http://nyoj.top/problem/715

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

题意分析：

定义一个函数是把一个01串的所有相邻位上的数相乘，问字符串长度为x，函数值为y的字符串共有多少种组成。

解题思路：

定义一个三维数组 dp[x][y][0];

其中x表示字符串长度，y表示函数的值，0或1表示字符串末尾为0或1；

转移方程：

dp[ i ] [ j ] [ 0 ] = dp[ i -1 ][ j ][ 0 ] + dp[ i -1  ][ j ][ 1 ];

dp[ i ] [ j ] [ 1 ] = dp[ i -1 ][ j ][ 0 ] + dp[ i -1  ][ j - 1 ][ 1 ];

#include <stdio.h>
#define N 120
int dp[N][N][2];
int main()
{
	int t,x,y,i,j;
	scanf("%d", &t);
	while(t--){
		scanf("%d%d", &x, &y);
		dp[1][0][0] = 1;
		dp[1][0][1] = 1;
		for(i=2; i<=x; i++)
			for(j=0; j<=i; j++)
			{
				dp[i][j][0] = dp[i-1][j][1] + dp[i-1][j][0];
				if(j)
					dp[i][j][1] = dp[i-1][j-1][1] + dp[i-1][j][0];	
				else
					dp[i][j][1] =  dp[i-1][j][0];
			}
		printf("%d\n", dp[x][y][0] + dp[x][y][1]);
	}
	return 0;
}