GNYR09F - Adjacent Bit Counts


For a string of n bits x1,x2,x3,...,Xn the adjacent bit count of the string (AdjBC(x)) is given by


X1*X2 + X2*X3 + X3*X4 + ... + Xn-1 * Xn


which counts the number of times a 1 bit is adjacent to another 1 bit. For example:
AdjBC(011101101) = 3
AdjBC(111101101) = 4
AdjBC(010101010) = 0

Write a program which takes as input integers n and k and returns the number of bit strings x of n bits (out of 2ⁿ) that satisfy AdjBC(x) = k. For example, for 5 bit strings, there are 6 ways of getting AdjBC(x) = 2:
11100, 01110, 00111, 10111, 11101, 11011

Input

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set is a single line that contains the data set number, followed by a space, followed by a decimal integer giving the number (n) of bits in the bit strings, followed by a single space, followed by a decimal integer (k) giving the desired adjacent bit count. The number of bits (n) will not be greater than 100 and the parameters n and k will be chosen so that the result will fit in a signed 32-bit integer.

Output

For each data set there is one line of output. It contains the data set number followed by a single space, followed by the number of n-bit strings with adjacent bit count equal to k.

Example

Input:
10
1 5 2
2 20 8
3 30 17
4 40 24
5 50 37
6 60 52
7 70 59
8 80 73
9 90 84
10 100 90
Output:
1 6
2 63426
3 1861225
4 168212501
5 44874764
6 160916
7 22937308
8 99167
9 15476
10 23076518

hide comments
Davit Safrastyan: 2016-12-10 18:16:32

Find the solution of this problem in xoptutorials.com/spoj.php

tanmaysachan: 2016-11-01 07:02:44

How to do this in 2d? took me a 3d dp

yash_18: 2016-10-14 21:42:44

Good question!!
took a lot time to solve ...feeling great! :D

siddharth_0196: 2016-10-08 20:38:36

A paper, a pen and a bit of observation will do the trick! ;)
Nice question, great confidence booster and can be done using 2D array! :D

hash7: 2016-06-22 10:24:26

NYC QSN...

karthik1997: 2016-06-18 19:59:47

For Beginners striving for the subcase : Solve @www.spoj.com/problems/PERMUT1 First. and You can easily figure out the sub cases ... SImple 3D Dp with O(N*N*2) complexity . PS My complexity itself suggests the required space complexity tooo :D ....

Ravi: 2016-03-24 21:51:24

precompute + dp
O(n^2)
Fantastic Dp :)

lakshay_v06: 2016-03-24 13:12:22

AC in one go! O(n^2) : 0.00 <- :D

anshal dwivedi: 2016-01-05 17:24:46

yo!AC in one go...! Nice One ..:)

Aswin Siva: 2015-12-30 06:48:00

Any other (trivial) approach other than 3D DP ? Any Hints...


Added by:Tamer
Date:2009-11-14
Time limit:3s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: NODEJS objc PERL 6 VB.net
Resource:ACM Greater New York Regional Contest 2009