边工作边刷题：70天一遍leetcode: day 41-2

Water and Jug Problem

要点：利用了Bézout’s theorem：mx+n*y对于任意m，n都是gcd(x,y)的整倍数。所以z是gcd整倍数并且z<=x+y

• gcd(a,b)：a一定<b做辗转相除。结束条件是a=0，返回b（和1没关系）。而gcd==0只可能a=b=0

class Solution(object):
    def canMeasureWater(self, x, y, z):
        """
        :type x: int
        :type y: int
        :type z: int
        :rtype: bool
        """
        def gcd(a, b): # assume a<b
            if a==0: return b # error 1: return b not return 0 
            return gcd(b%a, a)
        
        if x>y: x,y=y,x
        gval = gcd(x,y)
        #print gval
        if gval==0:
            return z==0
        return z%gval==0 and z<=x+y