[LeetCode] Water and Jug Problem 水罐问题

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 have z liters of water contained within one or both buckets by the end.

Operations allowed:

• Fill any of the jugs completely with water.
• Empty any of the jugs.
• 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

Credits:
Special thanks to @vinod23 for adding this problem and creating all test cases.

这道问题其实可以转换为有一个很大的容器，我们有两个杯子，容量分别为x和y，问我们通过用两个杯子往里倒水，和往出舀水，问能不能使容器中的水刚好为z升。那么我们可以用一个公式来表达：

z = m * x + n * y

其中m，n为舀水和倒水的次数，正数表示往里舀水，负数表示往外倒水，那么题目中的例子可以写成: 4 = (-2) * 3 + 2 * 5，即3升的水罐往外倒了两次水，5升水罐往里舀了两次水。那么问题就变成了对于任意给定的x,y,z，存不存在m和n使得上面的等式成立。根据裴蜀定理，ax + by = d的解为 d = gcd(x, y)，那么我们只要只要z % d == 0，上面的等式就有解，所以问题就迎刃而解了，我们只要看z是不是x和y的最大公约数的倍数就行了，别忘了还有个限制条件x + y >= z，因为x和y不可能称出比它们之和还多的水，参见代码如下；

class Solution(object):
    def canMeasureWater(self, x, y, z):
        dmax = max(x, y)
        dmin = min(x, y)
        if dmin ==0 and dmax ==z or z ==0:
            return True
        elif dmin == 0 and dmax != z or x+y<z:
            return False
        resid = self.gcd(dmax, dmin)
        return True if z % resid == 0 else False

    def gcd(self, x, y):
        return x if y == 0  else self.gcd(y, x%y)

x = 11
y = 13
z = 2
s = Solution()
print s.canMeasureWater(x,y,z)