思路：用二分，每次二分间距，判断需要的组数是否>k。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int L, n, k;
int a[N];bool check(int p)
{ // 看此时的间距所用的警力数满不满足<=kint cnt = 0;for (int i = 2; i <= n; i++){int temp = a[i] - a[i - 1];int a1 = temp / p; // 需要多少组int a2 = temp % p;if (a1 > 0){if (a2 > 0){cnt += a1;}else{ // 无余数，例4/2=2，只需2-1=1组即可cnt += a1 - 1;}}}if (cnt > k)return false;return true;
}int main()
{cin >> L >> n >> k;for (int i = 1; i <= n; i++){cin >> a[i];}sort(a + 1, a + n + 1);int l = 1, r = L, mid;while (l <= r){mid = (l + r) / 2;if (check(mid)){r = mid - 1;}elsel = mid + 1;}cout << l;
}

 下面是错误代码：（有没有好心人告诉我为什么错了/(ㄒoㄒ)/~~）

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int l, n, k;
priority_queue<int, vector<int>, less<int>> q;//大根堆
int b[N];
int main()
{cin >> l >> n >> k;for (int i = 1; i <= n; i++){cin >> b[i];}sort(b + 1, b + n + 1); // 升序for (int i = 2; i <= n; i++){q.push(b[i] - b[i - 1]);}for (int i = 0; i < k; i++){int pp = q.top();q.pop();int uu = pp / 2;q.push(uu);q.push(pp - uu);}cout << q.top();
}