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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 |
hide comments
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2016-10-14 21:42:44
Good question!! took a lot time to solve ...feeling great! :D |
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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 |
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2016-06-22 10:24:26
NYC QSN... |
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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 .... |
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2016-03-24 21:51:24 Ravi
precompute + dp O(n^2) Fantastic Dp :) |
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2016-03-24 13:12:22
AC in one go! O(n^2) : 0.00 <- :D |
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2016-01-05 17:24:46 anshal dwivedi
yo!AC in one go...! Nice One ..:) |
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2015-12-30 06:48:00 Aswin Siva
Any other (trivial) approach other than 3D DP ? Any Hints... |
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2015-10-09 17:17:52 Sarthak Munshi
First time AC . Generate pattern for n=1..10 and k=1..10 using brute force . Observe pattern . Then easy DP . |
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2015-08-25 18:35:53 sneh shikhar
3D DP :* BOTTOM UP!! |