365. Water and Jug Problem

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 = 4Output: True

Example 2:

Input: x = 2, y = 6, z = 5Output: False

思路：

每个x和y都有其最大公约数，如果z是这个最大公约数的倍数，则一定可以量出。考虑特殊情况，z=0肯定真，还有x或者y等于0，则z只要是y或者x的倍数即可。如果x和y的总容量都不到z，z必定不能由它们两个装着，返回假。

class Solution {     int minyinzi(int  x,int y)    {       if(y>x) return minyinzi(y,x);       else if(y==0) return x;       else return minyinzi(y,x%y);    }public:    bool canMeasureWater(int x, int y, int z) {        if(z==0) return true;       if((!x&&!y)||x+y<z) return false;       if(!x) return !(z%y);       if(!y) return !(z%x);        int t=minyinzi(x,y);//最大公约数        return z%t?false:true;            }};