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
arun: 2015-08-02 07:53:31

@sanchit 10111 & 11101

:.Mohib.:: 2015-06-25 01:57:48

Great que for beginners in dp like me...!!

TgTtv: 2015-02-19 01:50:19

Recursion fits in the constrains. I wonder, if there is formula based solution.

rishabh aggarwal: 2014-12-17 20:37:31

nice problem.....

BLANKRK: 2014-05-17 21:05:21

nice one :D

mystique_blue: 2013-10-31 08:32:34

This is what should be done:
1) Come up with a brute force solution..
2) Alanlyze the recurisve statement to be used..
3)Memoize it..

A few simpler test cases..
1 5 1 --> 10
2 7 3 --> 16
3 7 4 --> 8

Pierre Boix: 2013-07-23 14:37:29

And if i have 11111. It's a solution, isn't it?

Samer Dawalib: 2013-05-18 12:45:42

@Sanchit Manchanda
11011
11100
01110
00111
10111
11101

Balaji Ramasubramanian: 2013-01-14 03:24:43

@Sanchit Manchanda You can also have 11101 and 10111.

Vaishnavh: 2013-01-04 09:34:06

Seems like the value of K is at most 90 :)


Added by:Tamer
Date:2009-11-14
Time limit:3s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64 NODEJS OBJC PERL6 SQLITE VB.NET
Resource:ACM Greater New York Regional Contest 2009