public class Pi {

public static void main(String[] args) {

int max = 1_000_000;

int increment = max / 8;

long startTime = System.currentTimeMillis();

//int stopTime = System.currentTimeMillis() – startTime;

int counter = 0;

while (counter <= max) { String s = ""; for (int i = 0; i < counter; i++) { s += "0"; } if (s.equals(reverse(s))) { System.out.println(counter + "/" + max); } else { if (counter % increment == 0) { // Print every 8th number. This is just for debugging purposes, and you can remove this in your own code if you want. System.out.println(counter); // This line outputs the current number we're trying to reverse to see if it's a palindrome or not. You can comment this out or delete it if you want, but I like to leave it in for debugging purposes because it helps me get a feel for where my program is import java.io.*; import java.util.ArrayList; import java.util.Random; public class Main { static int n; static int[] a; static int[] b; static ArrayList arr = new ArrayList();

static Random r = new Random();

public static void main(String[] args) throws IOException {

n = Integer.parseInt(args[0]);

a = new int[n];

b = new int[n];

for (int i = 0; i < n; i++) { a[i] = r.nextInt(10); b[i] = r.nextInt(10); arr.add(a[i]);arr.add(b[i]); if (i != 0) { // not the first digit of pi, so we need to insert a comma or period into the current digit, as appropriate for its place in pi's decimal expansion if (i == 1) arr.add(1); // decimal point before second digit of pi: 3.,1415... else if ( public static void main(String[] args) { int j; for(j=1;j<10;j++) { System.out.print(""+j); if(j==9) { break; }else { System.out.print(" "); } }//end for loop for one side of Pi System.out.print(" 3.");//Print the decimal point after the 1st digit of Pi. //count from 0 to 9,999,999 while also checking if it is a palindrome or not. int palindromeCheck = 0;//counter for while loop while(palindromeCheck < 9999999)//will run through 9999999 numbers and see if it's a palindrome or not and then print it out if it is a palindrome. { //convert integer to string String integerToString = Integer.toString(palindromeCheck);//convert palindromeCheck (int) to (string) so we can check if it is a palindrome or not public class Palindrome { public static void main (String[] args) { int x = 999; int y = 999; int largest = 0; Boolean found = false; while (x>100 && y>100) {

if (isPalindrome(x*y)) {

largest = x*y;

found = true;

break;

} else if (x%11==0){ // if x is divisible by 11, no need to check all of the combinations. Only those that are divisible by 11 and vice versa. This optimizes the program to run faster.

y–; // move down to the next number.

} else { // otherwise check all of the combinations until we find the lowest number that is a palindrome. Then move down 1, and repeat.

x–; // move left to the previous number.

}

}

System.out.println(“The largest palindrome is: ” + largest); // output our result! } public static Boolean isPalindrome(int n) { // method to determine whether or not a number is a palindrome int i=0,j=0,k=0;

public class Palindrome

{

public static void main(String[] args) {

String str;

int len, i, palindrome, count = 0;

boolean isPalindrome = false;

//Read the string from the keyboard

System.out.print(“Enter any string : “);

str = new Scanner(System.in).nextLine();

//Convert the string into lowercase

str = str.toLowerCase();

//Remove all whitespaces from the string

str = str.replaceAll(“\\s”, “”);

//Find the length of the string

len = str.length();

for (i = 0; i < len; i++) { palindrome = 1; for (int j = 0; j < len/2 && palindrome == 1; j++) { if (str.charAt(i+j) != str.charAt(len-1-j)) { palindrome = 0; } else { palindrome = 1; } if (palindrome == 1) { import java.util.*; import java.lang.*; import java.io.*; class Ideone { public static void main (String[] args) throws java.lang.Exception { // your code goes here Scanner in=new Scanner(System.in); int n=in.nextInt(); int a[]=new int[n]; for(int i=0;i

Tomy

Tomy is a contributor at AskMeCode. We are committed to providing well-researched, accurate, and valuable content to our readers.

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