You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs. If z liters of water is measurable, you must…

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs. If z liters of water is measurable, you must…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 数学题 相似题目 参考资料 日期 题目地址:https://leetcode.com/problems/water-and-jug-problem/description/ 题目描述 You are given two jugs with capacities x and y litres. There is an infinite amount o…

题目描述: You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly zlitres using these two jugs. Operations allowed: Fill any of the jugs…

可以想象有一个无限大的水罐,如果我们有两个杯子x和y,那么原来的问题等价于是否可以通过往里面注入或倒出水从而剩下z. z =? m*x + n*y 如果等式成立,那么z%gcd(x,y) == 0. class Solution(object): def canMeasureWater(self, x, y, z): """ :type x: int :type y: int :type z: int :rtype: bool """ if z…

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs. If z liters of water is measurable, you must…

https://leetcode.com/problems/water-and-jug-problem/description/ -- 365 There are two methods to solve this problem : GCD(+ elementary number theory) --> how to get GCF, HCD,  BFS Currently, I sove this by first method 1. how to compute GCD recursive…

［抄题］: 简而言之:只能对 杯子中全部的水/容量-杯子中全部的水进行操作 You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs. If z li…

有两个容量分别为 x升 和 y升 的水壶以及无限多的水.请判断能否通过使用这两个水壶,从而可以得到恰好 z升 的水?如果可以,最后请用以上水壶中的一或两个来盛放取得的 z升 水.你允许:    装满任意一个水壶    清空任意一个水壶    从一个水壶向另外一个水壶倒水,直到装满或者倒空示例1: (From the famous "Die Hard" example)输入: x = 3, y = 5, z = 4输出: True示例2:输入: x = 2, y = 6, z = 5输出…

莫名奇妙找了个奇怪的规律. 每次用大的减小的,然后差值和小的再减,减减减减减减到差值=0为止.(较小的数 和 差值 相等为止,这么说更确切) 然后看能不能整除就行了. 有些特殊情况. 看答案是用GCD做的,2行就写出来了,但是不是很理解. Ax + By = z,A B为正负 int,就是有接..神奇. 放个自己写的垃圾代码. public class Solution { public boolean canMeasureWater(int x, int y, int z) { if(z>x+…