Recursive recursive combination number, Hanoi, palindrome number problem (Java)

Recursive recursive combination number:

1 Ideas : using functions to obtain n! , call the function calculation result flowchart

2.1 flowchart

3 . 1 Source code:

import Java.util.Scanner;

Public class N

Public Static void Main (String [] args) {

int n,k;

int S

System. out. println ("Please enter the values of N and K in C (n,k):");

Scanner num = new Scanner (System. in);

N=num.nextint ();

K=num.nextint ();

s= Fun (n)/(Fun (k) *Fun (n-k));

System. The result of out. println ("C (" +n+k+ ") is:" +s);

}

Static int Fun (int a)

{

int m=1;

for (int b=1;b<a;b++) {

m=m* (b+1);

}

return m;

}

}

4.1 Results

2.2 Flowchart

3.2 Source code:

import Java.util.Scanner;

Public class Fact {

Static int Fun (int a) {

if (a==0| | A==1) {

return A;

}

Else {

return a* Fun (A-1);

}

}

Public Static void Main (String[]args) {

int n,k;

System. out. println ("Please enter N and K");

Scanner num = new Scanner (System. in);

N=num.nextint ();

K=num.nextint ();

int C

C= Fun (n)/(Fun (k) *Fun (n-k));

System. out. println ("Result:" +c);

}

}

4.2 Results:

2.3 Flowchart:

3.3 Source code:

import Java.util.Scanner;

Public class Yanghui {

Public Static void Main (String []args) {

int n = 0,k;

int b[][];

System. out. println ("Please enter the value of N:");

Scanner num = new Scanner (System. in);

N=num.nextint ();

K=num.nextint ();

b = new int[n][];

for (int i=1;i<=n;i++) {

B[I-1] = new int[i];

}

for (int j=0;j<n;j++) {

for (int z=0;z<=j;z++) {

if (j==0| | z==0| | Z==J) {

B[j][z]=1;

continue;

}

Else {

B[J][Z] = b[j-1][z-1]+b[j-1][z];

}

}

}

for (int x=0;x<n;x++) {

for (int y=0;y<=x;y++) {

System. out. Print (b[x][y]+ "");

}

System. out. println ("");

}

int S

S=B[N-1][K-2]+B[N-1][K-1];

System. out. println ("And yes:" +s);

}

}

4.3 Results:

Hanoi: (Baidu's)

Flow chart:

Source:

import Java.util.Scanner;

Public class Hannuota {

Public Static void Main (string[] args)

{

Hannuota h=New Hannuota ();

H.go ();

}

Public void Go ()

{

System. out. println ("Please enter the number of plates N:");

@SuppressWarnings ("resource")

Scanner scanner=New Scanner (System. in);

int n=scanner.nextint ();

if (n>=0)//Determine whether n meets the requirements

{

System. out. println (the steps for "moving" +n+ "plates are:");

Hanoi (N, ' A ', ' B ', ' C ');

}

Else System. out. println ("The number of inputs does not meet the requirements");

}

Public void Hanoi (int m,Char one,char one, char three)

{

if (m==1)

Move (One,three);

Else

{

Hanoi (M-1,one,three,two);//The first step is to move a n-1 plate on a C to B

Move (One,three);//move a plate to C on the second step

Hanoi (M-1,two,one,three);//The third step is to move a n-1 plate on B to C

}

}

Public void Move (char x,char y)

{

System. out. println (x+ "-" +y);

}

}

3 results:

Palindrome:

1 design ideas:

Defines the string, gets the length, and uses the recursive function to determine whether the two ends are equal.

2 Flowchart:

3 Source code:

Import Java.io.BufferedReader;

Import java.io.IOException;

Import Java.io.InputStreamReader;

public class Huiwen {

public static void Main (string[] args)

{

Huiwen h=new Huiwen ();

try {

H.go ();

} catch (IOException e) {

E.printstacktrace ();

}

}

public void Go () throws IOException

{

String Input=null;

System.out.println ("Input string:");

BufferedReader is=new BufferedReader (New InputStreamReader (system.in));

Input=is.readline ();

if (!input.isempty ())

{

int [] array=new int[input.length ()];

for (int i=0;i<input.length (); i++)

{

Array[i]=input.charat (i);

}

System.out.println ("is a palindrome number?") "+ifhuiwen (array,0));

}

Else System.out.println ("The number of inputs does not meet the requirements");

}

public boolean Ifhuiwen (int[] a,int L)

{

if (L==A.LENGTH/2)

return true;

if (A[l]==a[a.length-l-1])

Return Ifhuiwen (a,l+1);

else return false;

}

}

4 results:

Recursive recursive combination number, Hanoi, palindrome number problem (Java)