/stdc++.h>
using namespace std;

vector<int> solve(vector<int>& a, int k, int x) {
    int n = a.size(), l = 0, r = n - 1;

    // a[l] 和 a[r] 哪个离 x 远就删哪个
    while (r - l + 1 > k) {
        if (abs(a[l] - x) > abs(a[r] - x)) l++;
        else r--;
    }

    vector<int> ans;
    for (int i = l; i <= r; i++) ans.push_back(a[i]);
    return ans;
}

int main() {	
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    
    int n, k, x;
    cin >> n >> k >> x;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];

    vector<int> ans = solve(a, k, x);
    for (int i = 0; i < k; i++) cout << ans[i] << " \n"[i == k - 1];

    return 0;
}