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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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2018-05-30 03:41:57
Finalmente fiz aeeeeeeee |
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2018-04-20 11:40:02
my 49th AC in go !!!! took me 3 hrs.. |
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2018-03-15 04:43:47
dp Last edit: 2018-03-22 12:07:45 |
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2018-03-03 13:30:08
nice problem.. took 3 hrs but AC in one go ;) |
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2018-02-19 07:01:48
My 50th ;-p AC in one Go !! Last edit: 2018-02-19 07:02:15 |
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2018-01-21 18:34:10
Nice question Last edit: 2019-10-13 22:05:53 |
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2017-12-20 09:06:11
2-d dp, along with bit, time complexity - (n^2)logn per test case. |
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2017-12-12 13:44:01
3D dp top down approach will work |
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2017-09-30 07:09:01 sharif ullah
3D dp. just draw 0/1 knapsack topdown approach of this problem with pen&&paper |
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2017-07-31 08:52:09
AC in one go!! in 0.00 time Nice Dp!!. |