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));
}
Related
So basically the assignment says i need to get a number 'n' from the user and for all the numbers between 1 to 'n' code the program to print all the numbers divided by 3 without residue && print ONLY the numbers that both (or one) of their digits equal to 5 or less, for example if the user give 22 the programs prints 3,12,21.
thats what ive done by now (the place i putted a question mark is where im having hard time to figure out what to do) so this code in not compiled yet :
public static void main(String[] args) {
Scanner get = new Scanner(System.in);
int num;
System.out.println("Enter A Random Number: ");
num = get.nextInt();
for (int i=1;i>0 && i<=num;i++) {
if (i%3==0 && ?)
System.out.println(i);
I'm a bit confused about the second part of your question but based on the first part and what I got from the second part ( that the sum of the digits of numbers that is equal or less to 5 should only be printed )
Here is the code for your program : (It should work perfectly, update me if you find any problems)
public static void main(String[] args) {
Scanner get = new Scanner(System.in);
int num;
System.out.println("User please enter a number of your choice : ");
num = get.nextInt();
for(int x = 1 ; x < num ; x++){
String number = ""+ x ;
int sum = 0 ;
for(int i = 0 ; i < number.length() ; i ++ ){
sum +=number.charAt(i)-'0' ; }
if(x % 3 == 0 && sum <= 5){
System.out.println(x) ; }
sum = 0 ;
}
}
divided by 3 without residue
If you understand what "without residue" means, then I assume you are familiar with modulo arithmetic. In programming, we have the modulo operator % which returns the remainder from a division. So 25 % 8 evaluates to 1. You can use this to get the digits of a number 21 % 10 evalutes to 1 which is exactly the ones digit. To get the the tens digit, we need to divide by 10 first 21 / 10 % 10 evaluates to 2. This works because integer division throws away the remainder.
This will do. So, you have to go through every digit so, I converted it into a string and then matched the regex for 1 to 5 on that character and then put it back in another string and it solves it.
public static void main(String[] args) {
Scanner get = new Scanner(System.in);
int num;
System.out.println("Enter A Random Number: ");
num =Integer.parseInt(get.nextLine());
for(int i =1;i<=num; i++){
if(i%3==0){
String input = Integer.toString(i);
String toPrint = "";
for(int j =0 ; j<input.length();j++){
if(Character.toString(input.charAt(j)).matches("^[1-5]$")){
toPrint+=Character.toString(input.charAt(j));
}
}
//check the length to avoid cases like 30,60 etc.
if(input.length()==toPrint.length()){
System.out.println(toPrint);
}
}
}
}
This is the question we were assigned :
Nine coins are placed in a 3x3 matrix with some face up and some face down. You can represent the state of the coins using a 3x3 matrix with values 0 (heads) and 1 (tails). Here are some examples:
0 0 0 1 0 1 1 1 0
0 1 0 0 0 1 1 0 0
0 0 0 1 0 0 0 0 1
Each state can also be represented using a binary number. For example, the preceding matrices correspond to the numbers:
000010000 101001100 110100001
There are a total of 512 possibilities, so you can use decimal numbers 0, 1, 2, 3,...,511 to represent all the states of the matrix.
Write a program that prompts the user to enter a number between 0 and 511 and displays the corresponding matrix with the characters H and T.
I want the method toBinary() to fill the array binaryNumbers. I realized that this does not fill in 0s to the left. I have to think that through but is that the only thing that is the problem?
//https://www.geeksforgeeks.org/java-program-for-decimal-to-binary-conversion/
import java.util.Scanner;
public class HeadsAndTails {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int num = input.nextInt();
int[] binaryNumbers = toBinary(num);
for (int i = 0; i < 9; i++) {
printArr(binaryNumbers);
System.out.print(binaryNumbers[1]);
}
}
public static int[] toBinary(int inputtedNumber) {
int[] binaryNum = new int[9];
int i = 0;
while (inputtedNumber > 0) {
binaryNum[i] = inputtedNumber % 2;
inputtedNumber = inputtedNumber/2;
inputtedNumber++;
} return binaryNum;
}
public static void printArr(int[] arr) {
for (int i = 0; i < 9; i++) {
if (arr[i] == 0) {
System.out.print("H ");
} else {
System.out.print("T ");
}
if (arr[i+1] % 3 == 0) {
System.out.println();
} System.out.print(arr[i]);
}
}
}
Looks like you are incrementing the wrong variable in your while loop:
while (inputtedNumber > 0) {
binaryNum[i] = inputtedNumber % 2;
inputtedNumber = inputtedNumber/2;
i++; // NOT inputtedNumber
} return binaryNum;
Also note, a new int[9] is probably already initialized to 0, but if not, you could just loop 9 times, rather than until the inputtedNumber is 0:
for (int i = 0; i < 9; i++) {
binaryNum[i] = inputtedNumber % 2;
inputtedNumber = inputtedNumber/2;
}
return binaryNum;
Finally, I think your array might be backwards when you're done, so you may need to reverse it or output it in reverse order
I realize this is a homework assignment so you should stick with your current approach. However, sometimes it can be fun to see what can be achieved using the built in features of Java.
The Integer class has a method toBinaryString that's a good starting point:
int n = 23;
String s1 = Integer.toBinaryString(n);
System.out.println(s1);
Output: 10111
But as we can see, this omits leading 0s. We can get these back by making sure our number has a significant digit in the 10th place, using a little bit-twiddling:
String s2 = Integer.toBinaryString(1<<9 | n);
System.out.println(s2);
Output: 1000010111
But now we have a leading 1 that we don't want. We'll strip this off using String.substring, and while we're at it we'll use String.replace to replace 0 with H and 1 with T:
String s3 = Integer.toBinaryString(1<<9 | n).substring(1).replace('0','H').replace('1','T');
System.out.println(s3);
Output: HHHHTHTTT
Now we can print this string in matrix form, again using substring to extract each line and replaceAll to insert the desired spaces:
for(int i=0; i<9; i+=3)
System.out.println(s3.substring(i, i+3).replaceAll("", " ").trim());
Output:
H H H
H T H
T T T
If we're up for a bit of regex wizardry (found here and here) we can do even better:
for(String sl : s3.split("(?<=\\G.{3})"))
System.out.println(sl.replaceAll(".(?=.)", "$0 "));
Putting it all together we get:
int n = 23;
String s3 = Integer.toBinaryString(1<<9 | n).substring(1).replace('0','H').replace('1','T');
for(String s : s3.split("(?<=\\G.{3})"))
System.out.println(s.replaceAll(".(?=.)", "$0 "));
I have a program that reads a list of integers, and then display the number of even numbers and odd numbers. We assume that the input ends with 0. Here is the sample run of the program.
Input numbers: 1 2 3 4 5 6 7 8 9 0
Odd: 5 Even: 4
However, my result is
Odd: 5 and Even: 5.
The problem is that 0 is counted as an even number. This is my code
public class Q75 {
public static void main(String[] args){
java.util.Scanner input = new java.util.Scanner (System.in);
double [] numbers = new double[10];
System.out.print("Enter numbers: ");
for(int i = 0;i<numbers.length;i++){
numbers[i] = input.nextDouble();
}
int Evens = 0;
int Odd = 0;
for(int i = 0;i<numbers.length;i++){
if(numbers[i]%2 == 0){
Evens++;
}else{
Odd++;
}
}
System.out.println("The number of odd numbers: " + Odd);
System.out.println("The number of even numbers: " + Evens);
}
}
There are two options
A) Adding another branch in your if statements i.e.
if(number[i] > 0) {
if(number[i] % 2 >0)
Odd++;
else
Evens++;
}
NB: changing the else branch to else if(number[i] >0), you can do without the outer if condition.
B) Since the list of number ends with 0 you can put this as a condition in your for loop i.e.
for(int i =0; i < numbers.length && numbers[i] > 0 ; i++)
Also as a rule of thumb variable names in java start with a small letter
Just don't check the last element: Use i < numbers.length - 1
for(int i = 0;i < numbers.length - 1; i++) {
//
}
I am stuck in my project where we have to show a string of line in the number line scale so later we can delete or add characters to that string. I am not sure how to print out the scale in 5s based on the length of the string.
Ex:
0 5 10 15 20
|----+----|----+----|-
This is the first line
Then, the user will choose the characters they want to delete from the string using from position and to position. It will show what position the user chose from the string and delete.
Ex:
from position: 12
to position: 18
0 5 10 15 20
|----+----|----+----|-
This is the first line
^^^^^^^ --> // this will be deleted
y/n: y
0 5 10 15
|----+----|----+
This is the ine
I was able to delete the characters but I do not know how to show the number line based on a string. Here is my code so far:
public void showNumberLine(String line)
{
int lineCount = line.length(); // getting the length of the string being passed in
String numberLine = "";
for(int i = 0; i <= lineCount; i++) //
{
numberLine = "" + i;
System.out.println("|----+----|----+----|-");
}
}
public void deleteSubString()
{
Scanner keyboard = new Scanner(System.in);
showNumberLine(textOfLine); // this will print out then number line and the line
System.out.print("from position: ");
int fromIndex = keyboard.nextInt();
System.out.print("to position: ");
int toIndex = keyboard.nextInt();
if(fromIndex < 0 || fromIndex > numOfChar || toIndex < 0 || toIndex > numOfChar)
{
System.out.println("Cannot delete at the given index: Index Out of Bounds");
}
/*
* Create a new number line where it shows what is going to be deleted
*/
String newLineOfString = textOfLine.substring(fromIndex, toIndex);
textOfLine = textOfLine.replace(newLineOfString, "");
System.out.println(newLineOfString);
}
I would recommend you to implement a method printScale or something like that which takes a String or an int as argument and prints these two lines for you.
You sad you already can remove the characters so if you have a String with the value "This is the ine" as you showed in your example you could call the method like this:
printScale(myNewString.length());
This method could look something like this (not perfect but works):
public void printLine(int amountOfCharacters) {
StringBuilder lineNumber = new StringBuilder();
StringBuilder lineScaleSymbols = new StringBuilder();
for (int i = 0; i < amountOfCharacters; i++) {
if (i % 10 == 0) {
if (i < 10) {
lineNumber.append(i);
} else {
lineNumber.insert(i -1, i);
}
lineScaleSymbols.append('|');
} else if (i % 5 == 0) {
if (i < 10) {
lineNumber.append(i);
} else {
lineNumber.insert(i -1, i);
}
lineScaleSymbols.append('+');
} else {
lineNumber.append(' ');
lineScaleSymbols.append('-');
}
}
System.out.println(lineNumber.toString());
System.out.println(lineScaleSymbols.toString());
}
Hope this helps.
You're on the right track with your showNumberLine method.
Let's outline exactly what you need to do:
determine the length of the string
generate a number line of the same length as the string
every character ending with 0 will be the special character |
every character ending in 5 will be the special character +
every other character will be -
You could make your loop like this, using the modulus operator to determine which character to write:
for(int i = 0; i < line.length(); i++) {
if(i % 10 == 0) {
// the number is divisible by 10 (ends in zero)
System.out.print("|");
} else if(i % 5 == 0 && i % 10 != 0) {
// the number is divisible by 5 and not divisible by 10 (ends in 5)
System.out.print("+");
} else {
System.out.print("-");
}
System.out.println();
}
Output:
|----+----|----+----|----+----|----+----|---
The quick brown fox jumped over the lazy dog
You'll need some more code to write out the digits (0, 5, 10, 15) above the number line, I'll leave that to you. It will be similar logic but there are subtle issues to consider as the length of the numbers is 1 character, then 2 characters, then 3 characters as they increase (0, 5, 10, 15, ... 100, 105). At some point you'll have to stop as the numbers won't fit in the space.
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.