HDU 1293 The Number of Paths

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1293

Problem Description

Let f (n) be the number of paths with n steps starting from O (0, 0), with steps of the type (1, 0), or (-1, 0), or (0, 1), and never intersecting themselves. For instance, f (2) =7, as shown in Fig.1. Equivalently, letting E=(1,0),W=(-1,0),N=(0,1), we want the number of words A1A2...An, each Ai either E, W, or N, such that EW and WE never appear as factors.

 

Input

There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of steps(1<=n<=1000).

 

Output

For each test case, there is only one integer means the number of paths.

 

Sample Input

12

 

Sample Output

37

 

Author

SmallBeer (CML)

题目大意：给出已知线段的段数（即前进的步数），线段的前进方向不可以向下，求有多少种走方。

要求第N步的走方数：

走第N步就必须走完N-1步，而在N-1步里面有方向向上，向左，向右三种情况。向左和向右的下一步只有俩种走法（即向上、同向），向上的下一步有三种走法。

所以需要统计在N-1步里面有多少向上的。

在N-2的步法里面每一种都只能产生一种向上的步法，即在N-1里面有向上的步法q[n-2]种。

递推式：q[n]=2*q[n-1]+q[n-2];

AC代码：

import java.math.BigInteger;import java.util.Scanner;public class Main { public static void main(String[] args) {    int n;  Scanner cin =new Scanner(System.in);  while(cin.hasNext())  {   BigInteger ans=BigInteger.valueOf(3);   BigInteger count=BigInteger.valueOf(1);   BigInteger sum;   n=cin.nextInt();   for(int i=2;i<=n;i++)   {    sum=ans;     ans=ans.multiply(BigInteger.valueOf(2));     ans=ans.add(count);     count=sum;   }   System.out.println(ans);  } }}

路途中。。。。。