water-and-jug-problem

以下这个解法也是参考了一些讨论：

https://leetcode.com/discuss/110235/c-solution-using-euclidean-algorithm

还有这个解释原理的，没有看懂：https://leetcode.com/discuss/110525/a-little-explanation-on-gcd-method

class Solution {
    int gcd(int a, int b) {
        // below suppose a <= b
        // and if b < a, then b%a == b
        return a==?b:gcd(b%a, a);
    }
public:
    bool canMeasureWater(int x, int y, int z) {
        // notice, operator priority: % > == > && > ||
        // so below is equivalent to
        // (z == 0) || ((z <= (x+y)) && ((z % gcd(x,y)) == 0))
        return z ==  || z <= (x+y) && z % gcd(x, y) == ;
    }
};

31 / 31 test cases passed.	Status: Accepted
Runtime: 0 ms