The offer-of the second chapter of the sword--the Fibonacci sequence (the frog jumps the steps)

Recursion and looping

Recursion: This function is called inside a function.

Essence: Divide a problem into two, or multiple small problems (overlapping parts of multiple small problems, there will be repeated computations)

Advantages: Simple, easy to implement.

Disadvantage: time and space consumption is serious, if the hierarchy of recursive calls too many, it will exceed the stack capacity.

Loop: Repeats an operation within a range by setting the initial value of the calculation and the termination condition.

Fibonacci sequence

Title One: Write a function, enter N, find the nth of the retracement (FIBONACCI) sequence, defined as follows:

The first solution: recursive algorithm:

Long long Fabonacci (unsigned int n) {if (n<=0) return 0;if (n==1) return 1;return Fabonacci (n-1) +fabonacci (n-2);}

When n=10 the call graph is as follows:

From the time we can see recursion, most of them have been repeatedly computed, which has a very negative effect on performance, and the time complexity of the algorithm is n exponent.

Second solution: Using loops (Time complexity is O (n))

#include <iostream>using namespacestd;Long LongFabonacci (unsignedintN) {    intarrary[2]={0,1}; Long Longfabn; if(n<2) Fabn=Arrary[n]; Long LongFabone=1; Long Longfabtwo=0;  for(unsignedintI=2; i<=n;++i) {fabn=fabone+Fabtwo; Fabtwo=Fabone; Fabone=fabn; }    returnfabn;}voidMain () {Long LongN=fabonacci ( the); cout<<n<<Endl;}

Title Two: A frog can jump up a step at a time, you can jump up to 2 steps, the frog jumped on the number of steps on the N-step method.

Idea: When there is only one step, the frog's jumping method is only one. When there are two steps, there are two kinds of frog jumping method (one is: Jump two steps, two: first-level jump). When there are n steps, the frog in the first take-off when jumping only a step, then there are still n-1 steps to jump, if the first jump on the two-step stage, then there is still the n-2 step of the jumping method. The whole question is exactly a Fibonacci sequence. The formula is as follows:

Topic Three: The cover of the Matrix, with eight 2*1 small matrix to cover a 2*8 large matrix. As shown in the following:

The first small matrix can have two coverage methods across, then at this time, it must be the second small matrix is also transverse, leaving the large matrix of 2*6, and then left the 2*7 of the large matrix needs to be covered. It is therefore possible to have: f (8) =f (6) +f (7).

The formula above the second question.

The offer-of the second chapter of the sword--the Fibonacci sequence (the frog jumps the steps)