Sword refers to the source series of offer-abnormal jumping steps

Topic 1389: Abnormal jumping steps
Time limit: 1 seconds memory limit: 32 trillion special problem: No submit: 1906 resolution: 1102
Title Description:
A frog can jump up to 1 steps at a time, or jump up to level 2 ... It can also jump on n levels. Ask the frog to jump on an n-level step with a total number of hops.
Input:
The input may contain multiple test samples, for each test case,
The input includes an integer n (1<=n<=50).
Output:
corresponding to each test case,
The output of the frog jumps to an n-level step with a total number of hops.
Sample input:
6
Sample output:

32

Solution Analysis:

Analysis: Using FIB (n) to indicate the frog jumps on the N-Step steps of the number of hops, the frog jumped on the N-Step step of the number of hops 1 (n-Step), 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:

#include <iostream> #include <stdio.h>using namespace std;//recursive long  long f (int n) {    if (n==0) return 1 ;    if (n==1) return 1;    Return 2*f (n-1);} int main () {    int n;    while (scanf ("%d", &n)!=eof) {        printf ("%ld\n", f (n));    }    return 0;}

Sword refers to offer source series-Abnormal jumping steps