$RT$，#3($N=M=0$)过不去，是只能特判吗

$code:$

#include<iostream>
#include<algorithm>
using namespace std;
int L, M, N, dis[50008];
bool check(int ans){
    int last = 0, cnt = 0;
    for(int i = 1; i <= N; ++i){
        if(dis[i] >= last + ans){
            ++cnt;
            last = dis[i];
        }
    }
    return N - cnt <= M;
}
int main(){
    ios::sync_with_stdio(false);
    cin >> L >> N >> M;
    for(int i = 1; i <= N; ++i)
        cin >> dis[i];
    dis[N + 1] = L;
    int l = 1, r = dis[N];
    while(l <= r){
        if(check((l + r) >> 1))
            l = ((l + r) >> 1) + 1;
        else
            r = ((l + r) >> 1) - 1;
    }
    cout << r << endl;
}