365. Water and Jug Problem

 

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.

 

这是一道脑筋急转弯题,我想很多人以前应该听过这道题目,有一个容量为3升和一个容量为5升的水罐,问我们如何准确的称出4升的水。我想很多人都知道怎么做,先把5升水罐装满水,倒到3升水罐里,这时5升水罐里还有2升水,然后把3升水罐里的水都倒掉,把5升水罐中的2升水倒入3升水罐中,这时候把5升水罐解满,然后往此时有2升水的3升水罐里倒水,这样5升水罐倒出1升后还剩4升即为所求。这个很多人都知道,但是这道题随意给我们了三个参数,问有没有解法,这就比较难了。这里我就照搬网上大神的讲解吧:

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 {
public:
    bool canMeasureWater(int x, int y, int z) {
        return z == 0 || (x + y >= z && z % gcd(x, y) == 0);
    }
    int gcd(int x, int y) {
        return y == 0 ? x : gcd(y, x % y);
    }
};

 

参考资料:

https://leetcode.com/discuss/111528/clear-explanation-of-why-using-gcd

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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