Python3代码 class Solution: def canMeasureWater(self, x: int, y: int, z: int) -> bool: # solution one: BFS from collections import deque queue = deque([[0, 0]]) visited = set([(0, 0)]) while queue: cur_x, cur_y = queue.pop() if z in [cur_x, cur_y, cur_x + cur...
jug handled a. 不匀称的, 片面的, 单方面的 jug eared adj. 有招风耳的 water n. 1. 水,雨水,海水,水位,水面,流体,水域 v.[T] 1. 浇水,供以水,注入水,使湿,加水稀释,给...水喝 v.[I] 1. 流泪 2. 流口水 problem free 毫无问题的 problem solver n. 善于解决问题的人,实干家 参考例...
Pour water from one jug into another till the other jug is completely full or the first jug itself is empty. Example 1: (From the famous"Die Hard"example) Input: x = 3, y = 5, z = 4 Output: True Example 2: Input: x = 2, y = 6, z = 5 Output: False [暴力解法]: 时间分...
英语解释 water-jug problem water-jug problem是什么意思、water-jug problem怎么读 water-jug problem汉语翻译 【计】 水和罐问题
The Water Jug problem is a famous problem in Artificial Intelligence, Problem solving, Recreational, Computer Programming and Psychology. The solution of t... RS Mary,RJ Therese,TM Prabaharan - World Congress on Computing & Communication Technologies 被引量: 0发表: 2017年 An Algorithmic Arithmetic...
class Solution: def canMeasureWater(self, x: int, y: int, z: int) -> bool: # solution one: BFS from collections import deque queue = deque([[0, 0]]) visited = set([(0, 0)]) while queue: cur_x, cur_y = queue.pop() if z in [cur_x, cur_y, cur_x + cur_y]: retur...
Pour water from one jug into another till the other jug is completely full or the first jug itself is empty. **Example 1: ** Input:x=3,y=5,z=4Output:True Example 2: Input:x=2,y=6,z=5Output:False 这道题实际上是道数学智力题,求解的关键就是判断z能否整除x与y的最大公约数。当然,...
英文: She poured some water into the earthenware jug .中文: 她向那个陶罐里倒了些水。英文: She poured some water into the earthenware jug.中文: 她向那个陶罐里倒了些水。英文: Cost: Bottled water can cost approximately $1 for a gallon jug, while tap water costs pennies on the dollar.中文...
class Solution: def canMeasureWater(self, x: int, y: int, z: int) -> bool: if z == 0 or (z <= x + y and z % self.__gcd(x, y) == 0): return True return False def __gcd(self, x: int, y: int) -> int:
A simple solution to group the jugs into pairs is to compare a red jug with each blue jug. Once we get the matching pair, group that pair, and repeat for the next red jug. This algorithm can be easily implemented, but it will perform at mostn2comparisons. ...