Develop Reference

Java:Do-while Loop, Even number then trying again

I can loop simply but it is hard for me to do even and odd numbers only. I want it like these for example:
Enter your number: 20
2 4 6 8 10 12 14 16 18 20
Do you want to do it again? Yes/No?
My code:
public static void main(String[] args) {
Scanner input=new Scanner(System.in);
String c = null;
do {
int i;
System.out.println("Enter a Number:");
int n = input.nextInt();
for(i=1; i<n; i++) {
System.out.println(i);
if(n%2==0) {
System.out.println(i + " " );
}
System.out.println("Try Again? Y/N");
c = input.next();
}
}while(c.equalsIgnoreCase("y"));
}

You need to check i % 2 == 0 instead of n % 2 == 0.
Demo:
import java.util.Scanner;
class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String c = null;
do {
System.out.print("Enter a Number: ");
int n = input.nextInt();
for (int i = 1; i <= n; i++) {
if (i % 2 == 0) {
System.out.print(i + " ");
}
}
System.out.print("\nTry Again? Y/N: ");
c = input.next();
} while (c.equalsIgnoreCase("y"));
}
}
A sample run:
Enter a Number: 20
2 4 6 8 10 12 14 16 18 20
Try Again? Y/N: y
Enter a Number: 25
2 4 6 8 10 12 14 16 18 20 22 24
Try Again? Y/N: n

Your code is saying if (n%2 == 0) but the loop iterates over i. This n will never change in each loop.
It's probably not the only problem in your code, but certainly you should be looking at what i does in the loop instead of n on that line.

Decreasing number pyramid, nested for loop, user input, java

Write a program to produce the following output for any given integer number between
1 and 9 inclusive.
Enter an integer value [1..9]: 6
1
12
123
1234
12345
123456
666666
66666
6666
666
66
6
I have done the top half but I can not figure out the bottom with the repeating user input.
package lab7;
import java.util.Scanner;
public class problem5 {
public static void main(String[] args) {
Scanner scan = new Scanner (System.in);
System.out.println("Input an integer between 1 and 9");
int input = scan.nextInt();
while (input <= 9) {
for (int i = 1; i <= input; i++) {
for (int j = 1; j <= i; j++) {
System.out.print(j);
}
System.out.println();
}
break;
}
}
}
Expected result: included at the top; actual result so far (input of 5):
1
12
123
1234
12345

You're pretty close. You have a for loop that covers the first half of the output you want. You can add a second for loop to handle the second half of the output.
This is pretty similar to the first loop, but has a few small differences:
instead of the loop variable starting at 1 and increasing, this one starts at input and decreases each time through (i-- instead of i++)
instead of printing any of the loop variables (i or j), it prints the input value ("6" in your example)
for (int i = input; i > 0; i--) {
for (int j = 1; j <= i; j++) {
System.out.print(input);
}
System.out.println();
}
If I run that code locally – so your for loop, then this for loop, then the break statement – this is the output:
Input an integer between 1 and 9
6
1
12
123
1234
12345
123456
666666
66666
6666
666
66
6

I would prefer a more efficient algorithm, your current approach is O(n2); consider the digits '1' - '9'; if we store them in a String then we can take a simple substring of that String for each line at the top (for example, "123456789".substring(0, 3) -> "123") that can be used to generate the top through successive calls to substring. We can use a similar approach to build the bottom; use an array of all possible rows and iteratively call substring. Finally, don't forget to validate that input is between one and nine inclusive. Something like,
Scanner scan = new Scanner(System.in);
String digits = "123456789";
String[] btm = { "1", "22", "333", "4444", "55555",
"666666", "7777777", "88888888", "999999999" };
System.out.println("Input an integer between 1 and 9");
int input = scan.nextInt();
if (input < 1 || input > 9) {
System.err.printf("Invalid input: %d%n", input);
System.exit(1);
}
for (int i = 0; i < input; i++) {
System.out.println(digits.substring(0, i + 1));
}
for (int i = input - 1; i >= 0; i--) {
System.out.println(btm[input - 1].substring(0, i + 1));
}

Java Based on this input, the program will display the number of integers in less than 50

for the task i need to Write a Java application that accepts 30 integer numbers from the user. The input should be in the range of 1-200. Error message needs to be displayed if user entered input which is not in this range. Based on this input, the program will display the number of integers entered in the following categories:
 Less than 50
 Between 50-100 (inclusive of 50 and 100)
Sample output :
Enter number 4: 211
Input error..Pls enter number between 1 to 200 only
Enter number 4: 20
..
..
Enter number 30: 90
Less than 50: 12
Between 50-100 (inclusive of 50 and 100): 8
Between 101-150 (inclusive of 101 and 150): 5
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// Keyboard Initialization
Scanner kbin = new Scanner(System.in);
// a.Declare an array to hold 5 Integer values
int list[] = new int[5];
int i = 0;
System.out.print("\n\tInput numbers from 1 to 200: \n");
while (i < 5) {
// b.Fill the array with intgers from the keyboard(range: 1 to 200).
System.out.print("Enter Integer" + (i + 1) + ":");
int value = kbin.nextInt();
if (value >= 1 && value <= 200) {
list[i] = value;
i++;
} else {
System.out.println("!! Error! Please Enter Value between 1 and 200 !!");
}
}
}
}

Here is an example using Java Streams. Here we simply read all the values first, then filter out each category later. Performance wise, it's better to just use a counter for each category if you don't care about the actual numbers being counted though.
public class Test {
public static void main(String[] args) {
int counter = 0;
Scanner scanner = new Scanner(System.in);
List<Integer> values = new ArrayList<>();
while (counter < 5) {
System.out.print("Enter integer (" + (counter + 1) + "): ");
int value = scanner.nextInt();
if (value >= 1 && value <= 200) {
counter++;
values.add(value);
} else {
System.out.println("Please enter a value between 1 and 200");
}
}
System.out.println("Between 1-50 : " + values.stream().filter(val -> val < 50).count());
System.out.println("Between 50-100 : " + values.stream().filter(val -> val >= 50 && val <= 100).count());
System.out.println("Between 101-150: " + values.stream().filter(val -> val > 100 && val <= 150).count());
System.out.println("Between 151-200: " + values.stream().filter(val -> val > 150 && val <= 200).count());
}
Output example:
Enter integer (1): 5
Enter integer (2): 55
Enter integer (3): 125
Enter integer (4): 175
Enter integer (5): -2
Please enter a value between 1 and 200
Enter integer (5): 201
Please enter a value between 1 and 200
Enter integer (5): 199
Between 1-50 : 1
Between 50-100 : 1
Between 101-150: 1
Between 151-200: 2

It is a very simple problem here #Dilip. Just introduce a flag variable which would mark if an input is wrong. If it is, then just accept the value again. Else proceed.
Following is the code snippet for the input part also considering the wrong and right input boundaries.
System.out.print("\n\tInput numbers from 1 to 200: \n");
for(i = 0; i < 5; i++) {
flag = 0;
while(flag == 0) {
list[i] = kbin.nextInt();
if(!(list[i] < 1 || list[i] > 200)){
flag = 1;
}
else {
System.out.println("INVALID INPUT!!! Enter a valid number.");
}
}
}
I have gone through the problem statement which you have told about in the comments section and have coded the program which would serve the purpose. I have also attached the output for confirmation and reference. Hope this helps.
CODE:
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// Keyboard Initialization
Scanner kbin = new Scanner(System.in);
// a.Declare an array to hold 5 Integer values
int list[] = new int[5];
int category1[] = new int[5], k1 = 0;
int category2[] = new int[5], k2 = 0;
int category3[] = new int[5], k3 = 0;
int i, flag;
System.out.print("\n\tInput numbers from 1 to 200: \n");
for(i = 0; i < 5; i++) {
flag = 0;
while(flag == 0) {
System.out.print("Enter the input number " + (i+1) + ": ");
list[i] = kbin.nextInt();
if(!(list[i] < 1 || list[i] > 200)){
flag = 1;
}
else {
System.out.println("INVALID INPUT!!! Enter a valid number.");
}
}
}
for(i = 0; i < 5; i++) {
if(list[i] < 50)
category1[k1++] = list[i];
else if(list[i] < 101)
category2[k2++] = list[i];
else
category3[k3++] = list[i];
}
System.out.print("Category 1(1 TO 49): ");System.out.println(k1);
System.out.print("Category 2(50 TO 100): ");
System.out.println(k2);
System.out.print("Category 3(greater than 100): ");
System.out.println(k3);
System.out.print("Category 4(151 to 200): ");
System.out.println(k4);
}
}
OUTPUT:
Input numbers from 1 to 200:
Enter the input number 1: 2
Enter the input number 2: 3
Enter the input number 3: 50
Enter the input number 4: 56
Enter the input number 5: 159
Category 1(1 TO 49): 2
Category 2(50 TO 100): 2
Category 3(101 to 150): 0
Category 4(151 to 200): 1

Using a while loop to compare to a value in a array

I need the program to run in a loop until 0 is entered. My code will end with 0 entered but when attempting to run the program with numbers entered it still ends the program. instead of running the numbers entered. The while loop is to keep the program running unless a 0 is entered.
import java.util.Scanner;
public class CountCompare {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter the integers between 1 and 100 (0 to end, 0 < to exit): ");
int[] counts = new int[100];
// Count occurrence of numbers
count(counts);
while(counts[0] > 0){
// Display results
for (int i = 0; i < counts.length; i++) {
if (counts[i] > 0)
System.out.println((i + 1) + " occurs " + counts[i] +
" time" + (counts[i] > 1 ? "s" : ""));
}
System.out.print("Enter the integers between 1 and 100 : ");
// Count occurrence of numbers
count(counts);
}
System.out.print("\nEnd of run");
}
/** Method count reads integers between 1 and 100
* and counts the occurrences of each */
public static void count(int[] counts){
Scanner input = new Scanner(System.in);
int num; // holds user input
do {
num = input.nextInt();
if (num >= 1 && num <= 100)
counts[num - 1]++;
} while (num != 0);
}
}
I have posted the entire program.
output looks like this
Enter the integers between 1 and 100 (0 to end, <0 to exit):
23 23 4 5 6 7 8
0
4 occurs 1 time
5 occurs 1 time
6 occurs 1 time
7 occurs 1 time
8 occurs 1 time
23 occurs 2 times
Enter the integers between 1 and 100:

Your program still ends because:
int[] counts = new int[100];
You have defined the limit of the counts here. This means your loop will run
for (int i = 0; i < counts.length; i++)// counts.length=100;
So as far as you code suggest you want to end the user input when user input 0. So you might do this:
int x=1;
int y;
Scanner sc= new Scanner(System.in);
while(x!=0){
System.out.println("Enter your values");
y=sc.nextInt();
if(y==0){
x=0;
}
else{
System.out.println("You entered "+y);
}

int count=new Scanner(System.in).nextInt();
int myarray[]=new int[count];
for(int tmp=0; tmp<count;)
myarray[tmp]=++tmp;
while(count != 0){
for(int inc=1; inc<=count; inc++){
System.out.println(inc + "times occur");
}
System.out.println("Enter 0 to exit");
count=new Scanner(System.in).nextInt();
}

You should take a look at the line
while(counts[0] > 0) {
and try to figure out what is the purpose of this while loop in the main method.

Testing primes in Java - adding additional conditions

Beginner here. So I want to write a program that prints out all the prime numbers up to the number the user entered. E.g., user enters 5, program prints out 2 and 3. That part I understand, however what I am struggling with, is what if I want the program to print out whether the number the user entered is a prime or not (simple yes or no) IF the entered number is bigger than, let's say, 50. Here is code for first part:
public class Primes {
public static void main(String args[]) {
System.out.println("All primes up to: ");
int num = new Scanner(System.in).nextInt();
System.out.println("Prime numbers from 1 to " + num + " are: ");
for(int number = 2; number<=num; number++){
if(isPrime(number)){
System.out.println(number);
}
}
}
public static boolean isPrime(int number){
for(int i=2; i<number; i++){
if(number%i == 0){
return false;
}
}
return true;
}
}
I honestly can't wrap my around as to what I should be doing next. My first program ever ("Hello world" does not count ;P).

Edit :
Your current code seems to work fine.
As per your doubt as mentioned in one of the comments : Yes, but where do I add if statement that does the following: if the number entered is below 50, then the program prints out all the prime numbers up to the entered number. If the number the user entered is bigger than 50, it tells only whether the entered number is prime or not ( simply "It's a prime" or "No, it's not a prime"). Hope that made things clearer
The check you need to put is after you take the input :
int num = new Scanner(System.in).nextInt();
if( number > 50 )
{
if(isPrime(number))
{
// print out is prime
}
// print out it is not prime
}
else
{
System.out.println("Prime numbers from 1 to " + num + " are: ");
for(int number = 2; number<=num; number++){
if(isPrime(number)){
System.out.println(number);
}
}
}
SUGESTIONS :
However, just to touch upon the algorithmic part, I would recommend using Sieve of Eratosthenes for picking out all the prime numbers within a given range, as you need in your case.
Example :
To find all the prime numbers less than or equal to 30, proceed as follows:
First generate a list of integers from 2 to 30:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Strike (sift out) the multiples of 2 resulting in:
2 3 5 7 9 11 13 15 17 19 21 23 25 27 29
The first number in the list after 2 is 3; strike the multiples of 3 from the list to get:
2 3 5 7 11 13 17 19 23 25 29
The first number in the list after 3 is 5; strike the remaining multiples of 5 from the list:
2 3 5 7 11 13 17 19 23 29
The first number in the list after 5 is 7, but 7 squared is 49 which is greater than 30 so the process is finished. The final list consists of all the prime numbers less than or equal to 30.
Here's the code attached for reference ( Disclaimer : I'm picking up this code here from this site. Just pasted it here for more immediate visibility).
Code :
public class PrimeSieve {
public static void main(String[] args) {
int N = Integer.parseInt(args[0]);
// initially assume all integers are prime
boolean[] isPrime = new boolean[N + 1];
for (int i = 2; i <= N; i++) {
isPrime[i] = true;
}
// mark non-primes <= N using Sieve of Eratosthenes
for (int i = 2; i*i <= N; i++) {
// if i is prime, then mark multiples of i as nonprime
// suffices to consider mutiples i, i+1, ..., N/i
if (isPrime[i]) {
for (int j = i; i*j <= N; j++) {
isPrime[i*j] = false;
}
}
}
// count primes
int primes = 0;
for (int i = 2; i <= N; i++) {
if (isPrime[i]) primes++;
}
System.out.println("The number of primes <= " + N + " is " + primes);
}
}

Try this..
int j = 2; //variable
int result = 0; //variable
int number = 0; //variable
Scanner reader = new Scanner(System.in); //Scanner object
System.out.println("Please enter a number: "); //Instruction
number = reader.nextInt(); //Get the number entered
while (j <= number / 2) //start loop, during loop j will become each number between 2 and
{ //the entered number divided by 2
if (number % j == 0) //If their is no remainder from your number divided by j...
{
result = 1; //Then result is set to 1 as the number divides equally by another number, hergo
} //it is not a prime number
j++; //Increment j to the next number to test against the number you entered
}
if (result == 1) //check the result from the loop
{
System.out.println("Number: " + number + " is Not Prime."); //If result 1 then a prime
}
else
{
System.out.println("Number: " + number + " is Prime. "); //If result is not 1 it's not a prime
}
this is more efficient way tough:-
public boolean isPrime(int n) {
// fast even test.
if(n > 2 && (n & 1) == 0)
return false;
// only odd factors need to be tested up to n^0.5
for(int i = 3; i * i <= n; i += 2)
if (n % i == 0)
return false;
return true;
}

however what I am struggling with, is what if I want the program to print out whether the number the user entered is a prime or not (simple yes or no).
Your current isPrime function seems to work, so just ask for a number and test it.
Scanner scanner = new Scanner(System.in);
while (scanner.hasNextInt()) {
System.out.println("Enter a number (is it prime): ");
int num = scanner.nextInt();
if (isPrime(num)) {
System.out.printf("%d yes%n", num);
} else {
System.out.printf("%d no%n", num);
}
}
Or with a ternary,
Scanner scanner = new Scanner(System.in);
while (scanner.hasNextInt()) {
System.out.println("Enter a number (is it prime): ");
int num = scanner.nextInt();
System.out.printf("%d %s%n", num, isPrime(num) ? "yes" : "no");
}
Edit Based on your comment, move your print up sequence to a method
public static void primesUpTo(int num) {
System.out.println("Prime numbers from 1 to " + num + " are: ");
for (int number = 2; number <= num; number++) {
if (isPrime(number)) {
System.out.println(number);
}
}
}
Then
Scanner scanner = new Scanner(System.in);
while (scanner.hasNextInt()) {
System.out.println("Enter a number (is it prime): ");
int num = scanner.nextInt();
if (num > 50) {
System.out.printf("%d %s%n", num, isPrime(num) ? "yes" : "no");
} else {
primesUpTo(num); // <-- call the method above.
}
}

If i understand the question right:
If user enteres number lesser than or equal to 50, then print all primes that are lesser than that number.
Otherwise, just write if inputted number is a prime.
With already existing isPrime() method:
int num = new Scanner(System.in).nextInt();
if (num <= 50) {
System.out.println("Prime numbers from 1 to " + num + " are: ");
for (int number = 2; number <= num; number++) {
if (isPrime(number)) {
System.out.println(number);
}
}
} else { //num > 50
if(isPrime(num)) {
System.out.println(num + " is prime.");
} else {
System.out.println(num + " isn't prime.");
}
}