对动态规划的一点理解 Adjacent Bit Counts

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 + … + 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

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