nyoj 715 Adjacent Bit Counts

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 + … + x_n-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

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

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
讲解：看了半天没有看出来，其实就是一个dp问题；看下代码：

 #include<algorithm>
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 long long  dp[][][];
 void fun()
 { int i,j;
     memset(dp,,sizeof(dp));
     dp[][][]=;dp[][][]=;
     for(i=;i<=;i++)
    {
       dp[i][][]=dp[i-][][]+dp[i-][][];
       dp[i][][]=dp[i-][][];
       dp[i][i-][]=;
    }
     for(j=;j<=;j++)
       for(i=j+;i<=;i++)
       {
           dp[i][j][]=dp[i-][j][]+dp[i-][j][];
           dp[i][j][]=dp[i-][j][]+dp[i-][j-][];
       }
 }
 int main()
 {
     fun();
     int t,m,n;
     cin>>t;
     while(t--)
     {
         cin>>m>>n;
         cout<<dp[m][n][]+dp[m][n][]<<endl;
     }
     return ;
 }