10633 - Rare Easy Problem

windows2k · Post by **windows2k** » Fri Apr 09, 2004 10:36 am

I thought it is an easy problem.
But get WA all the time.
What's wrong with my input/output
input

output

Code: Select all

1012
111109 111110
4566
8637

Or someone could give some tricky input/output, thx

rotoZOOM · Post by **rotoZOOM** » Fri Apr 09, 2004 12:12 pm

Check trailing spaces in output.
You should use unsigned long long instead of signed long long.

Input:
1000000000000000000

Output:
1111111111111111111

prince56k · Post by **prince56k** » Fri Apr 09, 2004 1:25 pm

i don't think that u should use unsigned long long. i used only long long.
if n=9 output 10
and if u use 10*n/9 then u will face overflow problem if n is 18 digit

Hope, it will help u

Eduard · Post by **Eduard** » Fri Apr 09, 2004 2:14 pm

Maybe 10^18 is not so big,but i use big numbers and got AC at one's.
I don't know how in C++ may be long long is bigger than 10^18 but in Pascal i don't think that INT64 is bigger than 10^18.That why i use big numbers.

PdR · Post by **PdR** » Fri Apr 09, 2004 3:36 pm

With some care no overflow will ever ocur in a 64 bit signed integer.

mlvahe · Post by **mlvahe** » Sun Apr 11, 2004 12:39 pm

In Pascal INT64 type can store integers, less than 2^63-1>10^18. So you could fill free to use int64.

10*a/9 = a + a/9.
if you write this way you won't meet any overflowing.

P.S. This is realy very easy problem.

Sajid · Post by **Sajid** » Fri Apr 16, 2004 9:38 pm

How can we detemine about the multiple answer...?
(we can find out a number easily by the formula)

shamim · Post by **shamim** » Sat Apr 17, 2004 11:58 am

The other answers lie (+-) 10 of the number calculated using formula.

Sajid · Post by **Sajid** » Sat Apr 17, 2004 8:12 pm

hmm, I guess, I did simple mistake in the contest. I checked only previous ten number of the calculated number....
anyway, got AC now

Larry · Post by **Larry** » Sat Apr 17, 2004 8:20 pm

Actually, there's an easier way to check if a number is capable of multiple numbers..

Sajid · Post by **Sajid** » Sat Apr 17, 2004 8:39 pm

Larry,
Can you explain the ways?

Larry · Post by **Larry** » Sun Apr 18, 2004 2:22 am

I'm not a good math prover, but either prove to yourself or believe that the most number of ways is 2.

Knowing this, most numbers will have only one solution. Try to find that special case where there are two.

Hope it helps without giving too much away.. =)

soyoja · Post by **soyoja** » Tue May 04, 2004 10:09 am

Just I'm so curious that why maximum answers are only two.
Of course, I think it's very natural, but I want to know the reason.

soyoja · Post by **soyoja** » Tue May 04, 2004 10:10 am

Here is general idea of this problem ^^

A = original number
B = Sample input

A - A/10 = B ( by problem description )
therefore , (10A - A)/10 = B, 9A = 10B, A = 10/9 * B
So we can find one answer.
And minus value is always made by ( A/10 ),
so another answer's range is - 10 ~ + 10 at original answer.
You can find that number by small range searching.

Larry · Post by **Larry** » Tue May 04, 2004 11:17 pm

soyoja:

sorry, as I said, I'm not a good math prover, and it was just an intuition (that worked), so I don't know why this is the case.

OJ Board

10633 - Rare Easy Problem

10633 - Rare Easy Problem

Re: 10633 rare easy problem