用3升杯取水,倒入5升杯;再取一杯水倒入5升杯,自剩1升,做一记号。 将5升杯倒空,3升杯倒入5升杯; 按记号自取一升 |
不许做记号,没给条件做记号。 |
I used to write a small program to solve 3 Water Jug problem. Some water jug problem needed minimum 98 steps to solve, I remember that. I could take the java program out to solve it. But here I want to challenge my mental ability here. |
Here I assume you only allow to pour the water not allow to do other tricky actions, such as marking, eye measuring, or bla bla bla I think if you pour a water from A jug to B jug, only two results will come out. Jug A is empty and Jug B must have all the water which originates from A and B. If water A + water B <= Capacity of Jug B. Not other possible results. The extended conclusion is that, if you pour the water between two jugs, one must either be empty or full afterward. Since Capacity A = 3, Capacity B = 5, Capacity C >= 10, if you need 1 in A, 1 in B and 8 in C, it is impossible to get such result, due to my above analysis. None of water jugs is either empty or full. So it is impossible to get such combination. |
看,没浪费一滴水吧,至于以后跑厕所的事,偶就不管了。 |
In my analysis, I already said assume no tricky thing. |
The original problem is here. I simply changed wine to water, and translated English into Chinese. I believe the only trick, if we call it a trick, is we should understand "without wasting any wine" correctly. Jean-Pierre has a big wine-vat with 10 liters of wine in it, and two wine-jugs which can hold exactly 3 and 5 liters of wine respectively. |
偶就讨厌那种答案明确,死算就出结果的题。所以捡些好玩的题与大家分享。别太认真呦。 |
(kidding) |
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