Adjacent Bit Counts
时间限制：1000 ms | 内存限制：65535 KB
难度：4
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
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
我简单描述下题：
用户输入正整数 n ,k
X1，X2...只能为0或1
输出满足
X1*X2+X2*X3+....+X(n-1)*Xn=k;
情况的个数
穷举法的话肯定会超时。
代码摘自： http://blog.csdn.net/air_one/article/details/11750509
dp[达到的和][位置][0或1];
| 1 | dp[101][101][2] |
初始化数组：
for(j = 2; j < MAXN; j++)
{
dp[0][j][0] = dp[0][j-1][1] + dp[0][j-1][0];
dp[0][j][1] = dp[0][j-1][0];
}
注意：dp[101][101][2]为全局变量，所以C语言自动把空白部分赋值为0；
dp[0][j][0] = dp[0][j-1][1] + dp[0][j-1][0]; 即总和为0，第j位为0的情况数=总和为0，第j-1位为1的情况数 + 总和为0，第j-1位为0的情况数
dp[0][j][1] = dp[0][j-1][0]; 同理
看计算过程：
for(i = 1; i < MAXN; i++)
for(j = 1; j < MAXN; j++)
{
dp[i][j][0] = dp[i][j-1][0] + dp[i][j-1][1];
dp[i][j][1] = dp[i-1][j-1][1] + dp[i][j-1][0];
}
return 0;
dp[i][j][0] = dp[i][j-1][0] + dp[i][j-1][1]; 达到i，且j位为0的情况数= 达到i，第j-1位为0的情况数 + 达到i，第j-1位为1的情况数
同理： dp[i][j][1] = dp[i-1][j-1][1] + dp[i][j-1][0]; 达到i，且j位为1的情况数= 达到i-1，第j-1位为1的情况数 + 达到i，第j-1位为0的情况数