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);
}
}
}
}
Related
You have to print a pattern using recursion. Given a input
N
the pattern looks like this
N
,
a
i
,
a
i
+
1
,
a
i
+
2
,.....,
N
. Where if
a
i
>
0
then
a
i
+
1
=
a
i
−
5
else
a
i
+
1
=
a
i
+
5
. It will be a decreasing sequence from
N
till
a
i
<=
0
and then an increasing sequence till
N
. (See sample test cases for better explanation)
Input format
First line contains an integer
T
denoting number of test cases.
For each of the next
T
lines, each line contains an integer
N
.
Output format
For each test case on a new line, print the required pattern.
Constraints
1
<=
T
<=
6
0
<=
N
<=
2000
Example
Input
2
16
10
Output
16 11 6 1 -4 1 6 11 16
10 5 0 5 10
Sample test case explanation
For the first test case
N=16, it will be a decreasing sequence till the printing number becomes <=0.
16 11 6 1 −4
After this point it will be a increasing sequence till the printing number becomes N
1 6 11 16
So the pattern is 16 11 6 1 −4 1 6 11 16.
My code is below but i got the output as 16,11,6,1,-4 only. Help me to correct this code
import java.util.Scanner;
public class Day3
{
public static void series(int n,boolean b)
{
int temp = n;
boolean flag=b;
System.out.println(temp+" ");
if(flag==true)
temp-=5;
else if(flag==false)
temp+=5;
if(temp<=0)
flag=false;
if(temp<=n)
series(temp,flag);
}
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t>0)
{
int n = sc.nextInt();
series(n,true);
t-=1;
}
}
}
Because 'n' changes in every cases. You must create another variable and keep first 'n' in that, for control if temp smaller than 'n'.
For example
import java.util.Scanner;
public class Day3
{
int firstN = 0; //added that line
public static void series(int n,boolean b)
{
int temp = n;
boolean flag=b;
System.out.println(temp+" ");
if(flag==true)
temp-=5;
else if(flag==false)
temp+=5;
if(temp<=0)
flag=false;
if(temp<=firstN) //changed that line
series(temp,flag);
}
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t>0)
{
int n = sc.nextInt();
firstN = n; //added that line
series(n,true);
t-=1;
}
}
}
Also a little tip;
you can use (flag) for (flag==true)
and (!flag) for (flag==false)
Just FYI : Although this can be solved using recursion, it can be solved more efficiently without using recursion. It is as simple as this:
private static void printSeries(int N) {
int T = N;
while ( T >= 0 ) {
System.out.print(T + " ");
T = T - 5;
}
while ( T <= N ) {
System.out.print(T + " ");
T = T + 5;
}
}
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));
}
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++) {
//
}
Write a program that prompts the user to input an integer and then outputs both the individual digits of the number and the sum of the digits. For example, the program should: output the individual digits of 3456 as 3 4 5 6 and the sum as 18, output the individual digits of 8030 as 8 0 3 0 and the sum as 11, output the individual digits of 2345526 as 2 3 4 5 5 2 6 and the sum as 27, and output the individual digits of 4000 as 4 0 0 0 and the sum as 4.
Moreover, the computer always adds the digits in the positive direction even if the user enters a negative number. For example, output the individual digits of -2345 as 2 3 4 5 and the sum as 14.
This is the question I'm having minor difficulties with, the only part I can't figure out is how can I print the single integers in the order that he wants, from what I learned so far I can only print them in reverse. Here's my code:
import java.util.*;
public class assignment2Q1ForLoop {
static Scanner console = new Scanner (System.in);
public static void main(String[] args) {
int usernum, remainder;
int counter, sum=0, N;
//Asaking the user to enter a limit so we can use a counter controlled loop
System.out.println("Please enter the number of digits of the integer");
N = console.nextInt();
System.out.println("Please enter your "+N+" digit number");
usernum = console.nextInt();
System.out.println("The individual numbers are:");
for(counter=0; counter < N; counter++) {
if(usernum<0)
usernum=-usernum;
remainder = usernum%10 ;
System.out.print(remainder+" ");
sum = sum+remainder ;
usernum = usernum/10;
}
System.out.println();
System.out.println("the sum of the individual digits is:"+sum);
}
}
You have to storeremainder variables in an array and then print them in the loop from last index to first as shown in this tutorial.
You can either store digits in array and then print them, or you can try something like that:
final Scanner console = new Scanner(System.in);
System.out.println("Please enter your number");
final int un = console.nextInt();
long n = un > 0 ? un : -un;
long d = 1;
while (n > d) d *= 10;
long s = 0;
System.out.println("The individual numbers are:");
while (d > 1) {
d /= 10;
final long t = n / d;
s += t;
System.out.print(t + " ");
n %= d;
}
System.out.println();
System.out.println("the sum of the individual digits is:" + s);
An idea would be : convert int to string and write a method
getChar(int index) : String
which gives you for example 4 from 3456 with
getChar(2);
See Java - Convert integer to string
Here, I wrote a code for your problem with using a stack. If you want a simple code, you can comment my solution and I will wrote another one.
Scanner c1 = new Scanner(System.in);
System.out.print("Enter the number: ");
int numb = c1.nextInt();
numb = Math.abs(numb);
Stack<Integer> digits = new Stack<Integer>();
while(numb>0){
int n = numb%10;
digits.push(n);
numb = numb/10;
}
int sum = 0;
while(!digits.isEmpty()){
int n = digits.pop();
sum+=n;
System.out.print(n+" ");
}
System.out.print(sum);