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问题 1491 --Adjacent Bit Counts

1491: Adjacent Bit Counts

时间限制: 1 Sec  内存限制: 65535 MB
提交: 2  解决: 2
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题目描述

For a string of n bits x1, x2, x3, …, xn,  the adjacent bit count of the string  is given by     fun(x) = x1*x2 + x2*x3 + x3*x 4 + … + xn-1*x n
which counts the number of times a 1 bit is adjacent to another 1 bit. For example:  
     Fun(011101101) = 3
     Fun(111101101) = 4
     Fun (010101010) = 0
Write a program which takes as input integers n and p and returns the number of bit strings x of n bits (out of 2) that satisfy  Fun(x) = p.
 
For example, for 5 bit strings, there are 6 ways of getting fun(x) = 2:
11100, 01110, 00111, 10111, 11101, 11011

输入描述

On the first line of the input is a single positive integer k, telling the number of test cases to follow. 1 ≤ k ≤ 10 Each case is a single line that contains a decimal integer giving the number (n) of bits in the bit strings, followed by a single space, followed by a decimal integer (p) giving the desired adjacent bit count. 1 ≤ n , p ≤ 100

输出描述

For each test case, output a line with the number of n-bit strings with adjacent bit count equal to p.

样例输入

2
5 2
20 8 

样例输出

6
63426

提示

  河南省第六届ACM程序设计大赛

来源

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