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 + ... + Xn1 * 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 32bit 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 nbit 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
vikash_2016:
20181220 15:34:18
What is max value of k ? 

leonardopaes:
20180530 03:41:57
Finalmente fiz aeeeeeeee


codehead1997:
20180420 11:40:02
my 49th


shivr1:
20180315 04:43:47
dp Last edit: 20180322 12:07:45 

pallindromeguy:
20180303 13:30:08
nice problem.. took 3 hrs but AC in one go ;) 

mohit_aggarwal:
20180219 07:01:48
My 50th ;p


mrpandey:
20180121 18:34:10
I did it in O(nk) time complexity. A 3line for loop is what it took. For me it was a pure maths question. No need of DP. Try using combinatorics. 

javafreak:
20171220 09:06:11
2d dp, along with bit, time complexity  (n^2)logn per test case. 

gaurav_76170:
20171212 13:44:01
3D dp top down approach will work 

sharif ullah:
20170930 07:09:01
3D dp. just draw 0/1 knapsack topdown approach of this problem with pen&&paper 
Added by:  Tamer 
Date:  20091114 
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 