阿里巴巴实习笔试

 参见：http://blog.csdn.net/doc_sgl/article/details/8888904

附：

1. 士兵两两通话，最小通话次数

比如说假设有k个种子（k个人进行消息融合）

算法就是：

f(n) = n-k + f(k) + n-k

      = 2n + f(k) -2k

就是说要找使f(k) - 2k最小的k

而最终f(k) - 2k肯定是个常数

f(n)随n单增，从k=1开始肯定能找到使f(k) - 2k最小的k

于是找到k=4时，f(k) - 2k = -4最小

但最终发现只要n, k >= 4，无论k等于几，f(n) = 2n - 4

所以与k为多少无关

 

A popular formulation assumes there are n people, each one of whom knows a scandal which is not known to any of the others. They communicate by telephone, and whenever two people place a call, they pass on to each other as many scandals as they know. How many calls are needed before everyone knows about all the scandals? Denoting the scandal-spreaders as A, B, C, and D, a solution for n=4  is given by {A,B},{C,D} , {A,C}, {B,D}. The solution can then be generalized to n>4 by adding the pair {A,X} to the beginning and end of the previous solution, i.e.,{A,E} , {A,B}, {C,D}, {A,C}, {B,D}, {A,E}.

 

Let f(n) be the number of minimum calls necessary to complete gossiping among n people, where any pair of people may call each other. Then f(1)=0, f(2) = 1, f(3) = 3, and

f(n) = 2n-4,  for n>=4.

 This result was proved by (Tijdeman 1971), as well as many others.

In the case of one-way communication ("polarized telephones"), e.g., where communication is done by letters or telegrams, the graph becomes a directed graph and the minimum number of calls becomes

f(n) = 2n-2, for n>=4.

 (Harary and Schwenk 1974).

 

http://wenku.baidu.com/view/7f1b25c54028915f804dc22c.html

Gossiping (which is also called total exchange or all-to-all communication)： http://mathworld.wolfram.com/Gossiping.html