MT3021 拦截罪犯

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<=k
    int 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;
        }
        else
            l = 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();
}