715-Adjacent Bit Counts


    内存限制:64MB 时间限制:1000ms 特判: No

    通过数:15 提交数:19 难度:4


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 2ⁿ) 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:

没有提示哦

上传者:ACM_赵铭浩

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