I'm trying to print a number series that divides by 2 which would result in an output of 128, 64, 32, 16, 8, 4, 2, 1
int num = 128;
for (int i = 1; i <= 8; i++)
{
int prevNum = 0;
prevNum = num / 2;
System.out.print(prevNum);
System.out.print(", ");
}
Clearly my current code isn't working as it is only outputting the number 64 eight times. I'm not sure how I can throw away the initial number after printing it and only use the stored number after the division
From the code that you have posted, I feel that you have initialised num as 128 somewhere in your code.
Now in your loop you are storing prevNum as num/2 in other words 128/2 each time in the loop. So you are getting 64 as output 8 times as your loop loops 8 times.
The correction I would suggest in this code is you do not require the second variable prevNum. Simply use num = num/2 and print the value of num variable. This will solve your problem.
A better code for this loop would be :
int num = 128;
while (num > 1) // This takes care of the loop.
//Also if you write num = 256, then also this loop would work.
{
num = num / 2;
System.out.print(num);
System.out.print(", ");
}
int num = 128;
int prevNum = num;
for (int i = 1; i <= 8; i++)
{
prevNum = prevNum / 2;
System.out.print(prevNum);
System.out.print(", ");
}
This is in logic to the way you are working with. But, there are more efficient ways to get the same result.
OP, you initialized prevNum=0 inside the for-loop and doing so, for each iteration, it was getting assigned to the value 0. What you actually wanted was to carry over the quotient of the previous division to the next iteration. So in order to do that, the prevNum was intilaized outside the loop with value set to num(i.e; 128). And in each iteration, prevNum is getting divided by 2, and the result is assigned back to prevNum.
Try to replace prevNum to num, this will change the value of num, but it works.
Or declare prevNum outside the loop this way: int preNum = num;, and replace line 4 to prevNum = prevNum / 2;
Without local variables outside of the loop:
for (int i = 128; i > 0; i = i / 2) System.out.print(i + (i > 1 ? ", " : ""));
We loop over i starting with the initial value 128 and dividing by 2 each iteration (i = i / 2).
Then the current value of i is printed and appended with , if i > 1 is satisfied.
Output:
128, 64, 32, 16, 8, 4, 2, 1
int num = 128;
int prevNum = num; // modify here.
for (int i = 1; i <= 8; i++)
{
prevNum = prevNum / 2;
System.out.print(prevNum);
System.out.print(", ");
}
Edit:
You are not reducing the actual number. On each iteration you are dividing same number again. But you should divide this prevNum by 2 on each iteration.
Hope this help.
Related
I'm trying to make a program that generates random two-digit integers until I get a 10 or a 20. To then find the mount of numbers, the sum of the numbers less than 10, the sum of the numbers equal to 15, the sum of the numbers greater than 70. Can someone help me, please?
This is my code:
// Variables
int numRandom = 0, less10, equal15, more70;
// Process
txtS.setText("Random numbers: " + "\n");
for (int mountNum = 0; numRandom == 40 || numRandom == 20; mountNum++) {
numRandom = (int) (99 * Math.random());
txtS.append(numRandom + "\n");
}
You could just store the values directly and create variables for each case.
This is an example for you less than 10 case. (The 'if statement' would be contained inside your for loop).
int sumLessThanTen = 0;
int countLessThanTen = 0;
...
if(numRandom < 10){
sumLessThanTen += numRandom;
countLessThanTen++
}
I'd like to know how to remove numbers from an integer. For example, if I have the number 23875326, I want to remove the odd numbers, and get the result of 2826.
I've been trying to break each number to check if it's even or odd using a while loop, but I don't know how to merge the numbers into one integer. One important thing is, I'd like to do it without using strings, as it doesn't teach me anything new that way.
I actually think that dealing with a string of numbers is preferable, not only from a code readability point of view, but also possibly from a performance view. That being said, if we absolutely cannot use strings here, then it is still possible to work directly with the integer input.
We can examine each tens digit of the input number using num % 10. Should that digit be even, we can add it to the output number, otherwise do not add it. Note that at each iteration we need to scale the digit by 10 to the correct exponent.
Here is a working code snippet:
int length = (int) (Math.log10(num) + 1); // finds the number of digits
int output = 0;
int counter = 0;
for (int i=0; i < length; ++i) {
if ((num % 10) % 2 == 0) {
output += (num % 10) * Math.pow(10, counter);
++counter;
}
num = num / 10;
}
System.out.println(output);
2826
Demo
I used #Tim Biegeleisen code and changed it a bit, removed some code and changed the for loop to while loop:
int output = 0;
int counter = 0;
System.out.println("Enter a number");
int num = s.nextInt();
while (num != 0) {
if ((num % 10) % 2 == 0) {
output += (num % 10) * Math.pow(10, counter);
++counter;
}
num = num / 10;
}
System.out.println(output);`
I want this while loop to print every multiple of two below the number submitted(ex. if 100 was submitted it would print 2 4 8 16 32 64). Here's what I have(I'm only going to include a portion of the class because there was other things in it that don't pertain to this part)
i = 1;
Scanner myScanner = new Scanner(System.in);
System.out.print("Would thoughst be inclined to enter a number fair sir/madam: ");
String answer = myScanner.nextLine();
int number = Integer.parseInt(answer);
System.out.print("Your number set is: ");
while(i <= number)
{
i = 2*i;
System.out.print(" " + i + " ");
}
What this prints if I enter 100 is: 2 4 8 16 32 64 128
How do I get rid of that last number?
You would get rid of that number by modifying your logic to match. Your code is doing precisely what it says. One option is to start at 2, and increase i at the end of the loop instead of just before printing it. You could also use a for loop:
for (int i = 2; i < 100; i *= 2)
...
If you want to save the last power, you have a few options, e.g.:
int k = 2;
for (int i = k; i < 100; i *= 2) {
k = i;
...
}
Or undo the last operation:
int i;
for (i = 2; i < 100; i *= 2)
...;
i /= 2;
Or check the next one:
int i;
for (i = 2; i * 2 < 100; i *= 2)
...;
Checking the next one, in your original form:
while (i * 2 <= number)
...;
Etc.
By the way, your title says "factors", your description says "multiples", and your code says "powers"...
In your code
while(i <= number)
{
i = 2*i;
System.out.print(" " + i + " ");
}
the problem is that i, when it is equal to 64, is indeed less than 100, so the loop continues.
If you change it to
i = 2*i;
while(i <= number)
{
System.out.print(" " + i + " ");
i = 2*i;
}
it does as you wish, because it pre-computes the value before being analyzed as the while-loop terminator.
Try
while( i <= number / 2)
Those are powers of 2, not factors of 2.
"thoughst" is not a word. It should be "thou".
Update the value of i after you print it.
I am writing a program to print out a user inputed integer into binary form.
When I run it and input, say the number 5, it crashes and gives me the error:
java.lang.ArrayIndexOutOfBoundsException: 30
at PrintBinaryDigitsFixed.main(PrintBinaryDigitsFixed.java:27)
i.e the line "digits[counter] = number % 2;"
Why am I getting an out of bounds exception? It should assign the remainder to the first element then move on to the second shouldn't it?
I feel like I'm making a glaringly obvious mistake but I can't tell what it is
final int MIN = 0;
final int MAX = (int) (Math.pow(2, 30) - 1);
int[] digits = new int[30]; //array to hold the digits
int number = readInput
("Enter an integer from " + MIN + " to " + MAX, MIN, MAX);
int counter = 0;
int modNumber = 2;
while(modNumber / 2 != 0)
{
digits[counter] = number % 2;
modNumber = number / 2;
counter++;
}
System.out.print(number + " in binary form is ");
listBackwardsFrom(digits, counter);
Thanks
You never change number in your loop, and you assign modNumber = number / 2 in the loop, so from the second iteration onward modNumber is a constant (for most of the first iteration it's 2, but then you assign number / 2 to it); if you reach that point at all, you'll stay there. So the loop continues until counter reaches 30, at which point digits[counter] throws the exception.
So I need to output a sum of factorials like 1!+2!...+n!=sum I found a way to get a single factorial but I don't know how to sum them together. This is my attempt at doing so:
System.out.println("Ievadiet ciparu");
Scanner in = new Scanner(System.in);
n = in.nextInt();
if ( n < 0 )
System.out.println("Ciparam jabut pozitivam.");
else
{
while (x>2){
for ( c = 1 ; c <= n ; c++ )
fact = fact*c;
sum=sum+fact;
n=n-1;
if (n==0) break;
}
System.out.println("Faktorialu summa "+sum);
Rather than have a loop 1-n and calculate each factorial elsewhere, I would accumulate the sum as you calculate the factorials - ie have two local variables; one for factorial and one for the sum:
long factorial = 1, sum = 0;
for (int i = 1; i <= n; i++) {
factorial *= i;
sum += factorial;
}
When tested with n = 5, sum is 153, which is correct: 1 + 2 + 6 + 24 + 120
Your problem was that the sum was outside the loop - you just needed braces like here.
Also, your while loop condition x < 2 will never change, so either the loop will never execute (if x > 1) or the loop will never terminate, because x is not changed within the loop.
hmmm my search for finding a recursive(via recursive method calling) version of these code still getting nowhere
`public static long factorialSum(long n){
long x = n;
for(int i = 1; i < n; i++){
x = (n-i)*(1+x);
}
return x;
}`
if you just look at the problem more closely you'll see you can do it in linear time, the trick is in (n-1)! + n! = (n-1)!*(1 + n), to understand this more deeply i recommend add (n-2)! just to see how it grows.