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
>>
>>
>>
>>
>>
>>
>>
>>
>
> >
>

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