Definition of recursion: in the process of calling a function, the function itself is called directly or indirectly, which is called the recursion call of the function. This function is called the recursive function. If the p function is called in the p function definition, it is called direct recursion. If the q function is called in the p function definition, and the p function is called in the q function definition, it is called indirect recursion.

Two elements of recursion:

- Recursive expression (recursive equation)
- Recursive end condition (boundary condition)

Problems that can be solved recursively should meet the following three conditions

- The problem to be solved can be transformed into one or more sub problems, and the solution method of these sub problems is exactly the same as the original problem, but different in quantity and scale
- The number of recursive calls must be limited
- There must be a condition to end recursion to terminate recursion

When to use recursion:

- The definition is recursive (factorial, Fibonacci sequence, etc.)
- The data structure is recursive (single linked list, binary tree, etc.)
- The solution method of the problem is recursive (Hanoi Tower, backtracking method, etc.)

Advantages and disadvantages of recursion:

- Advantages: clear structure, strong readability, and it is easy to prove the correctness of the algorithm by mathematical induction. Therefore, it brings great convenience to design the algorithm and debug the program
- Disadvantages: the operation efficiency of recursive algorithm is low, and it consumes more computing time and storage space than non recursive algorithm
- Solution: in some recursive algorithms, recursive calls can be eliminated and transformed into non recursive algorithms

Recursive expression of factorial:

Recursive expression of fiboracci sequence:

## Kimi sequence

Title Description

Kimi recently studied a series: * F(0) = 7 * F(1) = 11 * F(n) = F(n-1) + F(n-2) (n≥2) Kimi calls it the Kimi sequence. Please help confirm whether the nth number in the sequence is a multiple of 3.

input

The input contains multiple sets of data. Each group of data contains an integer n, (0 ≤ n ≤ 30).

output

There is one line of output corresponding to each group of inputs. If F(n) is a multiple of 3, "Yes" is output; Otherwise, output "No".

public class Main{ public static void main(String[] args) { // Method stub automatically generated by TODO Scanner num=new Scanner(System.in); while(num.hasNext()) { int n=num.nextInt(); if(func(n)%3==0) { System.out.println("Yes"); }else { System.out.println("No"); } } System.out.println(); } public static int func(int n) { if(n==0) { return 7; }else if(n==1) { return 11; }else { return func(n-1)+func(n-2); } } }

## Recursive count

Title Description

Write a recursive program that returns the number of uppercase letters in a string. For example, enter "AbcD" and output 2.

input

Multiple sets of inputs, each containing a string consisting only of uppercase and lowercase letters.

output

The number of uppercase letters in the output string.

sample input

AbcD

sample output

2

public class Main { public static void main(String[] args) { // Method stub automatically generated by TODO int a; Scanner str=new Scanner(System.in); while(str.hasNext()) { String str1=str.nextLine(); char [] c=str1.toCharArray(); a=c.length; System.out.println(func(c,a-1)); } } public static int func(char [] c,int x) { if(x==-1) { return 0; }else { if(c[x]>='A'&&c[x]<='Z') { return 1+func(c,x-1); }else { return func(c,x-1); } } } }

## Hive problem

Title Description

A trained bee can only climb to the adjacent hive on the right, not in the opposite direction. Please program to calculate the number of possible routes for bees to climb from hive a to hive b.

The structure of the hive is as follows.

input

Multiple groups of data input, each group of data contains two positive integers a, b, and a < B.

output

The number of possible routes for bees to climb from hive a to hive b.

sample input

1 2 3 4

sample output

1 1

public class Main { public static void main(String[] args) { // Method stub automatically generated by TODO Scanner input=new Scanner(System.in); while(input.hasNext()){ int a=input.nextInt(); int b=input.nextInt(); System.out.println(func(a,b)); } } public static int func(int x,int y){ if(x==(y-1)){ return 1; } else if(x==(y-2)){ return 2; } else return func(x,y-1)+func(x,y-2); } }

## Reverse order output

Title Description

Write a program using recursion and output a non negative integer in reverse order. For example, input 1234 and output 4321 (excluding leading 0).

input

Multiple groups of inputs, each group input a non negative integer.

output

Output the results in reverse order, and each result occupies one line.

sample input

12 1230 0

sample output

21 321 0

public class Main { public static int func(int n){ if(n >= 10){ System.out.print(n%10); n = n/10; return func(n); }else { return n%10; } } public static void main(String[] args) { Scanner input = new Scanner(System.in); while (input.hasNext()){ int n = input.nextInt(); if(n==0){ System.out.println(n); continue; } else if(n > 0){ while(n%10==0){ n = n/10; } System.out.println(func(n)); } } } }

## Dominoes cover

Title Description

Use size 1 × 2's dominoes are covered with a size of 2 × N rectangular grid, write a program, input n and output the total number of placement schemes. For example, enter n=3, that is, the size is 2 × 3 grid, output 3. Three dominoes laying schemes are shown in the figure below:

input

Multiple groups of test cases, each group is a positive integer.

output

Each group of output occupies one line.

You only need to output the total number of placement schemes, and you do not need to output specific placement schemes.

sample input

3

sample output

3

public class Main { public static void main(String[] args) { // Method stub automatically generated by TODO Scanner num=new Scanner(System.in); while(num.hasNext()) { int n=num.nextInt(); System.out.println(function(n)); } } public static int function(int n) { if(n==1) { return 1; }else if(n==2) { return 2; }else { return function(n-1)+function(n-2); } } }