Develop Reference

Fibonacci sequence - How to calculate the sum of the first 100 even-values Fibonacci numbers?

Fibonacci sequence is defined as a sequence of integers starting with 1 and 1, where each subsequent value is the sum of the preceding two I.e.
f(0) = 1
f(1) = 1
f(n) = f(n-1) + f(n-2) where n>=2
My goal is to calculate the sum of the first 100 even-values Fibonacci numbers.
So far I've found this code which works perfectly to calculate the sum of even numbers to 4million , however I'm unable to find edit the code so that it stops at the sum of the 100th value, rather than reaching 4million.
public class Improvement {
public static int Fibonacci(int j) {
/**
*
* Recursive took a long time so continued with iterative
*
* Complexity is n squared.. try to improve to just n
*
*/
int tmp;
int a = 2;
int b = 1;
int total = 0;
do {
if(isEven(a)) total +=a;
tmp = a + b;
b = a;
a = tmp;
} while (a < j);
return total;
}
private static boolean isEven(int a) {
return (a & 1) == 0;
}
public static void main(String[] args) {
// Notice there is no more loop here
System.out.println(Fibonacci(4_000_000));
}
}
Just to show the console from #mr1554 code answer, the first 100 even values are shown and then the sum of all is 4850741640 as can be seen below:
Any help is appreciated, thanks!

You need to use BigInteger because long easily overflows as Fibonacci's scales quite easily. BigInteger is also tricky to check whether is an odd or even number, but you can use BigInteger::testBit returning boolean as explained in this answer.
Here is some complete code:
BigInteger fibonacciSum(int count, boolean isOdd) {
int i = 0;
BigInteger sum = BigInteger.ZERO;
BigInteger current = BigInteger.ONE;
BigInteger next = BigInteger.ONE;
BigInteger temp;
while (i < count) {
temp = current;
current = current.add(next);
next = temp;
if ((current.testBit(0) && isOdd) || ((!current.testBit(0) && !isOdd))) {
sum = sum.add(current);
i++;
}
}
return sum;
}
Or you can have some fun with Stream API:
BigInteger fibonacciSum(int count, boolean isOdd) {
final BigInteger[] firstSecond = new BigInteger[] {BigInteger.ONE, BigInteger.ONE};
return Stream.iterate(
firstSecond,
num -> new BigInteger[] { num[1], num[0].add(num[1]) })
.filter(pair ->
(pair[1].testBit(0) && isOdd) ||
(!pair[1].testBit(0) && !isOdd))
.limit(count)
.map(pair -> pair[1])
.reduce(BigInteger.ZERO, BigInteger::add);
}
In any way, don't forget to test it out:
#Test
void test() {
assertThat(
fibonacciSum(100, false),
is(new BigInteger("290905784918002003245752779317049533129517076702883498623284700")));
}

You said.
My goal is to calculate the sum of the first 100 even-values Fibonacci numbers.
That number gets very large very quickly. You need to:
use BigInteger
use the mod function to determine if even
For this I could have started from (1,1) but it's only one term so ...
BigInteger m = BigInteger.ZERO;
BigInteger n = BigInteger.ONE;
BigInteger sumOfEven= BigInteger.ZERO;
int count = 0;
BigInteger t;
while( count < 100) {
t = n.add(m);
// check if even
if (t.mod(BigInteger.TWO).equals(BigInteger.ZERO)) {
sumOfEven = sumOfEven.add(t);
count++;
}
n = m;
m = t;
}
System.out.println(sumOfEven);
Prints
290905784918002003245752779317049533129517076702883498623284700
If, on the other hand, from your comment.
My aim is to calculate the sum of the first 100 even numbers
Then you can do that like so
sumFirstNeven = (((2N + 2)N)/2 = (N+1)N
so (101)100 = 10100 and the complexity is O(1)

as I figured, you want a program to sum 100 first even values of the Fibonacci series.
here is a sample code, when you run the program it will ask you to determine the number of the even values, you want 100 value e.g, type 100 in consul:
public static void main(String[] args) {
int firstNumber = 0;
int secondNumber = 2;
System.out.print("Enter the number of odd elements of the Fibonacci Series to sum : ");
Scanner scan = new Scanner(System.in);
int elementCount = scan.nextInt(); // number of even values you want
System.out.print(firstNumber + ", ");
System.out.print(secondNumber + ", ");
long sum = 2;
for (int i = 2; i < elementCount; i++) {
int nextNumber = firstNumber + secondNumber;
System.out.print(nextNumber + ", ");
sum += (nextNumber);
firstNumber = secondNumber;
secondNumber = nextNumber;
}
System.out.print("...");
System.out.println("\n" + "the sum of " + elementCount + " values of fibonacci series is: " + sum);
}

Palindrome Java: While and If/Else Statements

My code keeps printing out "true" no matter what I enter. My code does not show any errors, I enter 56 as the input and it still prints "true." I'm having a really hard time figuring out why. Also, I'm new to Java as well as Stack Overflow. Any help would be appreciated.
//Palindrome verification
temp = integer;
while (integer > 0)
{
remainder = integer%10;
reverse = reverse * 10 + remainder;
integer=integer /10;
}
if (integer==reverse)
System.out.println("True");
else
System.out.println("False");

In your program when while loop completely execute integer variable value will become zero and it is the problem.
you need to change if (integer==reverse) to if (temp==reverse).
complete code:
public class palindrom {
public static void main(String[] args) {
int integer = 122; //change this value according to your preference
int temp = integer;
int remainder;
int reverse = 0;
while (integer > 0){
remainder = integer%10;
reverse = reverse * 10 + remainder;
integer=integer /10;
}
if (temp==reverse)
System.out.println("True");
else
System.out.println("False");
}
}

You need compare START value with REVERSE. In your code (integer==reverse) will always be FALSE, because integer==0 after cycle in while.
int integer = 12344321;
int remainder, reverse = 0;
int start = integer;
while (integer > 0)
{
remainder = integer%10;
reverse = reverse * 10 + remainder;
integer = integer / 10;
}
if (start==reverse)
System.out.println("True");
else
System.out.println("False");
System.out.println(reverse);

Is there a function in java that works like SUBSTRING function but for integers?

is there a function in java that works like SUBSTRING function but for integers. Like for example the user's input is 456789, I want to break it into two part and put them into different variable. and divide them. for example,
user's input : 456789
the first 3 numbers will be in variable A.
the last 3 numbers will be in variable B.
pass = A/B;
can someone help me how can I do this,
thanks.

Use integer division and the modulus operator:
int input = 456789;
int a = input / 1000;
int b = input % 1000;

Here is a mathematical based implementation for positive and negative integers (probably can be optimized):
public static void main(String[] args) {
System.out.println(substring(2, 0, 1)); // prints "2"
System.out.println(substring(2523, 2, 2)); // prints "23"
System.out.println(substring(-1000, 0, 2)); // prints "-1"
System.out.println(substring(-1234, 0, 4)); // prints "-123"
System.out.println(substring(-1010, 2, 1)); // prints "0"
System.out.println(substring(-10034, 3, 2)); // prints "3"
System.out.println(substring(-10034, 2, 4)); // prints "34"
}
public static int substring(int input, int startingPoint, int length) {
if (startingPoint < 0 || length < 1 || startingPoint + length > size(input)) {
throw new IndexOutOfBoundsException();
}
if (input < 0 && startingPoint == 0 && length < 2) {
throw new IndexOutOfBoundsException("'-' can not be returned without a digit");
}
input /= (int) Math.pow(10, size(input) - length - startingPoint); // shift from end by division
input = input % (int) Math.pow(10, length); // shift from start by division remainder
if (input < 0 && startingPoint > 0) {
input = Math.abs(input); // update sign for negative input
}
return input;
}
private static int size(int input) {
int size = 1;
while (input / 10 != 0) {
size++;
input /= 10;
}
return input < 0 ? ++size : size; // '-'sign is a part of size
}

Splitting a "number" is not what you want to do. You want to split the NUMERAL, which is the string representing a quantity (number). You can split the numeral just like any other literal (string): don't parse the user's input and use substring. If you want to make sure the literal is an actual numeral, parse it to see if an exception is thrown. If not, you have a numeral. But don't hold on to the integers, keep the strings instead.

Here you go:
You give the method a number, start, and end indexes.
public static void main(String[] args) {
System.out.println(getPartOfInt(93934934, 3, 7));`
}
public static int getPartOfInt(int number, int start, int end){
Integer i = new Integer(number);
char[] chars = i.toString().toCharArray();
String str = "";
for(int j = start; j < end && j < chars.length; j++){
str += chars[j];
}
return Integer.parseInt(str);
}
OR:
public static int getPartOfInt(int number, int start, int end){
String str = new Integer(number).toString();
return Integer.parseInt(str.substring(start, Math.min(end, str.length())));
}

You could also first convert the number to String, like so:
int num = 456789;
String strNum = String.valueOf(num);
String part1 = strNum.substring(0, strNum.length() / 2 - 1);
String part2 = strNum.substring(strNum.length() / 2);
Then, you could convert it back to int, like so:
int part1Num = Integer.parseInt(part1);
int part2Num = Integer.parseInt(part2);
Now you can do all the arithmetic you want with those two int's.

#DeanLeitersdorf Here's the function ( #clcto s solution)
int subinteger(int input, int from, int size)
{
while (input > pow(10, size + from) - 1)
input /= 10;
return input % (int)pow(10, size);
}
This is a c++ solution but i hope that you can convert it to Java easily

Accessing single letters in String / digits in numbers - Java

I have numeric input (11 digits), and I need to perform some operations on each digit (example: multiply 1st by 5, 2nd by 3, etc.). How can I do so in Java? Is there a simple way to access single letter / digit? Is there another way to do it?

If you don't want to convert the number to a string, then there is a simple trick.
digit = number % 10
will give you the right most digit.
number = number / 10
Will remove the right most digit from the number.
So you can run in a loop until the number reaches 0.
while(0 < number)
{
int digit = number % 10;
number = number / 10;
// do here an operation on the digits
}
You can use a for loop to help you count. For example
for(int index = 0; 0 < number; ++index, number /= 10)
{
int digit = number % 10;
// do an operation on the number according to the 'index' variable
}
Here is a similar StackOverFlow question on a similar question

Well there are many ways you can do it like :
int a = 12345;
int b;
while(a>0)
{
b = a%10;
System.out.print(b + " ");
a = a/10;
}
Here it gives you the digits in reverse order like you will get b=5 then b=4....
You can just manipulate them
Other way
int d = 12345;
String str = String.valueOf(d);
for(int i=0;i<str.length();i++)
{
char c = str.charAt(i);
System.out.print(Character.getNumericValue(c) * 10 + " ");
}
Or
char c[] = str.toCharArray();
for(Character ch : c)
{
System.out.print(Character.getNumericValue(ch) * 2 + " ");
}

You can use .charAt() to get a character from a string. Then using Character.getNumericValue() you can convert the character to an integer.
Like this:
String string = "1434347633";
int digit1 = Character.getNumericValue(string.charAt(1));

Convert that number input to String data type so that you can interpret it as a String.
int numbers = 1122334455; // integer won't be able to save this much data,
// just for example ok,
String numberString = numbers.toString();
foreach (char number in numberString) {
// do work on each of the single character in the string.
}
You should work them out, depending on the some condition.

If you want to access the digits by index without converting to a string, you can use these two functions length and digitAt:
public class DigitAt {
public static int length(long v) {
return (int) Math.ceil(Math.log10(v));
}
public static int digitAt(long v, int digit) {
return (int) ((v / (long) Math.pow(10, length(v) - digit - 1)) % 10);
}
public static void main(String[] args) {
System.out.println("Digits: " + length(1234567));
System.out.println(digitAt(1234567, 0));
System.out.println(digitAt(1234567, 1));
System.out.println(digitAt(1234567, 6));
}
}

public String stringAfterOperations(String digits) {
ArrayList<Integer> z = new ArrayList<Integer>();
for(Character c: digits.toCharArray()) {
z.add(Character.getNumericValue(c));
}
//TODO Set your own operations inside this "for"
for(int i=0; i<z.size(); i++){
if(i == 1){
z.set(i, z.get(i)*4);
}
else if(i == 7){
z.set(i, z.get(i)/3);
}
else {
z.set(i, z.get(i)+2);
}
}
String newString = "";
for(Integer i: z){
newString += i;
}
return newString;
}

Number of digits and which digits it is in integer

Example int i=185;
Then I want to get that 'i' contains 3 digits and those digits are 1,8, and 5.

Hint: You need to take the modulus of the number by 10, to get the last digit. And then divide the same number by 10, do get the first two numbers. Repeat yourself as many times as required.

1st solution:
/**
* Using Integer/String classes functionality
*/
public class Shweta {
private static Integer i = 185;
public static void main(String... args) {
String iStr = i.toString();
for (char digit : iStr.toCharArray()) {
System.out.println(digit);
}
System.out.println("Length is: " + iStr.length());
}
}
2nd solution:
/**
* Doing that in a 'homework' way
*/
public class ShwetaNoCheats {
private static Integer i = 185;
public static void main(String... args) {
int length = 0;
while (i != 0) {
System.out.println(i - (i / 10) * 10);
i /= 10;
length++;
}
System.out.println("Length is: " + length);
}
}

The easy way to do this is by converting to a locale-agnostic string, then looking at each character in the string. I am not giving the final solution in case this is homework, but here are some important APIs...
Converting to string:
String stringForm = Integer.toString(number);
Handling negatives:
int nonNegative = Math.abs(number);
Length of a string:
int length = stringForm.length();
Getting the i-th character of a string:
char c = stringForm.charAt(i);

One way would be:
int i = 185;
int a = i / 100; // 1
int b = (i % 100) / 10; // 8
int c = i % 10; // 5
But i think you need something more generic? Try via string
int i = 185;
String iAsString = String.format("%d", i);
if(iAsString.contains("1")){
// do something...
}
And more advanced:
int i = 185;
String iAsString = String.format("%d", i);
HashSet<Integer> set = new HashSet<Integer>();
for(char c : iAsString.toCharArray()){
set.add(Integer.valueOf(String.valueOf(c)));
}
Then you can work on the set.

The number of decimal digits is also given by Math.ceil(Math.log10(i)), for integral i.