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; } };
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