I am trying to get 3 numbers separated by a space after user's input. I can get the first number and the last one dividing by 10, but I really have no idea how to get the middle number
I tried to take the remainder of the first two numbers and then divide them by ten, but IDEA says that the answer is always zero
public static void main(String[] args) {
System.out.println("Input the number");
int number = read.nextInt();
int a = number%10;
int b = (number%10)/10; // the answer is always 0
int c = number / 100;
System.out.println(c + " " + b + " " + a);
}
Anything modulo 10 will return a result in the range of 0 to 9, and (integer) dividing that by 10 will return 0. You need to reverse the order - first divide the 10 to remove the last digit, and then take the remainder from 10 to keep the middle digit:
int b = (number / 10) % 10;
I suggest using this instead of %. Because you can split and get any digit of number using this code:
int x=158;
char[] xValueInString = Integer.toString(x).toCharArray();
for(int i=0; i<xValueInString.length; i++){
System.out.println(xValueInString[i]);
}
You missed scanner in your code, add Scanner before reading number, try this with little change
public static void main(String[] args) {
System.out.println("Input the number");
Scanner read = new Scanner(System.in);
int number = read.nextInt();
int a = number%10;
int b = (number/10)%10; // the answer is always 0
int c = number/100;
System.out.println(c + " " + b + " " + a);
}
Related
I am trying to create a program where the user inputs a four digit code. I then need to separate the individual digits and apply some basic math separately.
For example user input: 1234
I need to grab numbers 1, 2, 3, 4, apply basic math to them, then return them as output as integers.
This is what I have so far:
public static void main(String[] args) {
int fourDigitPin;
int firstDigit;
int secondDigit;
int thridDigit;
int forthDigit;
Scanner keyInput = new Scanner(System.in);
System.out.print("Enter your 4 digit pin number: ");
fourDigitPin = keyInput.nextInt();
firstDigit = fourDigitPin.charAt(0);
As you can see, I havent gotten to the math portion yet. I am attempting to use charAt to grab the numbers, but cannot as they are integers. Should I set the set input variable "fourDigitPin" as a string or char? Any help would be greatly appreciated.
1st method:
public static void main(String[] args) {
String fourDigitPin;
int firstDigit;
int secondDigit;
int thirdDigit;
int forthDigit;
Scanner keyInput = new Scanner(System.in);
System.out.print("Enter your 4 digit pin number: ");
fourDigitPin = keyInput.next();
firstDigit = fourDigitPin.charAt(0) - '0';
secondDigit = fourDigitPin.charAt(1) - '0';
thirdDigit = fourDigitPin.charAt(2) - '0';
forthDigit = fourDigitPin.charAt(3) - '0';
System.out.println(firstDigit + " " + secondDigit + " " + thirdDigit + " " + forthDigit);
}
2nd method:
public static void main(String[] args) {
String[] fourDigitPin;
int firstDigit;
int secondDigit;
int thirdDigit;
int forthDigit;
Scanner keyInput = new Scanner(System.in);
System.out.print("Enter your 4 digit pin number: ");
fourDigitPin = keyInput.next().split("");
firstDigit = Integer.parseInt(fourDigitPin[0]);
secondDigit = Integer.parseInt(fourDigitPin[1]);
thirdDigit = Integer.parseInt(fourDigitPin[2]);
forthDigit = Integer.parseInt(fourDigitPin[3]);
System.out.println(firstDigit + " " + secondDigit + " " + thirdDigit + " " + forthDigit);
}
3rd method:
public static void main(String[] args) {
int fourDigitPin;
int firstDigit;
int secondDigit;
int thirdDigit;
int forthDigit;
Scanner keyInput = new Scanner(System.in);
System.out.print("Enter your 4 digit pin number: ");
fourDigitPin = keyInput.nextInt();
forthDigit = fourDigitPin % 10;
fourDigitPin /= 10;
thirdDigit = fourDigitPin % 10;
fourDigitPin /= 10;
secondDigit = fourDigitPin % 10;
fourDigitPin /= 10;
firstDigit = fourDigitPin % 10;
fourDigitPin /= 10;
System.out.println(firstDigit + " " + secondDigit + " " + thirdDigit + " " + forthDigit);
}
You should convert it to a string because you're expecting it to be four characters (not one).
Also, minor UI point: right now you don't verify that the user actually entered a four-digit number. You should do so before you get the individual digits so that you don't get an exception.
One more thing to be careful of: make sure that you don't "directly" cast the char back to an integer at any point because then it'll be cast to the equivalent ASCII value (not the actual value of the number).
Take the input as a String rather than an int.
String number = keyInput.next();
Then
firstDigit = number.charAt(0);
secondDigit = number.charAt(1);
thirdDigit = number.charAt(2);
fourDigit = number.charAt(3);
For this, you would have to use as many variables as individual digits.
Or you can use a loop.
for(int i=0;i<number.length();i++){
digit = number.charAt(i);
//More code
}
But a better soln would be to take modulo.
int n = keyInput.nextLine();
while(n>0){
lastDigit=n%10;
n/=10;
} //you get digits from the rear end as modulo returns remainder
You better use "modulo" operator approach to get individual digit and then perform math on that digit. This operator will give you reminder.
Example:
int remainder = a % b;
Refer javadoc operators for more details.
Other approaches like charAt might work in this case, because you know the pin is of size 4, but if the size is not known upfront and want to use your code for 5 digits or 3 digits, it will fail.
final char[] chars = String.valueOf(fourDigitPin).toCharArray();
int firstDigit = Integer.parseInt(String.valueOf(chars[0]));
int secondDigit= Integer.parseInt(String.valueOf(chars[1]));
int thridDigit=Integer.parseInt(String.valueOf(chars[2]));
int forthDigit= Integer.parseInt(String.valueOf(chars[3]));
tl;dr
String
.valueOf( 1_234 ) // Convert `int` to `String`.
.codePoints() // `IntStream` of Unicode code points, one integer for each character.
.filter( Character :: isDigit ) // Remove any non-digit.
.mapToObj( Character :: toString ) // Convert code point back to character of that digit.
.map( Integer :: valueOf ) // Parse that textual digit back to an integer. In this case, an `Integer` object.
.toList() // Make a list of our `Integer` objects, each element being a single digit from our original number.
.toString()
[1, 2, 3, 4]
Code points
The char type in Java is legacy, and is essentially broken. As a 16-bit value, it cannot represent most characters.
Instead, use code point integer numbers.
Convert your int to a String.
String inputString = String.valueOf( inputInt ) ;
Verify the length.
if( inputString.length() != 4 ) { … }
Get an IntStream of code point numbers.
IntStream codePoints = inputString.codePoints() ;
Loop those code point numbers, verifying it is a digit. If so, get the character. Convert that character to an int. Add it to our collection of integers.
List< Integer > digits =
codePoints
.filter( Character :: isDigit )
.mapToObj( Character :: toString )
.map( Integer :: valueOf )
.collect( Collectors.toList() ) // In Java 16+, replace with .toList()
;
See this code run live at IdeOne.com.
[1, 2, 3, 4]
It will be easy to improve your code's robustness by reading a string and testing its worthiness for further processing. Additionally, it will probably be easier to read your code if you store the digits in an array instead of enumerating each digit in a separate variable name. One indication that an array is more appealing is the fact that you have fewer variable names to misspell. You misspelled "third" as "thrid". Another is the fact that you have fewer parameters to pass if you have to pass parameters to some function. Yet another benefit is "rectangularization", which I'll define (redefine?) as being vertically aligned so as to spot typos more easily.
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
int digit[] = new int[4];
Scanner keyInput = new Scanner(System.in);
String pinstr;
do {
System.out.print("Enter your 4 digit pin number: ");
pinstr = keyInput.next();
if (pinstr.matches("^[0-9]{4}$"))
break;
System.err.println("You did not enter precisely 4 decimal digits");
} while (true);
digit[0] = pinstr.charAt(0) - '0';
digit[1] = pinstr.charAt(1) - '0';
digit[2] = pinstr.charAt(2) - '0';
digit[3] = pinstr.charAt(3) - '0';
for (int i=0; i<4; i++)
System.out.println(String.format("Digit at offset %d is %d", i, digit[i]));
}
}
I have been tasked with the assignment of creating a method that will take the 3 digit int input by the user and output its reverse (123 - 321). I am not allowed to convert the int to a string or I will lose points, I also am not allowed to print anywhere other than main.
public class Lab01
{
public int sumTheDigits(int num)
{
int sum = 0;
while(num > 0)
{
sum = sum + num % 10;
num = num/10;
}
return sum;
}
public int reverseTheOrder(int reverse)
{
return reverse;
}
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
Lab01 lab = new Lab01();
System.out.println("Enter a three digit number: ");
int theNum = input.nextInt();
int theSum = lab.sumTheDigits(theNum);
int theReverse = lab.reverseTheOrder(theSum);
System.out.println("The sum of the digits of " + theNum + " is " + theSum);
}
You need to use the following.
% the remainder operator
/ the division operator
* multiplication.
+ addition
Say you have a number 987
n = 987
r = n % 10 = 7 remainder when dividing by 10
n = n/10 = 98 integer division
Now repeat with n until n = 0, keeping track of r.
Once you understand this you can experiment (perhaps on paper first) to see how
to put them back in reverse order (using the last two operators). But remember that numbers ending in 0 like 980 will become 89 since leading 0's are dropped.
You can use below method to calculate reverse of a number.
public int reverseTheOrder(int reverse){
int result = 0;
while(reverse != 0){
int rem = reverse%10;
result = (result *10) + rem;
reverse /= 10;
}
return result;
}
Essentially I'm trying to create a program that counts up the sum of the digits of the number, but every time a number that is over 1000 pops up, the digits don't add up correctly. I can't use % or division or multiplication in this program which makes it really hard imo. Requirements are that if the user inputs any integer, n, then I will have to be able to compute the sum of that number.
I've already tried doing x>=1000, x>=10000, and so forth a multitude of times but I realized that there must be some sort of way to do it faster without having to do it manually.
import java.util.Scanner;
public class Bonus {
public static void main(String[] args) {
int x;
int y=0;
int u=0;
Scanner s = new Scanner(System.in);
System.out.print("Enter the number:");
x = s.nextInt();
int sum = 0;
{
while(x >= 100) {
x = x - 100;
y = y + 1;
}
while(x>=10) {
x = x - 10;
u = u + 1;
}
sum = y + u + x;
System.out.println("The sum of the digits in your number is" + " " + sum);
}
}
}
So if I type in 1,000 it displays 10. And if I type in 100,000 it displays 100. Any help is appreciated
Convert the number to a string, then iterate through each character in the string, adding its integer value to your sum.
int sum = 0;
x = s.nextInt();
for(char c : Integer.toString(x).toCharArray()) {
sum += Character.getNumericValue(c);
}
I have seen this question asked a few times, but all of the responses have included functionality that I haven't learned yet in this class and am I sure there must be a way to do it with only what I have learned. No arrays, etc... just loops and prior. I am not really looking for the answer, but just some direction. I have included the code I have already done. The program needs to be able to hand negative numbers, the sum and then print in the proper order. Right now my code does everything except print in the proper order. I understand why it is printing in reverse order (because the loop gets rid of and then prints the last number in the int), but I can't seem to figure out a way to change it. I have tried converting it to a string, char and just can't get it. Please take a look, and provide some guidance if you don't mind. thank you in advance.
public static void main(String[] args) {
int num;
int sum;
int temp;
System.out.print("Enter an integer, positive or negative: ");
num = keyboard.nextInt();
System.out.println();
if (num < 0)
num = -num;
sum = 0;
while (num > 0) {
temp = num;
sum = sum + num % 10; //Extracts the last digit and adds it to the sum
num = num / 10; //removes the last digit
System.out.print(temp % 10 + " ");
}
System.out.println(" and the sum is " + sum);
}
}
Not knowing what they've taught you in class so far, an easy albeit inefficient thing to do is to recreate the number as string.
String numbers = "";
while (num > 0) {
temp = num;
sum = sum + num % 10; //Extracts the last digit and adds it to the sum
num = num / 10; //removes the last digit
// System.out.print(temp % 10 + " ");
numbers = (temp % 10) + " " + numbers;
}
System.out.println(numbers + "and the sum is " + sum);
You were already grabbing the ones digit with (temp % 10) and then right shifting the original number with num = num / 10. There are other data structures you could use like a Stack or a LinkedList that are more natural to use in a situation like yours, or you could use a StringBuilder to append the digits to the end and then use the reverse() method to get them back in the correct order, but those data structures probably come after Arrays which you mentioned you didn't know.
Given those constraints, I used String concatenation. In general here is what happens:
String numbers = "";
num = 123;
digit = num % 10; // digit=3
num /= 10; // num=12
numbers = digit + " " + numbers; // numbers="3 " uses old value on right side of the equals
// next iteration
digit = num % 10; // digit=2
num /= 10; // num=1
numbers = digit + " " + numbers; // numbers="2 3 " see how the digit is put to the left of the old value
// last iteration
digit = num % 10; // digit=1
num /= 10; // num=0
numbers = digit + " " +numbers; // numbers="1 2 3 " notice there is an extra space at the end which is ok for your example
Set a counter, loop num/10, if result>0 counter++. In the end, counter+1 will be the number of digits
System.out.println("Please enter numbers: ");
int number_entered = input.nextInt();
int sum = 0;
String reserve = "";
if (number_entered < 0 ) {
number_entered = number_entered * -1;
}
for (number_entered = number_entered; number_entered > 0; number_entered/=10){
int lastdgt = number_entered%10;
sum += lastdgt;
reserve = lastdgt + " " + reserve + " ";
}
System.out.println(reserve);
System.out.println("The sum is = " + sum );
}
}
This question already has answers here:
Java reverse an int value without using array
(33 answers)
Closed 8 years ago.
I've looked around and came up with this solution, but it doesn't seem to be working. Does anyone have an idea? I need to get a number from the user that is only 3 digits and positive. after that, to reverse the 3 digits. what i wrote below only give me the last digit out of the three that I need.
int reversedNum=0;
Scanner scan = new Scanner (System.in);
System.out.println("Please enter a 3 digit positive number whose first and last digits are different: ");
int userNumber = scan.nextInt();
if (userNumber >= 100 && userNumber <= 999)
{
System.out.println("User number is: " + userNumber);
reversedNum = (reversedNum*10) + (userNumber%10);
userNumber = userNumber/10;
System.out.println("Difference "+reversedNum);
}
else
System.out.println("The number you entered is not a 3 digit positive number");
When you do
reversedNum = (reversedNum*10) + (userNumber%10);
userNumber = userNumber/10;
reversedNum is 0 so you end up with only userNumber%10.
You need something like this:
int hundreds = (int)(userNumber/100);
int remaining = userNumber-100*hundreds;
int dec = (int)(remaining /10);
remaining -= 10*dec;
int reversed = 100*remaining + 10*dec + hundreds
System.out.println("Reversed: " + reversed);
System.out.println("Difference " + (userNumber-reversed);
You can use % operator in order to print the last digit of input and then use / between int operand to get remaining digits
while(input%10 != input) {
int mod = input % 10;
System.out.print(mod);
input /= 10;
}
String result = "" + Integer.toString(userNumber).charAt(2) + Integer.toString(userNumber).charAt(1) + Integer.toString(userNumber).charAt(0);
int reversedNum = Integer.valueOf(result);
That will reverse your integer.
To reverse the number the logic should be as below. Use % and / operator to find the individual digit.
if (userNumber >= 100 && userNumber <= 999)
{
System.out.println("User number is: " + userNumber);
int unitdigit = userNumber%10;
userNumber = userNumber/10;
int tenthdigit = userNumber%10;
int lastdigit = userNumber/10;
reversedNum = (unitdigit*100) + (tenthdigit*10) + lastdigit;
System.out.println("reversed numnber "+reversedNum);
}