Discussion @ 1407. One-two, One-two @ Timus Online Judge

Please, help!!! [SSAU_#617]snipious_#1 aka Pimenov Sergey Nikolaevich 9 Jan 2007 00:52

The biggest integer, that I find is 11111212122112, but I do not find any sequense or algo, what can I do???

Edited by author 09.01.2007 00:55

Re: Please, help!!! KirillB(Arkhangelsk - PomorSU) 9 Jan 2007 00:57

86 digits left for solution
Use long arithmetics

Re: Please, help!!! it4.kp 9 Jan 2007 01:55

why 86?

my solution has 98 digits.

BTW, does anyone know what is the shortest answer?

Re: Please, help!!! [SSAU_#617]snipious_#1 aka Pimenov Sergey Nikolaevich 9 Jan 2007 08:43

My helpprogram work very long... %(((

Edited by author 09.01.2007 08:43

Re: Please, help!!! Burunduk1 9 Jan 2007 08:47

It's very easy to constract answer with length N... (induction)

Re: Please, help!!! KirillB(Arkhangelsk - PomorSU) 9 Jan 2007 11:41

Burunduk1 wrote 9 January 2007 08:47

It's very easy to constract answer with length N... (induction)

You mean that it can be solved without long numbers?
I can not find nothing better than divide number on 2^N
on every step

Reccurence! svr 11 Jan 2007 22:57

Let H[N]- number under considaration.
Let we define integer K[N]=H[N]/(2^N);
Then K[N+1]=(m*(5^(N-1))+K[N-1])/2; where m=1 or 2;
and K[1]=1;
We solve this reccurence in long arithmetics.
In my module was bad long dividing, thank to this problem
that bug removed now.

Who can explain, how can I find the number, wich devide on 2^100??? [SSTtU]snipious_#0 aka Pimenov Sergey Nikolaevich 29 Jan 2007 18:44

I find 2^100 and than add 2^100 until all digits will be 1 and 2... My programm worked 2 hours, but didn't give me the answer!!!

Re: Who can explain, how can I find the number, wich devide on 2^100??? [SSTtU]snipious_#0 aka Pimenov Sergey Nikolaevich 6 Feb 2007 04:34

Am I right???

Re: Who can explain, how can I find the number, wich devide on 2^100??? Tralfamador 6 Feb 2007 18:06

Some number is divisible on 2^n if number which has been written down in his last n digits is divisible on 2^n.

Means we can consistently select digits in the answer.

Discussion of Problem 1407. One-two, One-two