2014-03-19 06:40

题目:有20瓶药,其中19瓶装的都是1.0克的药片,只有1瓶装了1.1克的药。给你一个能称出具体克数的电子秤,只允许你称一次,怎么找出那瓶不一样的?

解法:如果药片管够,从每个瓶子里取出数量各不相同的药片,根据质量的总和减去“期望的”质量总和,就知道哪瓶有问题了。

代码:

 1 // 6.1 There are 20 bottles of pills, all of which have pills of 1g, except one with 1.1g.
 2 // Given a balance that can provides exact measurement, can you devise a method to find out the special bottle with just one try?
 3 // Answer:
 4 //    If you pick n[i] pills from bottle i, the total weight should be sigma(n[i] * 1.0).
 5 //    But there is one with 1.1g pills, that would cause a little deviation to the sum.
 6 //    For example, if we pick i pills from bottle i, the sum should be (20 + 1) * 20 / 2 = 210.
 7 //    The real weight w should be greater than 210, (w - 210) / (1.1 - 1.0) will be the number.
 8 int main()
 9 {
10     return 0;
11 }

 

 posted on 2014-03-19 06:40  zhuli19901106  阅读(431)  评论(0编辑  收藏  举报