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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 |
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2015-06-25 01:57:48 :.Mohib.:
Great que for beginners in dp like me...!! |
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2015-02-19 01:50:19 TgTtv
Recursion fits in the constrains. I wonder, if there is formula based solution. |
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2014-12-17 20:37:31 rishabh aggarwal
nice problem..... |
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2014-05-17 21:05:21 BLANKRK
nice one :D |
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2013-10-31 08:32:34 mystique_blue
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 |
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2013-07-23 14:37:29 Pierre Boix
And if i have 11111. It's a solution, isn't it? |
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2013-05-18 12:45:42 Samer Dawalib
@Sanchit Manchanda 11011 11100 01110 00111 10111 11101 |
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2013-01-14 03:24:43 Balaji Ramasubramanian
@Sanchit Manchanda You can also have 11101 and 10111. |
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2013-01-04 09:34:06 Vaishnavh
Seems like the value of K is at most 90 :) |
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2012-12-25 17:19:26 Sanchit Manchanda
how n=5 and k=2 gives ans=6?? isn't it should be 4?? 11011 11100 01110 00111 for these, 01111=3,11110=3. not 2. can someone pls explain this test case? |