#Padovan string, C/C++ Programming

A Padovan string P(n) for a natural number n is defined as: P(0) = ‘X’ P(1) = ‘Y’ P(2) = ‘Z’ P(n) = P(n-2) + P(n-3), n>2 where + denotes string concatenation. For a string of the characters ‘X’ , ‘Y’ and ‘Z’ only, and given value of n, write a program that counts the number of occurrences of the string in the n-th Padovan string P(n)


program in java

// aakash , suraj , prem sasi kumar kamaraj college
program 1 :

package test.padovanstring;

public class PadovanString {
    public int stringOccurrences(int n, String str){
    if(n >= 40)
    return -1;
    return -1;
    String res=pad(n);
    return(res.length() - res.replaceAll(str,"").length())/(str.length());
public String pad(int n){
    if(n == 0) return "X";
    if(n == 1) return "Y";
    if(n == 2) return "Z";
    else return pad(n-2) + pad(n-3);
    public static void main(String []args)
        PadovanString p = new PadovanString();




Program set 2 :

import java.util.Scanner;

import java.util.ArrayList;


public class PadovanSeries


    public static void main(String[] arg)


        Scanner read = new Scanner(System.in);

        System.out.println("Enter starting no. : ");

        int start = read.nextInt();

        System.out.println("Enter ending no. : ");

        int end = read.nextInt();

        int[] ans = getSeries(start, end);

        System.out.println("Padovan series : ");

        for (int a : ans)

            System.out.print(a + " ");



    public static int[] getSeries(int s, int e)


        ArrayList list = new ArrayList();

        int i, j = 0;

        for (i = s; i <= e; i++, j++)


        int[] ans = new int[j];

        for (i = 0; i < j; i++)

            ans[i] = list.get(i);

        return ans;



    public static int getPadovan(int p)


        if (p == 0 || p == 1 || p == 2)

            return 1;

        return (getPadovan(p - 2) + getPadovan(p - 3));






Posted Date: 9/3/2012 3:34:55 AM | Location : United States

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