CountZeroes in java - java

I don't know where am I going wrong. I want to count zeroes via recursion but I am not getting it:
public class countzeroes {
public static int countZerosRec(int input){
int count=0;
return countZerosRec(input,count);
}
private static int countZerosRec(int input ,int count){
if (input<0) {
return -1;
}
if(input==0) {
return 1;
}
int m = input%10;
input = input/10;
if(m==0){
count++;
}
countZerosRec(input,count);
return count;
}
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
System.out.println(countZerosRec(n));
}
}

Put return count in if(input == 0) statement and instead of
countZerosRec(input, count); return count; put return countZerosRec(input, count);.

The correct method would be:
public class countzeroes {
private static int countZerosRec(int input){
if (input<0) {
return -1;
}
if (input==0) {
return 1;
}
if(input < 10) {
return 0;
}
int m = (input%10 == 0)? 1: 0;
input = input/10;
return m + countZerosRec(input);
}
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
System.out.println(countZerosRec(n));
}
}
Let me explain 2 problems in your code:
1- First of all, your second if statement (if(input == 0)) ruin everything. Consider 1200100 as an example. In the 6th round of recursion the input would be 1, and if you divide it on 10 the result is 0 (which is the input of next recursion round) and therefore all the answers would be 1.
2- Secondly, it would be nice if you don't change the input parameter in your code. because it's completely error-prone (In complicated codes, you can not trace the changes happen on a parameter and it makes debugging hard). So, I just removed the count parameter from the input.
And finally, It is better to name your classes in CamelCase form. (CountZeroes)

Change your method as below. Return count always
private static int countZerosRec(int input ,int count){
if (input <= 0) { // check if input is negative or zero
return count;
}
int m = input % 10;
input = input / 10;
if (m == 0) {
count++; // increment if current digit is zero
}
return countZerosRec(input,count);
}

public static int zeroCount(int num)
{
if(num == 0)
return 0;
if(num %10 ==0)
return 1 + zeroCount(num / 10);
else
return zeroCount(num/10);
}
this would work

You can use Streams:
System.out.println("11020304".chars().filter(c -> c == '0').count());
Result: 3

Your count logic is excellent.
in below line ... you are making logic mistake.. just fix it.
private static int countZerosRec(int input, int count) {
if (input < 0) {
return -1;
}
if (input == 0) {
return count;
//return 1; /// you need to change your code here, in last its getting zero as (num < 10 )/10 is 0
// its entering here everytime, and returning one.
// since its the base condition to exit the recursion.
// for special case of 0 (zero) count, handle your logic when it is //returned.
//...... rest of your code
}

Related

reversing an integer in java without a loop

This is an homework problem
Is there a way tor reverse a number in Java without using any loops? The only solution I can think of is reversing it using String and then casting it back to an integer.
If you want to reverse a number withour using any loop you can use Recursion method call. Following program is doing same
public static void reverseMethod(int number) {
if (number < 10) {
System.out.println(number);
return;
} else {
System.out.print(number % 10);
reverseMethod(number / 10);
}
}
public static void main(String args[]) {
int num = 4567;
reverseMethod(num);
}
Even if you were to reverse the number by casting it into a String, you would still need a loop if you want the program to work when having ints of different sizes. If I were to make a method to reverse a number but could not do it with loops, I would probably do it with recursion (which still uses loops indirectly). The code will look something like this:
class Main {
public static void main(String[] args) {
String input = "1234"; // or scanner to take in input can be implemented
System.out.println(Integer.parseInt(reverseInt(input)));
}
public static String reverseInt(String x) {
if (x.length() == 1) {
return x;
} else {
return x.substring(x.length() - 1) + reverseInt(x.substring(0, x.length() - 1));
}
}
}
Hope this helps!
By using reverse() of StringBuilder:
int number = 1234;
String str = String.valueOf(number);
StringBuilder builder = new StringBuilder(str);
builder.reverse();
number = Integer.parseInt(builder.toString());
System.out.println(number);
will print:
4321
if you want reverse method without loop and recursion then use this code
int a=12345;
int b,c,d,e,f;
b=a%10;
c=a%100/10;
d=a%1000/100;
e=a%10000/1000;
f=a%100000/10000;
System.out.println(b+","+c+","+d+","+e+","+f);
you can go like :
public int reverse(int x) {
String o = "";
if (x < 0) {
x *= -1;
String s = Integer.toString(x);
o += "-";
o += new StringBuilder(s).reverse().toString();
}
else {
String s = Integer.toString(x);
o += new StringBuilder(s).reverse().toString();
}
try {
int out = Integer.parseInt(o);
//System.out.println(s);
return out; }
catch (NumberFormatException e) {
return 0;
}
}
This is a solution using recursive method call
public class Tester{
public static int findReverse(int num, int temp){
if(num==0){
return temp;
}else if(num<10){
return temp*10 + num; //up to this is stopping condition
}else{
temp = temp*10 + num%10;
return findReverse(num/10, temp);
}
}
public static void main(String args[]){
int num = 120021;
int reverseNum = findReverse(num, 0);
System.out.println(reverseNum);
if(num == reverseNum)
System.out.println(num +" is a palindrome!");
else
System.out.println(num +" is not a palindrome!");
}
}
This will be fast.
static int reverseNum(int num) {
StringBuilder sb = new StringBuilder(String.valueOf(num));
sb.reverse();
return Integer.parseInt(sb.toString());
}

how to count many times a character occurs in a string without using s loop

the code below is meant to count each time character 'x' occurs in a string but it only counts once ..
I do not want to use a loop.
public class recursionJava
{
public static void main(String args[])
{
String names = "xxhixx";
int result = number(names);
System.out.println("number of x: " + result);
}
public static int number (String name)
{
int index = 0, result = 0;
if(name.charAt(index) == 'x')
{
result++;
}
else
{
result = result;
}
index++;
if (name.trim().length() != 0)
{
number(name);
}
return result;
}
}
You could do a replacement/removal of the character and then compare the length of the resulting string:
String names = "xxhixx";
int numX = names.length() - names.replace("x", "").length(); // numX == 4
If you don't want to use a loop, you can use recursion:
public static int number (String name)
{
if (name.length () == 0)
return 0;
int count = name.charAt(0)=='x' ? 1 : 0;
return count + number(name.substring(1));
}
As of Java 8 you can use streams:
"xxhixx".chars().filter(c -> ((char)c)=='x').count()
Previous recursive answer (from Eran) is correct, although it has quadratic complexity in new java versions (substring copies string internally). It can be linear one:
public static int number(String names, int position) {
if (position >= names.length()) {
return 0;
}
int count = number(names, position + 1);
if ('x' == names.charAt(position)) {
count++;
}
return count;
}
Your code does not work because of two things:
Every time you're calling your recursive method number(), you're setting your variables index and result back to zero. So, the program will always be stuck on the first letter and also reset the record of the number of x's it has found so far.
Also, name.trim() is pretty much useless here, because this method only removes whitespace characters such as space, tab etc.
You can solve both of these problems by
making index and result global variables and
using index to check whether or not you have reached the end of the String.
So in the end, a slightly modified (and working) Version of your code would look like this:
public class recursionJava {
private static int index = 0;
private static int result = 0;
public static void main(String[] args) {
String names = "xxhixx";
int result = number(names);
System.out.println("number of x: " + result);
}
public static int number (String name){
if(name.charAt(index) == 'x')
result++;
index++;
if(name.length() - index > 0)
number(name);
return result;
}
}
You can use StringUtils.countMatches
StringUtils.countMatches(name, "x");

Find Max Number in String of Numbers

I have run into an interesting problem and wanted to know if anyone had any leads on how to solve it. When given a string of numbers, I want to find the highest number. So if the string is "2836", then the output should be 8, if the string is "12345" then the output should be 5 and so on. Here is the method I am working on:
public static void main(String[] args) {
max("215");
}
public static void max(String number) {
if (number.isEmpty()) {
System.out.println("The string is empty");
System.exit(0);
}
int compCount = 1;
int max = number.charAt(0);
int compare = number.charAt(compCount);
for (int i = 0; i < number.length(); i++) {
if (max > compare) {
compCount++;
} else if (compare > max) {
max = compare;
} else {
System.out.print(max);
}
}
System.out.print(max);
}
When this code executes it gives me 50 and I want 5
Well, you have a lot of wrong stuff going on. Firstly, you never change compare, Second, you take the int in ascii, and don't convert it to its Integer representation. Thirdly, you need only one if statement. All combined, it gives
int max = Character.getNumericValue(number.charAt(0));
for (int i = 1; i < number.length(); i++) {
int compare = Character.getNumericValue(number.charAt(i));
if (max < compare) {
max = compare;
}
}
return max;
I've used Character#getNumericValue to convert the char to int
Since ASCII codes of digits are sorted in ascending order you can easily do it with Java 8 like this:
public static void max(String number) {
if (number == null || number.isEmpty()) {
return;
}
int max = number.chars().max().getAsInt();
System.out.println(Character.getNumericValue((char) max));
}
If your input can has mix of digits and other symbols you also able to handle it:
public static void max(String number) {
if (number == null || number.isEmpty()) {
return;
}
OptionalInt max = number.chars().filter(Character::isDigit).max();
if (max.isPresent()) {
System.out.println(Character.getNumericValue((char) max.getAsInt()));
} else {
System.err.println("Provided string doesn't contain digits");
}
}
you have to modify your max method as given below
public static void max(String number) {
if (number.isEmpty()) {
System.out.println("The string is empty");
System.exit(0);
}
int max = Integer.parseInt(number.charAt(0)+"");
for (int i = 1; i < number.length(); i++) {
int compare = Integer.parseInt(number.charAt(i)+"");
if (compare > max) {
max = compare;
}
}
System.out.print(max);
}

Loop in Credit Card Validation in java

I am a high school student in an introductory Computer Science course. Our assignment was the following:
The last digit of a credit card number is the check digit, which protects against transcription errors such as an error in a single digit or switching two digits. the following method is used to verify actual credit card numbers but, for simplicity, we will describe it for numbers with 8 digits instead of 16:
Starting from the rightmost digit, form the sum of every other digit. For example, if the credit card number is 4358 9795, then you form the sum 5+7+8+3 = 23.
Double each of the digits that were not included in the preceding step. Add all the digits of the resulting numbers. For example, with the numbers given above, doubling the digits, starting with the next-to-last one, yields 18 18 10 8. Adding all the digits in these values yields 1+8+1+8+1+0+8=27.
Add the sums of the two preceding steps. If the last digit of the result is 0, the number is valid. In our case, 23 + 27 = 50, so the number is valid.
Write a program that implements this algorithm. The user should supply an 8-digit number, and you should print out whether the number is valid or not. If it is not valid, you should print out the value of the check digit that would make the number valid.
I have everything done except for the part in bold. My code is listed below:
public class CreditCard
{
private String creditCardNumber;
private boolean valid;
private int checkDigit;
int totalSum;
/**
* Constructor for objects of class CreditCard
*/
public CreditCard(String pCreditCardNumber)
{
creditCardNumber = pCreditCardNumber;
checkDigit = Integer.parseInt(pCreditCardNumber.substring(creditCardNumber.length() - 1));
int sumOfDigits = checkDigit + Integer.parseInt(pCreditCardNumber.substring(6,7)) + Integer.parseInt(pCreditCardNumber.substring(3,4)) + Integer.parseInt(pCreditCardNumber.substring(1,2));
int dig7 = Integer.parseInt(pCreditCardNumber.substring(7,8));
int dig5 = Integer.parseInt(pCreditCardNumber.substring(5,6));
int dig3 = Integer.parseInt(pCreditCardNumber.substring(2,3));
int dig1 = Integer.parseInt(pCreditCardNumber.substring(0,1));
String string7 = Integer.toString(dig7);
int doubledDig7a = Integer.parseInt(string7.substring(0));
int doubledDig7b = 0;
if (dig7 * 2 >= 10)
{
doubledDig7a = Integer.parseInt(string7.substring(0));
doubledDig7b = 0;
}
String string5 = Integer.toString(dig5);
int doubledDig5a = Integer.parseInt(string7.substring(0));
int doubledDig5b = 0;
if (dig5 * 2 >= 10)
{
doubledDig5a = Integer.parseInt(string5.substring(0));
doubledDig5b = 0;
}
String string3 = Integer.toString(dig3);
int doubledDig3a = Integer.parseInt(string3.substring(0));
int doubledDig3b = 0;
if (dig3 * 2 >= 10)
{
doubledDig3a = Integer.parseInt(string3.substring(0));
doubledDig3b = 0;
}
String string1 = Integer.toString(dig1);
int doubledDig1a = Integer.parseInt(string1.substring(0));
int doubledDig1b = 0;
if (dig1 * 2 >= 10)
{
doubledDig1a = Integer.parseInt(string1.substring(0));
doubledDig1b = 0;
}
int doubleDigits = doubledDig1a + doubledDig1b + doubledDig3a + doubledDig3b + doubledDig5a + doubledDig5b + doubledDig7a + doubledDig7b;
totalSum = sumOfDigits + doubleDigits;
if (totalSum % 10 == 0)
{
valid = true;
}
else
{
valid = false;
}
}
public void makeItValid()
{
while (totalSum % 10 != 0)
{
checkDigit--;
if (totalSum % 10 == 0)
{
break;
}
}
}
public boolean isItValid()
{
return valid;
}
}
The loop is what I am having issues with. I always end up in an infinite loop whenever it compiles. It looks like everything should work, though. It's supposed to decrease the value of the check Digit (not increase so I don't end up with a check digit of 10 or higher), and then add that number back into the total sum until the total sum is divisible by 10, and then the loop would end. Is the type of loop I'm using wrong? Any advice would be appreciated.
Your problem is that both of your loop conditions involve totalSum but you only change checkDigit.
while (totalSum % 10 != 0)
{
checkDigit--;
if (totalSum % 10 == 0)
{
break;
}
}
You either need to recalculate totalSum or change the condition to be based on checkDigit. If you want to loop and decrement like you are doing you will need to add a method that performs the algorithm and call it every time. The way you have your class outlined makes this very inconvenient because you don't convert the numbers.
public static int[] cardToNumbers(String cardText) {
// \D is regex for non-digits
cardText = cardText.replaceAll("\\D", "");
int[] cardNumbers = new int[cardText.length()];
// convert unicode to corresponding integers
for (int i = 0; i < cardText.length(); i++)
cardNumbers[i] = cardText.charAt(i) - '0';
return cardNumbers;
}
public static int calcTotalSum(int[] cardNumbers) {
int sum = 0;
/* "every other one" loops
*
* I recommend against the "mod 2 index" scheme
* i % 2 relies on the card number being even
* you can't have your code blow up with unusual inputs
*
*/
for (int i = cardNumbers.length - 1; i >= 0; i -= 2) {
sum += cardNumbers[i];
}
for (int i = cardNumbers.length - 2; i >= 0; i -= 2) {
int dig = cardNumbers[i] * 2;
while (dig > 0) {
sum += dig % 10;
dig /= 10;
}
}
return sum;
}
Now you can do something like:
public void makeItValid() {
int[] invalidNumbers = cardToNumbers(creditCardNumber);
int sum = calcTotalSum(invalidNumbers);
while ((sum = calcTotalSum(invalidNumbers)) % 10 != 0)
invalidNumbers[invalidNumbers.length - 1]--;
totalSum = sum;
checkDigit = invalidNumbers[invalidNumbers.length - 1];
}
But you should be able to just subtract the difference to find the valid check digit:
if (totalSum % 10 != 0) checkDigit -= totalSum % 10;
Or something like:
public void makeItValid() {
int[] invalidNumbers = cardToNumbers(creditCardNumber);
checkDigit = invalidNumbers[invalidNumbers.length - 1] -= totalSum % 10;
totalSum = calcTotalSum(invalidNumbers);
valid = true;
}
Some asides,
I would recommend storing the digits as a field and have checkDigit represent an index in the array. This would simplify some of the operations you are doing.
I would also suggest not to be "silently" changing fields internally IE like in your makeItValid method unless this is a specification of the assignment. I think a better form is to let the "owning" code make the changes itself which is more clear externally. A somewhat complete implementation would look like this:
public class CreditCard {
public static void main(String[] args) {
if (args.length == 0) return;
CreditCard card = new CreditCard(args[0]);
if (!card.isValidNumber()) {
card.setCheckDigit(card.getValidCheckDigit());
}
}
private final String cardText;
private final int[] cardDigits;
private final int cdIndex;
public CreditCard(String ct) {
cardDigits = cardToNumbers(cardText = ct);
if ((cdIndex = cardDigits.length - 1) < 0) {
throw new IllegalArgumentException("# had no digits");
}
}
public boolean isValidNumber() {
return calcTotalSum(cardDigits) % 10 == 0;
}
public void setCheckDigit(int dig) {
cardDigits[cdIndex] = dig;
}
public int getValidCheckDigit() {
int sum = calcTotalSum(cardDigits);
if (sum % 10 != 0) {
return cardNumbers[cdIndex] - sum % 10;
} else {
return cardNumbers[cdIndex];
}
}
// above static methods
}
The best form IMO would be to disallow creation of a credit card object at all unless the check digit is valid. As an OOP principle it should not make sense to create invalid credit cards. The constructor should throw an exception if the card is invalid and have a static method to correct the number.
I would do something like the following (shortened):
public class CreditCard {
public CreditCard(String number) {
if (!validateCheckDigit(number)) {
throw new IllegalArgumentException("check digit failure");
}
}
}
public static void main(String[] args) {
String number = args[0];
CreditCard card = null;
boolean valid = false;
do {
try {
card = new CreditCard(number);
valid = true;
} catch (IllegalArgumentException e) {
number = CreditCard.correctCheckDigit(number);
}
} while (!valid);
}
I guess that's more or less doing your homework for you but I'm sure you can learn from it.
Unless I'm missing something major on how the validation works your makeitvalid method wont work in the way you are approaching it.
It makes more sense (at least to me) to extract everything you have in your constructor into a method ie.
boolean isValid(String cardNumber);
which would do everything that your constructor does except set the valid flag. your constructor then becomes
public CreditCard(String pCreditCardNumber){
valid = isValid(pCreditCardNumber);
}
and then to find what change would make it valid your check valid method does something like
change the value of check digit
if (isValid(Changed String))
return checkdigit
else
continue
repeat until you either find one that works or until you determine that it can't work.
Something along these lines should do. You'll still need to implement a few methods on your own.
public static void main(String[] args) {
String creditCardNumber = readCreditCardNumber();
String correctCreditCardNumber = getCorrectCreditCardNumber(creditCardNumber);
if (creditCardNumber.equals(correctCreditCardNumber)) {
System.out.println("Credit Card Valid");
} else {
System.out.println("Credit Card Invalid. Did you mean " + correctCreditCardNumber + "?");
}
}
public static String getCorrectCreditCardNumber(String creditCardNumber) {
int[] creditCardDigits = getCreditCardDigits(creditCardNumber);
int sum = 0;
for (int i = creditCardDigits.length - 2; i >= 0; i--) {
if (isOdd(i)) {
sum += creditCardDigits[i];
} else {
sum += digitSum(creditCardDigits[i] * 2);
}
}
int last = creditCardDigits.length - 1;
int remainder = sum % 10;
if (remainder != 0) {
creditCardDigits[last] = 10 - remainder;
}
return getCreditCardNumberAsString(creditCardDigits);
}
This program is very dynamic. I did not add too much error handling. You can enter any number that is divisible by 8.
Code in action:
Enter a card number: 4358 9795
Number is valid?: true
Continue? (y/n): y
Enter a card number: 4358 9796
Number is valid?: false
Continue? (y/n): y
Enter a card number: 43-58 97-95
Number is valid?: true
Continue? (y/n): n
Exiting...
CreditCardValidator.java
import java.text.ParseException;
import java.util.Scanner;
public class CreditCardValidator {
Integer[] digits;
public CreditCardValidator(String numberSequence) {
parseNumber(numberSequence);
}
private void parseNumber(String numberSequence) {
try {
String sequence = numberSequence.replaceAll("[\\s-]+", "");
int length = sequence.length();
if (length % 8 != 0) {
throw new IllegalArgumentException("Number length invalid.");
}
digits = new Integer[length];
int pos = 0;
for (Character c : sequence.toCharArray()) {
if (Character.isDigit(c)) {
digits[pos++] = Character.getNumericValue(c);
} else {
throw new ParseException("Invalid digit.", pos);
}
}
} catch (Exception e) {
e.printStackTrace();
}
}
private boolean validateNumber() {
int sum = 0;
for (int i = digits.length - 1; i >= 0; i--) {
if (i % 2 == 1) {
sum += digits[i];
} else {
sum += NumberUtils.sumDigits(digits[i] * 2);
}
}
return sum % 10 == 0;
}
public static void main(String[] args) {
boolean stop = false;
CreditCardValidator c;
while (!stop) {
System.out.print("Enter a card number: ");
c = new CreditCardValidator(new Scanner(System.in).nextLine());
System.out.println("Number is valid?: " + c.validateNumber());
System.out.print("\nContinue? (y/n): ");
if (new Scanner(System.in).next().charAt(0) == 'n') {
stop = true;
}
System.out.println();
}
System.out.println("Exiting...");
System.exit(0);
}
}
I wrote a separate digit summation utility:
public class NumberUtils {
public static void main(String[] args) {
for(int i = 0; i < 2000; i+=75) {
System.out.printf("%04d: %02d\n", i, sumDigits(i));
}
}
public static int sumDigits(int n) {
if (n < 0)
return 0;
return sumDigitsRecursive(n, 0);
}
private static int sumDigitsRecursive(int n, int total) {
if (n < 10)
return total + n;
else {
return sumDigitsRecursive(n / 10, total + (n % 10));
}
}
}

Uva's 3n+1 problem

I'm solving Uva's 3n+1 problem and I don't get why the judge is rejecting my answer. The time limit hasn't been exceeded and the all test cases I've tried have run correctly so far.
import java.io.*;
public class NewClass{
/**
* #param args the command line arguments
*/
public static void main(String[] args) throws IOException {
int maxCounter= 0;
int input;
int lowerBound;
int upperBound;
int counter;
int numberOfCycles;
int maxCycles= 0;
int lowerInt;
BufferedReader consoleInput = new BufferedReader(new InputStreamReader(System.in));
String line = consoleInput.readLine();
String [] splitted = line.split(" ");
lowerBound = Integer.parseInt(splitted[0]);
upperBound = Integer.parseInt(splitted[1]);
int [] recentlyused = new int[1000001];
if (lowerBound > upperBound )
{
int h = upperBound;
upperBound = lowerBound;
lowerBound = h;
}
lowerInt = lowerBound;
while (lowerBound <= upperBound)
{
counter = lowerBound;
numberOfCycles = 0;
if (recentlyused[counter] == 0)
{
while ( counter != 1 )
{
if (recentlyused[counter] != 0)
{
numberOfCycles = recentlyused[counter] + numberOfCycles;
counter = 1;
}
else
{
if (counter % 2 == 0)
{
counter = counter /2;
}
else
{
counter = 3*counter + 1;
}
numberOfCycles++;
}
}
}
else
{
numberOfCycles = recentlyused[counter] + numberOfCycles;
counter = 1;
}
recentlyused[lowerBound] = numberOfCycles;
if (numberOfCycles > maxCycles)
{
maxCycles = numberOfCycles;
}
lowerBound++;
}
System.out.println(lowerInt +" "+ upperBound+ " "+ (maxCycles+1));
}
}
Are you making sure to accept the entire input? It looks like your program terminates after reading only one line, and then processing one line. You need to be able to accept the entire sample input at once.
I faced the same problem. The following changes worked for me:
Changed the class name to Main.
Removed the public modifier from the class name.
The following code gave a compilation error:
public class Optimal_Parking_11364 {
public static void main(String[] args) {
...
}
}
Whereas after the changes, the following code was accepted:
class Main {
public static void main(String[] args) {
...
}
}
This was a very very simple program. Hopefully, the same trick will also work for more complex programs.
If I understand correctly you are using a memoizing approach. You create a table where you store full results for all the elements you have already calculated so that you do not need to re-calculate results that you already know (calculated before).
The approach itself is not wrong, but there are a couple of things you must take into account. First, the input consists of a list of pairs, you are only processing the first pair. Then, you must take care of your memoizing table limits. You are assuming that all numbers you will hit fall in the range [1...1000001), but that is not true. For the input number 999999 (first odd number below the upper limit) the first operation will turn it into 3*n+1, which is way beyond the upper limit of the memoization table.
Some other things you may want to consider are halving the memoization table and only memorize odd numbers, since you can implement the divide by two operation almost free with bit operations (and checking for even-ness is also just one bit operation).
Did you make sure that the output was in the same order specified in the input. I see where you are swapping the input if the first input was higher than the second, but you also need to make sure that you don't alter the order it appears in the input when you print the results out.
ex.
Input
10 1
Output
10 1 20
If possible Please use this Java specification : to read input lines
http://online-judge.uva.es/problemset/data/p100.java.html
I think the most important thing in UVA judge is 1) Get the output Exactly same , No Extra Lines at the end or anywhere . 2) I am assuming , Never throw exception just return or break with No output for Outside boundary parameters.
3)Output is case sensitive 4)Output Parameters should Maintain Space as shown in problem
One possible solution based on above patterns is here
https://gist.github.com/4676999
/*
Problem URL: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=36
Home>Online Judge > submission Specifications
Sample code to read input is from : http://online-judge.uva.es/problemset/data/p100.java.html
Runtime : 1.068
*/
import java.io.*;
import java.util.*;
class Main
{
static String ReadLn (int maxLg) // utility function to read from stdin
{
byte lin[] = new byte [maxLg];
int lg = 0, car = -1;
String line = "";
try
{
while (lg < maxLg)
{
car = System.in.read();
if ((car < 0) || (car == '\n')) break;
lin [lg++] += car;
}
}
catch (IOException e)
{
return (null);
}
if ((car < 0) && (lg == 0)) return (null); // eof
return (new String (lin, 0, lg));
}
public static void main (String args[]) // entry point from OS
{
Main myWork = new Main(); // create a dinamic instance
myWork.Begin(); // the true entry point
}
void Begin()
{
String input;
StringTokenizer idata;
int a, b,max;
while ((input = Main.ReadLn (255)) != null)
{
idata = new StringTokenizer (input);
a = Integer.parseInt (idata.nextToken());
b = Integer.parseInt (idata.nextToken());
if (a<b){
max=work(a,b);
}else{
max=work(b,a);
}
System.out.println (a + " " + b + " " +max);
}
}
int work( int a , int b){
int max=0;
for ( int i=a;i<=b;i++){
int temp=process(i);
if (temp>max) max=temp;
}
return max;
}
int process (long n){
int count=1;
while(n!=1){
count++;
if (n%2==1){
n=n*3+1;
}else{
n=n>>1;
}
}
return count;
}
}
Please consider that the integers i and j must appear in the output in the same order in which they appeared in the input, so for:
10 1
You should print
10 1 20
package pandarium.java.preparing2topcoder;/*
* Main.java
* java program model for www.programming-challenges.com
*/
import java.io.*;
import java.util.*;
class Main implements Runnable{
static String ReadLn(int maxLg){ // utility function to read from stdin,
// Provided by Programming-challenges, edit for style only
byte lin[] = new byte [maxLg];
int lg = 0, car = -1;
String line = "";
try
{
while (lg < maxLg)
{
car = System.in.read();
if ((car < 0) || (car == '\n')) break;
lin [lg++] += car;
}
}
catch (IOException e)
{
return (null);
}
if ((car < 0) && (lg == 0)) return (null); // eof
return (new String (lin, 0, lg));
}
public static void main(String args[]) // entry point from OS
{
Main myWork = new Main(); // Construct the bootloader
myWork.run(); // execute
}
public void run() {
new myStuff().run();
}
}
class myStuff implements Runnable{
private String input;
private StringTokenizer idata;
private List<Integer> maxes;
public void run(){
String input;
StringTokenizer idata;
int a, b,max=Integer.MIN_VALUE;
while ((input = Main.ReadLn (255)) != null)
{
max=Integer.MIN_VALUE;
maxes=new ArrayList<Integer>();
idata = new StringTokenizer (input);
a = Integer.parseInt (idata.nextToken());
b = Integer.parseInt (idata.nextToken());
System.out.println(a + " " + b + " "+max);
}
}
private static int getCyclesCount(long counter){
int cyclesCount=0;
while (counter!=1)
{
if(counter%2==0)
counter=counter>>1;
else
counter=counter*3+1;
cyclesCount++;
}
cyclesCount++;
return cyclesCount;
}
// You can insert more classes here if you want.
}
This solution gets accepted within 0.5s. I had to remove the package modifier.
import java.util.*;
public class Main {
static Map<Integer, Integer> map = new HashMap<>();
private static int f(int N) {
if (N == 1) {
return 1;
}
if (map.containsKey(N)) {
return map.get(N);
}
if (N % 2 == 0) {
N >>= 1;
map.put(N, f(N));
return 1 + map.get(N);
} else {
N = 3*N + 1;
map.put(N, f(N) );
return 1 + map.get(N);
}
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
try {
while(scanner.hasNextLine()) {
int i = scanner.nextInt();
int j = scanner.nextInt();
int maxx = 0;
if (i <= j) {
for(int m = i; m <= j; m++) {
maxx = Math.max(Main.f(m), maxx);
}
} else {
for(int m = j; m <= i; m++) {
maxx = Math.max(Main.f(m), maxx);
}
}
System.out.println(i + " " + j + " " + maxx);
}
System.exit(0);
} catch (Exception e) {
}
}
}

Categories

Resources