A24009. #include <vector> #include <algorithm> #include <iostream> using namespace std; bool f0(vector<int> &a, int m, int k) { int s = 0; for (int i = 0, j = 0; i < a.size(); i++) { while (a[i] - a[j] > m) …

#include <vector>
#include <algorithm>
#include <iostream>

using namespace std;

bool f0(vector<int> &a, int m, int k) {
    int s = 0;
    for (int i = 0, j = 0; i < a.size(); i++) {
        while (a[i] - a[j] > m)
            j++;
        s += i - j;
    }
    return s >= k;
}

int f(vector<int> &a, int k) {
    sort(a.begin(), a.end());

    int g = 0;
    int h = a.back() - a[0];
    while (g < h) {
        int m = g + (h - g) / 2;
        if (f0(a, m, k)) {
            h = m;
        }
        else {
            g = m + 1;
        }
    }

    return g;
}

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

假设输入总是合法的且 1 ≤ a_i ≤ 10⁸, n≤ 10000, 1≤ k ≤ n(n−1)/2，完成下面的判断题和单选题:

将第 24 行的 m 改为 m - 1，输出有可能不变，而剩下情况为少 1。（    ）