(这几天转载的题目,主要来自田苗)
一个男孩、一个女孩和一条狗,他们行走(跑)的速度分别是每小时4里,3里和10里。他们从一笔直路上的同一起点,朝同一方向同时出发。出发后,狗则在男孩和女孩之间徘徊,即狗追上男孩,就往回走,与女孩碰头就往前走。一小时后,狗离起点有多远?朝什么方向(朝前还是朝后)?
一小时和一小时后是两个不同的时间概念,叫人怎理解啊 |
This is not a well defined problem. The dog can be at any position between 3 and 4 li, and can be running toward either the boy or the girl. Just think it backwards... |
The problem is that the dog bounced back and forth between the boy and the girl infinite many times at the begining. You may think the problem backward: at the begining, the boy start at the 4 li position, the girl at the 3 li position, and you can put the dog at any position toward either direction. Now, let the boy and the girl walk back to the origin..., and then reverse the process... |
的确,严格说来,该题有问题,也就是在起点的时候。 但是如果我们不考虑这个边界问题。假定狗调头的不需要时间。 或者这么说吧,开始的时候,男孩在女孩前1里,狗跟女孩在同一起跑线上。这就顺利地解决的边界问题。 |
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