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

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

hide comments
2017-05-10 16:18:32
Nice one :)
2017-03-30 18:16:27
TOP-DOWN+MEMO+3D DP
DP ROCKS!!!!
2017-02-28 16:11:09 gautam
nailed it...;)
2017-02-23 05:03:39
nice problem
2017-02-01 03:05:13
Hello, I can solve it just fine... in around 1 minute.

Last edit: 2017-02-01 03:05:54
2017-01-24 16:54:36


Last edit: 2017-01-24 16:56:24
2017-01-16 14:18:31 Shubham Agrawal
Easy DP problem. Just try to observe a pattern and you will get the answer...
2017-01-15 06:53:48
Easy dp :)
2016-11-01 07:02:44
How to do this in 2d? took me a 3d dp
2016-10-14 21:42:44
Good question!!
took a lot time to solve ...feeling great! :D
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