难度：中等

思路源于题目标签 “二分”：

1. 二分的上界应该是所有车都给修车能力值最小的人修，下界我设为0
2. 每次搜索时判断当前时间下，每位机械工总共能修 n 辆车，n >= cars 则右边界左移，否则左边界右移

class Solution:
    def repairCars(self, ranks: List[int], cars: int) -> int:
        l, r = 0, min(ranks) * (cars ** 2)
        while l < r:
            mid = (l + r) >> 1
            n = 0
            for i in ranks:
                n += int((mid//i) ** 0.5)
                if n >= cars:
                    break
            if n >= cars:
                r = mid
            else:
                l = mid + 1
        return r

if n >= cars:break 提前结束循环可以快很多很多
官解一句话完成转移判断

sum([floor(sqrt(m // x)) for x in ranks]) >= cars