Frog Jumping Step problem

(1) A frog can jump up to 1 steps at a time, or jump up to 2 levels. Ask the frog to jump on an n-level step with a total number of hops.

(2) A frog can jump up to 1 steps at a time, can also jump on 2 levels ... It can also jump on the N-level, at which point the frog jumps up the level of the N-step total number of hops?

Analysis: 1) when n = 1, there are only 1 hops; when n = 2 o'clock, there are two kinds of jumps; when n = 3 o'clock, there are 3 kinds of jumps; when n = 4 o'clock, there are 5 kinds of jumps; when n = 5 o'clock, there are 8 kinds of jumping method;

The law is similar to the Fibonacci sequence

The code is as follows:

[CPP]View PlainCopy

2. int Fib (int n)
4. {
6. if (n <= 0)
8. {
10. cout << "error!" << Endl;
12. return-1;
14. }
18. if (1 = = N)
20. {
22. return 1;
24. }
26. Else if (2 = = N)
28. {
30. return 2;
32. }
34. Else
36. {
38. return fib (n-1) + fib (n-2);
40. }
42. }

2) using FIB (n) to indicate that the frog jumps on the N-Step steps of the number of hops, the frog jumps on the N-Step step number 1 (n-step Jump), set fib (0) = 1;

When n = 1 o'clock, there is only one method of jumping, that is, the 1-Step Jump: Fib (1) = 1;

When n = 2 o'clock, there are two ways of jumping, one-step and second-order jumps: fib (2) = FIB (1) + fib (0) = 2;

When n = 3 o'clock, there are three ways to jump, the first step out of one, followed by FIB (3-1) in the Jump method, the first second out of the second, there is a fib (3-2) in the Jump method, after the first jump out of third order, followed by FIB (3-3) in the Jump method

FIB (3) = FIB (2) + fib (1) +fib (0) = 4;

When n = n, there is a total of n jumps, the first step out of one order, followed by FIB (n-1) in the Jump method, after the first second out of the second, there is a fib (n-2) in the Jump method ..... ..... After the first step out of the N-order, there is a Fib (n-n) jump in the back.

FIB (n) = fib (n-1) +fib (n-2) +fib (n-3) +..........+fib (n-n) =fib (0) +fib (1) +fib (2) +.......+fib (n-1)

And because Fib (n-1) =fib (0) +fib (1) +fib (2) +.......+fib (n-2)

Two-type subtraction: FIB (n)-fib (n-1) =fib (n-1) ===== "fib (n) = 2*fib (n-1) n >= 2

The recursive equation is as follows:

Frog Jumping Step problem