365. Water and Jug Problem ---------M ~~~~~~~~

 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

 

Constraints:

• 0 <= x <= 10^6
• 0 <= y <= 10^6
• 0 <= z <= 10^6

A:                 不是我做的，直接抄的答案

Forget about two jugs pouring between each other, which may make you confused.

Let's make it simple: assuming we have one big enough bucket and two cups with volume x and y, respectively. Now we want to perform a series of operation -- pouring water in and out only by those two cups with exactly amount x or y. Somehow, there will be only z water left in this big bucket eventually. Then the equation will be:

z = m * x + n * y

m means using cup-x m times. If m is positive, it means pouring in. Otherwise, it's pouring out.
n is similar.

For example, 4 = (-2) * 3 + 2 * 5, which means you pour in water twice with cup-5 and pour out water twice with cup-3. Talk back to the question, it's like we first fill jug-5, pour water to jug-3 from jug-5, empty jug-3, pour the remaining 2 water into jug-3 from jug-5, fill jug-5 again, pour water into jug-3 from jug-5, empty jug-3, then we have only 4 water left in jug-5. It's exactly fill jug-5 twice and empty jug-3 twice.

Now the question is, can we find those two m and n exist?

The answer is YES. Here we need a little math -- Bezout's identity, which is:

We can always find a and b to satisfy ax + bx = d where d = gcd(x, y)

So, everything is clear, if z % d == 0, then we have (a*z/d)*x + (b*z/d)*y = z, which means m and n exist.

Below is the code:

class Solution {
public:
    bool canMeasureWater(int x, int y, int z) {
        int a = min(x,y);
        int b = max(x,y);
        if(z>a+b) return  false;
        if(z==a+b) return true;
        if(z==a) return true;
        if(z==b) return true;
        if(a==0) return false;
       int gcd = __gcd(a,b);
        if(gcd!=0 && z%gcd!=0) return false;
        return true;
    }
};