Well, for 9999999999999999999999999999999999999999 (40 ‘9’s) -> 3240 -> 29. I think it doesn’t matter what or how big the integer is, it will eventually go into [1,162]. :-)
_____ 发件人: [email protected] [mailto:[email protected]] 代表 Iuri 发送时间: 2009年7月23日 11:27 收件人: [email protected] 主题: Re: 答复: 答复: Clear Numbers exercise Reading the question, I guess we can define n = 10^10 and m = 10^15 - 1, and it will broke this idea, because the maximum is not 10^15 - 1 anymore. 2009/7/23 Hawston LLH <[email protected]> "long long" arithmetic calculation (+ - x /) is fully supported by most machines, so this should not be a problem. :) 2009/7/23 周游弋 <[email protected]> Because any number bigger than 162 can be eventually summed back to a value less than 162, 999 -> 243 -> 29, 9999 -> 324 -> 29 etc. Thus if we could find determine the clear numbers in [1,162], then all integers can be deducted eventually by (or “to”) these determined numbers. As for finding S(n), it is just a counting problem. Starting from n, try n+1, n+2, n+2 …. And for S(n, m), just start from n+m, try n+m+1, n+m+2… As I said, no matter how big the number is, it will eventually fall in between [1,162] in four or five steps. As for the concrete implementation, there is an issue. The data scale is too big for long integer. Though a “double” variable could do, there will be precision problems with “double” values. In GNU extension we could use a “long long” type but it destroy the portability of the solution. So, we may need to write a function to sum two big integers together, digit by digit, to compute n+m when n or m are rather large. For this specific problem, first build a look-up table T[i] (i = 1, 2, 3, … , 1215) using the method we discussed, T[i] denotes whether i is a clear number. Now compute n+m by the “big integer adding function”. Starting from n+m+1, sum all its digits up and look it up in T. _____ 发件人: [email protected] [mailto:[email protected]] 代表 Hawston LLH 发送时间: 2009年7月22日 15:58 收件人: [email protected] 主题: Re: 答复: Clear Numbers exercise i think either you misunderstood the S(n) which is the minimum clear number greater than n or i have misunderstood your purpose of "the largest possible sum of all digits". Could you elaborate your purpose to find "162"? On Wed, Jul 22, 2009 at 3:35 PM, 周游弋 <[email protected]> wrote: For the largest possible value is 10^15, thus the largest possible sum of all digits comes from 999999999999999, which gives 15*81 = 1215 For 1215, the largest possible sum of all digits comes from 999 --> 243. For 243, the largest possible sum of all digits comes from 199 --> 163 For 163, it's 162. And it ends here. Finding the clear numbers in [1,162] is trival. -----邮件原件----- 发件人: [email protected] [mailto:[email protected]] 代表 FameofLight 发送时间: 2009年7月22日 15:00 收件人: google-codejam 主题: Re: Clear Numbers exercise Anybody please give me some clue on How to Solve this problem. On Jul 22, 5:26 am, khanh le <[email protected]> wrote: > Dear people, > i have a exercise, so i want to you solve it. you must code by C++ langguage > now, read the exercise below: > > Peter has just found a definition of *clear numbers* as the following: for > each positive integer n, we form another number by summing the squares of > the digits of n. We repeat this procedure. If at some step, we obtain the > number 1 then n is called a *clear number*. For example, for n=19, we have: > > 19 → 82 (= 12 +92) → 68 → 100 → 1 > > Thus, 19 is a clear number. > > Not all numbers are clear numbers. For example, for n=12, we have: > > 12 → 5 → 25 → 29 → 85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89 → 145 > > Peter is very interested in this definition of clear numbers. He issued a > challenge to the landlord: given a positive integer n, find S(n), the clear > number succeeding n, i.e. S(n) is the minimum clear number greater than n. > However, this question is so easy for the landlord that he challenged Peter > with another problem: given two positive integers n and m (1 ≤ n, m ≤ 1015), > find the number Sm(n)=S(S(…S(n) )) which is the mth clear number succeeding > n. > > Please help Peter to solve the task! > Input > > The first line contains t (0 < t ≤ 20) , the number of test cases. > > Each line in the next t lines contains two positive integers n and m. > Output > > Print t lines, each line contains the result of the corresponding test case. > Example > > *Input* > 2 > 18 1 > 1 145674807 > > *Output* > 19 > 1000000000 > > Notes > > There are 50% of the test cases in which 1 ≤ n, m ≤ 107. > > -- > Regard! > Khanh --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "google-codejam" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~----------~----~----~----~------~----~------~--~---
