python【力扣LeetCode算法题库】365- 水壶问题(裴蜀等式)

李响Superb 2021-07-20 03:46:00
Python python开发


  1. 水壶问题
    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: False
若a,b是整数,且gcd(a,b)=d,那么对于任意的整数x,y,ax+by都一定是d的倍数,特别地,一定存在整数x,y,使ax+by=d成立。
对任意两个整数a、b设d是它们的最大公约数。那么关于未知数x和y的线性丢番图方程(称为裴蜀等式):
ax + by = m
有整数解(x,y)当且仅当m是d的倍数。裴蜀等式有解时必然有无穷多个解。

针对任意给出的x,y,z,可以假设用x与y分别往一个很大的容器里倒水和倒出水,那么如果z=mx+ny,找到合适的正整数m与n即为所求,其中m与n为负时,往外倒水,为正时,往里倒水。

若z<=x+y且z%d==0,说明存在m、n使得最终剩余水为z。根据贝祖等式,d为x、y的最

版权声明
本文为[李响Superb]所创,转载请带上原文链接,感谢
https://blog.51cto.com/u_14013325/2891411

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