Stair climbing--recursive and function self-invocation algorithm

Title Description Description

The tree teacher climbed the stairs, he can walk 1 or 2 levels, enter the number of stairs, to find different ways to go
For example: The staircase has a total of 3 levels, he can go one level at a time, or the first step, the second walk two levels
Can also be the first time to go two levels, the second step, a total of 3 methods. input/output format input/output Input Format:
The input contains several rows, each containing a positive integer N, representing the stair series, 1 <= N <= 30
output Format:
Different walk count, each line input corresponding line output input and Output sample sample Input/output sample Test point # #

Input Sample:

5

8

10

Sample output:

8

34

89

Ideas:

Use enumerations to abstract out general expressions:

When N=0, f (n) =1

N=1,f (n) =1

N=2,f (n) =2

N=3,fn=3

N=4,fn=5

F (2) =f (1) +f (0)

F (3) =f (2) +f (1)

F (4) =f (3) +f (2)

.......

From the above analysis can be summed up the general expression:f (n) =f (n-2) +f (n-1)

So the problem is the transformed Fibonacci sequence!

The code is as follows:

1#include <stdio.h>2 intDfsintN)3 {4     if(n<3)//if it is 1 or 2, there are 1, 2 methods to return directly.5     {6         returnN;7     }8     Else//otherwise, recursion! 9     {Ten         returnDFS (N-1) +dfs (n2);//use a general expression to solve (in fact, the Fibonacci sequence) One     }     A } - intMain () - { the     intN; -scanf"%d",&n); -printf"%d\n", DFS (n)); -     return 0; +}

Stair climbing--recursive and function self-invocation algorithm