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
", dp[x][y][0] + dp[x][y][1]);
}
return 0;
}