我的逻辑很简单，就是枚举相邻最大距离，check一下判断需要插入的路标数量

#include <iostream>
using namespace std;

const int N = 1e5 + 2;
int l, n, k, a[N];

int check(int x){
    int cnt = 0; //路标
    for(int i = 1, last = 0; i < n; i++){
        if(last) continue;
        cnt += (a[i] - a[i - 1]) % x;
        last = i;
    }
    return cnt;
}

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> l >> n >> k;
    for(int i = 0; i < n; i++) cin >> a[i];
    int s = 0, t = l;
    while(s + 1 != t){
        int mid = (s + t) / 2;    
        if(check(mid) >= k) t = mid;
        else s = mid;
    }
    cout << t;
    return 0;
}