LibreOJ 2423 跳石头

题目链接：LibreOJ 2423 跳石头

题目大意：

题解：
对最短跳跃距离进行二分答案即可。

#include <iostream>
using namespace std;

int res[60000];
int l, n, m, ans;

int get(int x) {
    int t = 0;
    int last = res[0];
    for (int i = 1; i <= n + 1; i++) {
        if (res[i] - last < x) {
            t++;
        } else {
            last = res[i];
        }
    }
    return t;
}

int main() {
    cin >> l >> n >> m;
    for (int i = 1; i <= n; ++i) {
        cin >> res[i];
    }
    res[n + 1] = l;
    int lift = 0, right = l;
    while (lift <= right) {
        int mid = (lift + right) / 2;
        if (get(mid) <= m) {
            ans = mid;
            lift = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    cout << ans;
    return 0;
}