>     事实上，早在昨天之前我已经有在网上看过这个题的啦，提供题目的网站上也给出了这题的解法分析，标程跟测试数据。看了下，今天A题的report，发现跟网站上的一样，只是翻译了一下。若此题是poj上注的作者所创，奇怪作者提供给月赛的题目为什么会在月赛前出现。希望出题人给出解释。
> 
> 
> 下面是此题在网站分析：
> In order to solve this problem, we should search in two more or less same sequences of numbers,
> we only need to find the kth digit in the consecutive integers sequence and the kth digit in
> the squares sequence.
> 
> In the consecutive integers sequence there are 9 one-digit integers, 90 two-digit integers, 
> 900 three-digit integers and so on. In the squares sequance there are floor (sqrt (10)) one-digit
> squares, floor (sqrt (100)) - floor (sqrt (10)) two-digit squares and so on. Knowing this (slicing 
> the sequence and counting them in each interval) is far enough to determine the addition of them.
> 
> There is only one thing left. That is to determine whether we have an overflow from right or not.
> That'll be determinable by doing the same with k + 1, (i.e. finding the result of k + 1th digits
> addition). If it is more than 9 surely we won't have any overflow, if it is less than 9 we don't
> have overflow, and finally if it is exactly 9 we should repeat this process for k + 2. Continuing
> this will help us to determine the overflow value
> 
> 至于网站地址，Google下第一个就可以找到。

• http://www.iaumccc.com/info/r2m1/radd.pdf (0) sothink 2005-08-01 13:10:25
  • Re:http://www.iaumccc.com/info/r2m1/radd.pdf (144) windbells 2005-08-01 13:30:02