I am new to java and i want to make a program where i want to get the integers the user has input, and look at what different numbers they are. and move them around as well as add stuff between them.
For example:
Enter number
User input: 123456
after that i want to see what the first number is, which is 1 or what the third number is which is 3. i want to be able to get those different information. and play around with them, for example add the sixth and the first number and print it out for the user.
Which technique in java allows me to do that, what is the name so i could study it more.
if there is a video online teaching that, it would also be great.
Thanks.
ps. i know the basics, i dont need to know how to do the second part playing with the numbers i just dont know how to tell what the first or second number in the lists of number is.
Eg: 123456
Convert the int to string keep the int also
1) String.valueOf(number)
2) Integer.toString(number)
get the length of the string using String.length();
the if you want the 3rd number
Use the int value for the following operation
Eg: 123456 (3rd number 3) Since you have the length of the string you know what will be the divisor
divide using 10 * 10 * 10 this will 123456/1000 = 123
Then % it with 10 always you can put these within a methos to get any position
Try the below code :You are taking an integer input and converting it into a string
public class InputNumberProgram {
public static void main(String[] args) {
Integer i = 5678;
// Convert your input to a string
String s = Integer.toString(i);
System.out.println("Length of input:" + s.length()); System.out.println(Integer.parseInt(s.valueOf(s.charAt(0)))+Integer.parseInt(s.valueOf(s.charAt(1))));
}
}
If you read it as a String and do what you want with String methods
For example you have String number. To get i-th symbol you can use number.charAt(i)
Related
Hi I want to convert string to some unique number in java.
Exmple: "Production-0-1" to 100021
"Process-23-30" to 12310
And all return number has to be unique.
I dont wanted to use hashCode as they can return duplicate like "Aa" and "BB" has same has code.
Let me know math logic to create this is no method available.
String random = "Production-0-1";
String bi = new BigInteger(random.getBytes("UTF-8")).toString();
BigInteger numBig = new BigInteger(bi);
System.out.println(numBig);
Based on #markspace comments, I tried the following and every time it produces random unique number but beware if you have a very large String and a limited memory space then the output may go out of bound.
Say we have a number
int number = 1234
and I want to use formatting to extract the final two digits as a string 34. How can one accomplish that?
I attempted:
String extraction = String.format("%d", number)
but this simply copies the entire number. I'm new to the syntax used by the formatter and can't seem to figure out the syntax that can go inside the characters of a number (or a String, for that matter) and pull them out. I've found a solution using charAt methods but am particularly curious about whether it's possible to accomplish it using formatting techniques.
Going via a String is inefficient and unnecessary to extract the last two digits of an integer.
Simply:
int lastTwoDigits = number % 100;
If you do want to go via a String, you can use:
String s = Integer.toString(number);
s = s.substring(s.length() - 2);
int lastTwoDigits = Integer.parseInt(s);
(note that this handles -ve numbers slightly differently to the first suggestion).
For example
first number = 123456.....40 digits
second number = 123456.....40digits
Then I need to store the number
third = first * second;
after that I need to print third and again I need to perform operation like
fourth = third * third;
and print fourth. So how can I handle that much long integers which data type I need to use?
Use BigInteger class in java.math, then use BigInteger.multiply to multiply them together.
Check here for more on how to use it:
https://www.tutorialspoint.com/java/math/biginteger_multiply.htm
See this question its similar Arbitrary-precision arithmetic Explanation
the answer explain it quite good.
The basic is that you work with smaller parts. Just remember how you learned to work with big numbers in school (2-3 grade) you wrote down two numbers and
2351
*12
-----
4702
2351
------
28212
You just do small operations and store them somewhere you can put them in string or better in some array of integers. Where for example
number 123456789123456789 can be
number[0] = 6789
number[1] = 2345
number[3] = 7891
number[4] = 2345
number[5] = 1
String numberToShow = "";
for(int i = 0; i
There are some links for computer arighmetics
https://en.wikipedia.org/wiki/Category:Computer_arithmetic
and for adders
https://en.wikipedia.org/wiki/Category:Adders_(electronics)
In your computer you have basically also just some adders which can work only with some size of numbers and if you need to work with bigger you need to split it in smaller parts.
Some of this parts can be done parallel, so you can speed up your algorithm. These are usually more sophisticated.
But the basic principe is similar to working with big numbers on your primary school.
I don't want the answer, I just don't understand how to scan the first number to tell the program how many pairs there are. If you could nudge me in the right direction I would greatly appreciate it.
"Most programs should be able to make some choices and decisions. And we are going to practice conditional programming now.
This is usually done by a kind of if ... else statements which may look like:
IF some_condition THEN
do_something
ELSE
do_other_thing
ENDIF
Depending on your programming language syntax could be different and else part is almost always optional. You can read more in wikipedia article on Conditional statements.
Of two numbers, please, select one with minimum value. Here are several pairs of numbers for thorough testing.
Input data will contain number of test-cases in the first line.
Following lines will contain a pair of numbers to compare each.
For Answer please enter the same amount of minimums separated by space, for example:
data:
3
5 3
2 8
100 15
answer:
3 2 15 "
Firstly, you might want to format your example data a bit. I understood it, but mostly only because I've seen that question format before.
Well, to answer your question but not the question's question (heh), note this:
Following lines will contain a pair of numbers to compare each.
Note the "lines" (plural) and the "each." We're going to need a loop.
We also know each line is a test case.
So modify the instructions:
Loop over the following test cases, comparing each pair
But how many times do we loop?
Input data will contain number of test-cases in the first line
That's the first number.
So here's our code skeleton:
//We can use a Scanner for convenience, it has a readInt() method
Scanner input = new Scanner(/*your input*/);
int numCases = input.readInt();
for(int i = 0; i < numCases; i++) {
int first = input.readInt(); //readInt() will also skip newlines, just a tip.
int second = input.readInt();
/* Compare two inputs, do stuff*/
}
For the record, you could also simply ignore the first input and just loop until there is no more input, but that's sloppy.
First you need to create a Scanner. If you're reading from the console, then this will work:
Scanner scan = new Scanner(System.in);
If you need to read from a file, then you can add this line as well.
System.setIn(new FileInputStream("inputFileName"));
For your specific case, you can do something like:
int numPairs = scan.nextInt();
You can find out more about Scanner and its methods from the Oracle documentation here.
I was wondering if there is a way to compare strings in android with greater than or >.
Lets say I have this:
String numbers = number.getText().toString();
if (numbers.equals("9")){
output.setText("50");}
so if you enter 9 in the number EditText field the output TextView will display 50.
I have quite a few different numbers that will then = a different number but what can I do if I want 10,11,12,13,etc to = 100?
Is there a way to do this by using something like this?
if (numbers.equals("9"++))
or is there some kind of wildcard in android like
if (numbers.equals("1"+"*"))
i know if i replace the * with zero it will be 10 so if this is possible I could make one for 1, one for 2, one for 3, etc. and this would still save me so much code.
If this is not possible let me know if you have any ideas.
Thanks guys!
You'll need to convert the String to a number first. Something like this:
int number = Integer.parseInt(number.getText().toString());
if (number > 9)
{
output.setText("50");
}
If the String is not guaranteed to be a valid integer you'll have to handle NumberFormatException too.
Is there a reason you can't use
Integer.valueOf("9");