10081 - Tight Words

[Chameleon]

I have written code for 10081, but the judge doesn't like it.
Any ideas?

[10153EN]

Any more information? e.g. what's the judge's reply?

and what's your algorithm in solving it?

[Chameleon]

"Incorrect output".

I was using a recursive way that counted the number of ways to generate the word, and then for efficiency converted to iterative.

Result = Tight words / Total words

where total words = (k+1)^n

I had to put in some tricks to overcome overflow... perhaps that is the problem.

[junjieliang]

While counting the number of tight words and total words use double if not I think you could get integer overflow.

[Master]

Would you pls tell me what the problem states and How can I solve it?
If you know then Pls tell me....

Thanks in advance
M H Rasel
Cuet Old Sailor

[sohel]

This problem is hard to understand if you read it only once, but if you read it 3-4 times then you will probably get the hang of it.

consider the second case in the sample:
2 5

here k is 2 and n is 5.

you can form 3^5 numbers.... ( allowed digits are 0,1,2 and of length 5).

Of these 3^5 numbers, the numbers whose adjacent digits do not differ by one are tight words (numbers).
Just output the percentage.

Hint: Use DP as manual genertions will exceed Mem limit && time limit.

Hope it helps.

[Digit]

I'm trying to solve 10081, Tight words, but I'm getting WA all the time.

Can anybody give me the number of tight words for input '9 100' ?

I know that this is not asked, but this is part of the solution.
My program says 1069966138. Is this correct?

Or maybe someone has some input/output for testing?

Thanks.

[Eduard]

Here are some tests.
Input

Code: Select all

0 8
5 10
2 20
2 30
2 40
2 50
2 60
2 70
9 13

Output

Code: Select all

100.00000
0.09657
1.56615
0.17839
0.02032
0.00231
0.00026
0.00003
0.00012

[jagadish]

How do u acheive precision for this problem ? I used double and got quite different results for the above case :

Code: Select all

100.00000
0.09657
1.56615
0.17839
0.02032
0.00231
0.00026
0.00003
0.00003

for 9 100 i get 0.00000 ..

[chunyi81]

I was trying to solve this problem but could not figure out how to use DP here. Here is what I observed: For a string of length i ending with digit j, then the second last digit will be j - 1, j or j + 1 if 1 <= j <= k - 2, 0 or 1 if j = 0, and k - 2 or k - 1 if j = k - 1. At first I thought of using memoization(using a lookup table), after trying to work out the recursion tree to calculate the number of tight words for a small example, say the first test data in the sample input: 2 5. The lookup table would become too big because there are too many places in the recursion tree where the same value is calculated more than once.
Can anyone give me a hint how to use DP for this problem? Thanks.

[jagadish]

Got AC despite some lossy precision..

For a string of length i ending with digit j, then the second last digit will be j - 1, j or j + 1 if 1 <= j <= k - 2, 0 or 1 if j = 0, and k - 2 or k - 1 if j = k - 1.

formally that would look like :
f(n,t) = f(n-1,t-1)+f(n-1,t)+f(n-1,t+1)
take advantage of the fact that f(n) needs only f(n-1)

[chunyi81]

Thanks jagadish for the recurrence equation you posted. I figured out how to solve this problem using DP.

[migichen]

How do u acheive precision for this problem ? I used double and got quite different results for the above case :

Code: Select all

100.00000
0.09657
1.56615
0.17839
0.02032
0.00231
0.00026
0.00003
0.00003

for 9 100 i get 0.00000 ..

My program which got AC has the same result

[Alberto]

for k=9 and n=100
the number of tight words:
tights = 100512267936480837475180750161082113997249340218
And the number of possible words:
possibles = (k+1)^n = 10^100
Then
% = 100.0*(tights/possibles)
% = 0.00000

[Mukit Chowdhury]

Astonished to see, for computing 10^100 pow() can be used !!!! Would anybody please tell me, what is the actual range of returned value of pow() in C++ ??? It'll really help....
Thanks in advance...