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How to add zeros to the end of value of Long object [duplicate]

for(i=0; i<array.length; i++){
sum = 4 * 5;
}
What I'm trying to do is add ((array.length - 1) - i) 0's to the value of sum. For this example assume array length is 3. sum equals 20. So for the first iteration of the loop i want to add ((3 - 1) - 0) 0's to the value of sum, so sum would be 2000. The next iteration would be ((3 - 1) - 1) 0's. so sum would equal 200 and so on. I hope what I am trying to achieve is clear.
So my questions are:
Is it possible to just shift an int to add extra digits? My search thus far suggests it is not.
If not, how can i achieve my desired goal?
Thankyou for reading my question and any help would be greatly apreciated.

You can just multiply it by 10 however many times.
200 * 10 = 2000
etc
So in your case, you'd have to use a for loop until the end of the array and multiply sum every iteration. Be careful though, because the max value of an int is 2^31, so it of surpasses that, it will roll back to 0

You can add n zeroes to the end of a number, sum by multiplying sum by 10 * n.
int sum = 20;
for (int i = 0; i < ary.length; ++i) {
int zeroesToAdd = ary.length - 1 - i
sum *= (zeroesToAdd > 0) ? zeroesToAdd * 10 : 1
}
System.out.println("Sum after loop: " + sum);

for(int i=array.length; i>0; i--){
sum = 20;
for(int j=0; j < (i - 1); j++)
{
sum *= 10;
}
}
Use inner loop to multiply by 10 the number of times i is for that iteration. You would need to reset sum in your outer loop each time.

You will want to check your for-loop condition: i>array.length. Since i starts at 0, this loop will not run unless the array's length is also 0 (an empty array). The correct condition is i < array.length.
This "shift" you want can be achieved by creating a temporary variable inside the loop that is equal to the sum times 10i. In Java's Math library, there is a pow(a,b) function that computes ab. With that in mind, what you want is something like this:
int oldSum = 4 * 5;
for (int i = 0; i < array.length; i++) {
int newSum = oldSum * Math.pow(10,i);
}

Multiply by 10 instead, and use < (not >) like
int sum = 20;
int[] array = { 1, 2, 3 };
for (int i = 0; i < array.length; i++) {
int powSum = sum;
for (int j = array.length - 1; j > i; j--) {
powSum *= 10;
}
System.out.println(powSum);
}
Output is (as requested)
2000
200
20

For array of length = n; you will end up adding (n - 1) + (n - 2) + ... + 2 + 1 + 0 zeros for i = 0, 1, ... n-2, n-1 respectively.
Therefore, number of zeros to append (z) = n * (n-1) / 2
So the answer is sum * (10 ^ z)
[EDIT]
The above can be used to find the answer after N iteration. (I miss read the question)
int n = array.length;
long sum = 20;
long pow = Math.pow(10, n-1); // for i = 0
for (int i = 0; i < n; i++) {
System.out.println(sum*pow);
pow /= 10;
}

Why is my arr.length not consistently printing out values and instead always printing out "0"?

From my research, my first error was populating the array in the switch case. I fixed it so it is populated outside. I did a few tests and arr.length will give me the output of 1000 as expected, but it should be 500 (if I put 500 as the upperbound, how would I make values between 1-1000?). Case 5 works for some reason which uses arr.length as well.
I want to have 500 integers between the value of 1-1000 print out various outcomes. There are no errors in the code. All cases except case 2 & 7 work.
import java.util.Scanner;
import java.util.Random;
class ArrayMenu {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Pick an option 1-7, 0 to exit");
byte menu;
Random num = new Random();
int[] arr = new int[1000];
for (int i = 0; i <= 500; i++) {
arr[i] = num.nextInt(500);
}
do {
menu = scan.nextByte();
switch (menu) {
case 1:
for (int i = 0; i <= 500; i++) {
System.out.println(arr[i]);
}
break;
case 2:
int mean = 0;
for (int i = 0; i <= 500; i++) {
int sum = 0;
int z = arr[i];
sum = sum + z;
mean = sum / arr.length;
}
System.out.println("The mean is " + mean);
break;
case 3:
for (int i = 0; i <= 500; i++) {
int y = arr[i];
if (y % 2 != 0) System.out.println(arr[i]);
}
break;
case 4:
for (int i = 0; i <= 500; i++) {
int x = arr[i];
if (x % 2 == 0) System.out.println(arr[i]);
}
break;
case 5:
System.out.println("Median is " + arr[arr.length / 2]);
break;
case 6:
System.out.println("First is " + arr[0]);
break;
case 7:
System.out.println("Last is " + arr[arr.length - 1]);
System.out.println(arr.length);
break;
}
} while (menu != 0);
scan.close();
}
}
Bonus question: how would I randomly fill with negative values instead of only positive?

This is your code right now:
case 2:
int mean = 0;
for (int i = 0; i <= 500; i++) {
int sum = 0;
int z = arr[i];
sum = sum + z;
mean = sum / arr.length;
}
System.out.println("The mean is " + mean);
break;
However, there are a couple of mistakes here. First of all, the mean could be a decimal, so you should declare it as a double, not an int. Also, there's no need to assign it in the loop; you can just assign it once after the loop. You're also redeclaring sum every loop, which means the sum never gets incremented. Here's the fixed code:
case 2:
double mean;
int sum = 0;
for (int i = 0; i <= 500; i++) {
sum = sum + arr[i];
}
mean = (double) sum / arr.length;
System.out.println("The mean is " + mean);
break;
Also, for this line:
int[] arr = new int[1000];
You allocate an array of 1000 elements, but only fill from 0 to 500 inclusive (which is 501 elements). If you want arr.length to work properly, only allocate what you need:
int[] arr = new int[501];
If you don't want to manually allocate, then use ArrayLists instead.
If you want a negative number between -500 and 500, then generate one between 0 and 1000 then subtract 500:
rand.nextInt(1000) - 500

Note that case 5 also doesn't work. The mean is not defined as 'the middle in a random assortment', there is nothing special about the 500th element. It's defined as: "Once you've sorted it all, the middle element". You aren't sorting anything.
Without resorting to calling Arrays.sort (I don't know if your homework assignment allows you to do this), case 5 is in fact the one that's the most complicated to write.
Furthermore, you have an array of 1000 slots (so, array.length is 1000), and you wrote a bunch of code that assumes 1000 is in fact how large your array actually is - such as your 'print the last' code returning the [999] element. However, at the start, you only fill the first 500 with random numbers, the remaining 500 remain unfilled, which means they retain their initial value, which is 0.
Thus, we get to:
All cases except case 2 & 7 work.
No, they don't.
Programming is really, really hard. Before concluding that your stuff works just because it compiled and it ran, check that the outputs you see actually make sense.
for (int i = 0; i <= 500; i++) {
arr[i] = num.nextInt(500);
}```
The first use of '500' is how many times you want to loop. If you want to fill the whole array, use 1000 there. The second use of 500 decides from amongst how many numbers you want to pick random numbers. 500 means that you get any number, from as low as 0 to as large as 499 (which is a total span of 500 numbers).
You've used <= and that's the mistake; use <. An array of size 1000 has valid indices [0] all the way up to and including [999], but not [1000]. Think about it: if 0 is the first element, then 999 is the 1000th element, and [1000] would be the 1001st element, which isn't there. Thus:
for (int i = 0; i < 1000; i++) {
arr[i] = num.nextInt(500);
// or if you want uniform from -500 to 500 all inclusive:
arr[i] = num.nextInt(1001) - 500;
}
why 1001? Because that's how many different numbers you can generate: 500 negative numbers, 500 positive numbers.. and 0. 1001. (That goes some way into showing you why java is 0-based, the math tends to be less weird that way).

Write a program that will calculate the number of trailing zeros in a factorial of a given number. N! = 1 * 2 * 3 * ... * N

This is my code for the Codewars problem (Java) yet I cannot make it work. I'm pretty sure I've made a stupid mistake somewhere because of my lack of experience (coding for 4 months)
public static int zeros(int n) {
int f = 1;
int zerocount = 0;
for(int i = 2 ; i <= n; i++){
f *= i;
}
String factorial = String.valueOf(f);
String split [] = factorial.split("");
for(int i = 0; i < split.length; i++){
String m = split[i];
if(m.equals( "0")){
zerocount ++;
}
else {
zerocount = 0;
}
}
return zerocount;
}
}

In fact, you do not need to calculate the factorial because it will rapidly explode into a huge number that will overflow even a long. What you want to do is count the number of fives and twos by which each number between 2 and n can be divided.
static int powersoffive(int n) {
int p=0;
while (n % 5 == 0) {
p++;
n /= 5;
}
return p;
}
static int countzeros(int n) {
int fives = 0;
for (int i = 1; i <= n; i++)
fives += powersoffive(i);
return fives;
}
Note: Lajos Arpad's solution is superior.

As pointed out by other users your solution will probably not be accepted because of the exploding factorial you are calculating.
About the code you wrote there are two mistakes you have made:
You are calculating the factorial in the wrong way. You should start with i = 2 in the loop
for(int i = 2; i <= n; i++){
f *= i;
}
Also in Java you cannot compare strings using ==. This is not valid
if(m == "0")
You should compare them like this
if(m.equals("0"))
Anyway this is how I would have resolved the problem
public static int zeros(int n) {
int zerocount = 0;
for (int i = 5; n / i > 0; i *= 5) {
zerocount += n / i;
}
return zerocount;
}

A zero in a base-10 representation of a number is a 2*5. In order to determine the number of trailing zeroes you will need to determine how many times can you divide your number with ten, or, in other words, the minimum of the sum of 2 and 5 factors. Due to the fact that 5 is bigger than 2 and we go sequentially, the number of fives will be the number of trailing zeroes.
A naive approach would be to round down n/5, but that will only give you the number of items divisible with 5. However, for example, 25 is divisible by 5 twice. The same can be said about 50. 125 can be divided by 5 three times, no less.
So, the algorithm would look like this:
int items = 0;
int power = 5;
while (power < n) {
items += (int) (n / power);
power *= 5;
}
Here small numbers are in use in relative terms, but it's only a proof of concept.

You do need to use brute force here and you integers will overflow anyway.
With multiplication trailing zero appears only as the result of 2*5.
Now imagine the factorial represented by a product of it's prime factors.
Notice that for every 5 (five) we will always have 2 (two).
So to calculate the number of zeroes we need to calculate the number of fives.
That can be implemented by continuously dividing N by five and totaling results
In Java code that will be something like this:
static int calculate(int n)
{
int result = 0;
while (n > 0 ) {
n /= 5;
result += n;
}
return result;
}

Add 0's to the end of an integer

for(i=0; i<array.length; i++){
sum = 4 * 5;
}
What I'm trying to do is add ((array.length - 1) - i) 0's to the value of sum. For this example assume array length is 3. sum equals 20. So for the first iteration of the loop i want to add ((3 - 1) - 0) 0's to the value of sum, so sum would be 2000. The next iteration would be ((3 - 1) - 1) 0's. so sum would equal 200 and so on. I hope what I am trying to achieve is clear.
So my questions are:
Is it possible to just shift an int to add extra digits? My search thus far suggests it is not.
If not, how can i achieve my desired goal?
Thankyou for reading my question and any help would be greatly apreciated.

You can just multiply it by 10 however many times.
200 * 10 = 2000
etc
So in your case, you'd have to use a for loop until the end of the array and multiply sum every iteration. Be careful though, because the max value of an int is 2^31, so it of surpasses that, it will roll back to 0

You can add n zeroes to the end of a number, sum by multiplying sum by 10 * n.
int sum = 20;
for (int i = 0; i < ary.length; ++i) {
int zeroesToAdd = ary.length - 1 - i
sum *= (zeroesToAdd > 0) ? zeroesToAdd * 10 : 1
}
System.out.println("Sum after loop: " + sum);

for(int i=array.length; i>0; i--){
sum = 20;
for(int j=0; j < (i - 1); j++)
{
sum *= 10;
}
}
Use inner loop to multiply by 10 the number of times i is for that iteration. You would need to reset sum in your outer loop each time.

You will want to check your for-loop condition: i>array.length. Since i starts at 0, this loop will not run unless the array's length is also 0 (an empty array). The correct condition is i < array.length.
This "shift" you want can be achieved by creating a temporary variable inside the loop that is equal to the sum times 10i. In Java's Math library, there is a pow(a,b) function that computes ab. With that in mind, what you want is something like this:
int oldSum = 4 * 5;
for (int i = 0; i < array.length; i++) {
int newSum = oldSum * Math.pow(10,i);
}

Multiply by 10 instead, and use < (not >) like
int sum = 20;
int[] array = { 1, 2, 3 };
for (int i = 0; i < array.length; i++) {
int powSum = sum;
for (int j = array.length - 1; j > i; j--) {
powSum *= 10;
}
System.out.println(powSum);
}
Output is (as requested)
2000
200
20

For array of length = n; you will end up adding (n - 1) + (n - 2) + ... + 2 + 1 + 0 zeros for i = 0, 1, ... n-2, n-1 respectively.
Therefore, number of zeros to append (z) = n * (n-1) / 2
So the answer is sum * (10 ^ z)
[EDIT]
The above can be used to find the answer after N iteration. (I miss read the question)
int n = array.length;
long sum = 20;
long pow = Math.pow(10, n-1); // for i = 0
for (int i = 0; i < n; i++) {
System.out.println(sum*pow);
pow /= 10;
}

Simple program using loops (Beginning Java)

I have to write a program using loops that calculates the sum of all odd numbers between a and b (inclusive), where a and b are inputs.
I made this (below) and it works fine, but I noticed one problem with it: when i enter a larger number followed by a smaller number for the inputs, it returns 0, but when i enter the smaller number first it works perfectly. Any quick fixes for this? :)
import java.util.Scanner;
public class ComputeSumAAndB
{
public static void main (String[] args)
{
Scanner in = new Scanner(System.in);
System.out.print("Please enter 2 integers: "); //prompts user for ints
int a = in.nextInt();
int b = in.nextInt();
int sum = 0;
for (int j = a; j <= b; j++)
{
if (j % 2 == 1)
sum += j;
}
System.out.println("The sum of all odd numbers (inclusive) between " + a + " and "+ b + " is " + sum);
}
}

int temp;
if(a > b) {
temp = a;
a = b;
b = temp;
}
Put this right before your for loop starts.
The if checks whether a (the first number entered) is larger than b. If it is, it swaps a and b. Now your for loop will always start with the smallest number and iterate up to the larger number (because after this if, a will always be the smaller number).
Using this method has the added side effect of making your output make sense. Your output will now always say: "between [smaller number] and [larger number]".
rolfl's answer is more elegant and works perfectly fine, but when the user enters the larger number first, your output may look kind of weird: "between [larger number] and [smaller number]", etc.

You can get the smaller and larger inputs by using the Math.min() and Math.max functions....
for (int j = Math.min(a,b); j <= Math.max(a,b); j++) {
if (j % 2 == 1) {
sum += j;
}
}

It's not working because A is larger than B in the for loop, you have it iterate while A is less than or equal to B.
You could do what nhgrif says but it's changing your data.. But he is correct.

That's because you are first expecting for the a input (inferior limit) and then the b (superior). When your program reaches the for, j = a so the condition a <= b is False, if the first input is larger. In other words it never enters the for loop.

Actually you should do the following 2 things:
1 It is just just like rolfl mentioned above. You need to put the min and max in the right place in loop.
for (int j = Math.min(a,b); j <= Math.max(a,b); j++)
{
if (j % 2 == 1) {
sum += j;
}
}
2 Use if (j % 2 == 1) it is not enough to check whether the num is odd.
e.g.
a = -5, b =0;
What will be the result?
int sum = 0;
for(int j=-5;j<0;j++)
{
if(j%2 == 1)
{
sum+=j;
}
}
The value for sum will be 0.
We need to change the condition to if(!(j%2 == 0)) Then you will get the expected result.

That's because you are first expecting for the a input (inferior limit) and then the b (superior). When your program reaches the for, j = a so the condition a <= b is False, if the first input is larger. In other words it never enters the for loop.