题意:一直青蛙过河,河宽l米,上面有n个石头,给出每个石头的位置,然后要求最多m步跳过去,问他最小最远一步可以跳多远。
题解:最小的最大值的计算用二分,二分出距离然后拿去判断,在这个距离内贪心的跳,判断能否跳到对岸。

#include <stdio.h>
#include <algorithm>
using namespace std;
const int N = 500005;
int l, n, m, sto[N];

bool judge(int x) {
    int k = 0, cnt = 0;
    for (int i = 1; i <= n + 1; i++) {
        if (sto[i] - sto[k] > x) {
            if (k == i - 1)
                return false;
            cnt++;
            k = i - 1;
            i--;
            if (cnt >= m)
                return false;
        }
    }
    return true;
}

int main() {
    while (scanf("%d%d%d", &l, &n, &m) == 3) {
        for (int i = 1; i <= n; i++)
            scanf("%d", &sto[i]);
        sort(sto + 1, sto + n + 1);
        sto[0] = 0;
        sto[n + 1] = l;
        int left = 0, right = l + 1;
        while (left < right) {
            int mid = (left + right) / 2;
            if (judge(mid))
                right = mid;
            else
                left = mid + 1;
        }
        printf("%d\n", left);
    }
    return 0;
}