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My last step involves getting the integer percent of the sum.
So if I enter:
2
1
1
2
The output should be:
2, which is 33.333% of the sum.
1, which is 16.666% of the sum.
1, which is 16.666% of the sum.
2, which is 33.333% of the sum.
Since I am very new to arrays, I am extremely puzzled. I don't understand how to get the percent since user can enter any amount of integers. If it was only 2 integers, just say 2 and 2, they each would be 50 percent
import java.util.Scanner;
public class Integers {
/* program 7-1*/
public static void main(String[] args)
{
Scanner keyboard = new Scanner(System.in);
System.out.println("How many numbers will you enter?");
int size = keyboard.nextInt();
int[] entry = new int[size];
System.out.println("Enter " + entry.length + " integers, one per line:");
int sum = 0;
for (int index = 0; index < entry.length; index++)
{
entry[index] = keyboard.nextInt();
sum += size;
}
System.out.println("The sum is " + sum + "." + "\nThe numbers are:" );
}
}
Now that you have all entries and the sum, you are almost there:
You need another loop to go over entries one at a time
To calculate percentage, multiply the entry by 100.0, and then divide by the sum. Note the dot zero at the end of 100.0 - it's there on purpose
If you are on Java 5 or later, a very convenient way of printing out a number followed by percentage sign is printf. Note, however, that the percent sign % needs to be escaped.
What answer are you getting that's incorrect? That's what is puzzling. I looked at your code, but I don't see any division, percentage calculation, or output.
I compiled and ran your code. Here's the output I got. So far, so good. What's wrong? What's your question?
"C:\Program Files\Java\jdk1.7.0_02\bin\java" -Didea.launcher.port=7533 "-Didea.launcher.bin.path=C:\Program Files (x86)\JetBrains\IntelliJ IDEA 120.11\bin" com.intellij.rt.execution.application.AppMain cruft.Integers
How many numbers will you enter?
4
Enter 4 integers, one per line:
2
1
1
2
The sum is 16.
The numbers are:
Process finished with exit code 0
Write more code. Be careful to remember that integer division isn't what you want; percentages need to be doubles.
int x = 1/2; // x will equal zero. know why?
You've already calculated the sum. You know that if you take in ten numbers, no matter how many or what their values are, the percentage of the sum that each one will represent is the number divided by the sum. Is that what you're asking?
1 hour later:
It's been a whole hour, and you seem to think that writing comments is more educational than actually writing the four lines of code that you need. Okay, I'll bite - here's your solution. I'll risk the wrath of all those who will be outraged by someone who does homework. I want you to see how ridiculous it is that you wouldn't even attempt four lines of code:
import java.util.Scanner;
public class Integers {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.println("How many numbers will you enter?");
int size = keyboard.nextInt();
int[] entry = new int[size];
System.out.println("Enter " + entry.length + " integers, one per line:");
int sum = 0;
for (int index = 0; index < entry.length; index++) {
entry[index] = keyboard.nextInt();
sum += entry[index]; // this was wrong - I fixed it.
}
// This is all you had to do.
for (int anEntry : entry) {
System.out.println(String.format("value: %d %6.2f%%", anEntry, anEntry * 100.0 / sum));
}
System.out.println(String.format("total: %d %6.2f%%", sum, 100.0));
}
}
Since this is not homework, I am not going to give you an answer, but give you an approach that you can figure out to implement (and learn from :-D)
create two arrays of the length the user enters and one counter for the second. the first array will be the one you currently have. The second will store unique values and the counter will represent how many unique values are in that array. When the user enters a value, check to see if its in the current array, if not add it to unique and increment the counter. Then iterate through the unique array at the end of your program performing the calculation.
--cheers
If you have an unknown number of integers separated by a ' ' space, you could take them in as a String and use the String.split function - you can find documentation at http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html#split(java.lang.String) - which will give you an array which is as large as you need it to be, and whose length is stored in the length field of the array.
Related
I am new to Java and programming all together.. I am trying to make a program that ciphers a number for the user. The user inputs 5 digits separately so I add them together to get a total. I need to pull the first digit and second digit of the total and enter it into (firstDigit+key)%10 and (secondDigit+key)%10. Then need to combine the answer to each equation together.
My total needs to be two digits, so even if the user enters all 1's which would total to be 5, I need it to be displayed and seen as 05 so that both equations have a digit to use. It needs to be two digits. I cant seem to figure how to enter a place holder. I was thinking about trying to use:
if (total < 10)
but then what?
Secondly, the method I used below seems like a terrible way to pull a single digit from a number. I think I changed the int total into a string so I can use .substring to pull the digits, then converted back to an int. Surprisingly it works. Is there a better way to do this knowing that the number is random?
String totalString = Integer.toString(total);
String stringDigit1 = totalString.substring(0,1);
String stringDigit2 = totalString.substring(1,2);
int firstDigitInt1 = Integer.parseInt(stringDigit1);
int firstDigitInt2 = Integer.parseInt(stringDigit2);
int encodedDigit1 = (firstDigitInt1+key)%10;
int encodedDigit2 = (firstDigitInt2+key)%10;
System.out.println("Your encoded Number is: " + encodedDigit1 + encodedDigit2);
Your method for obtaining the individual digits is good, and if you want to maintain it I believe your intuition is correct, it should suffice to say:
if (total < 10) {
firstDigitInt1 = 0
}
This will work out with your following math.
Your method with substrings is far from terrible, but in case you wanted something more sleek you can instead say:
int Digit1 = total / 10;
int Digit2 = total % 10;
Here, you can take advantage of integer truncation (where an integer won't remember decimal places) to get the first digit, which also solves the first problem: 5 / 10 = 0 (in terms of ints). For the second digit, it suffices to say modulo 10, as it is the remainder once the total is divided by 10.
Here, I am finding number of perfect square numbers in given range.
But I am dealing with 'for' loop execution that takes much time for execution.
The index/key traverses from two numbers, lets say A to B, and does some operation in for loop.
The problem arises when there's large difference between A and B (e.g. A = 2 & B = 100000)
Can u suggest how can I reduce or optimize the execution time?
Scanner in = new Scanner(System.in);
int A = in.nextInt();
int B = in.nextInt();
int cnt = 0;
for(int number =A ; number<= B; number++){
int sqrt = (int) Math.sqrt(number);
if(sqrt*sqrt == number) {
cnt++;
}
}
System.out.println(cnt);
Or is it because of Math class operations that takes too much time to execute?
Can you suggest any alternate approach to find the square numbers between given range?
Thanks in advance!
I found an alternate way to find the count of perfect square numbers between given range.
This can be simply achieve by using Math.floor and Math.ceil operations.
Math.floor(Math.sqrt(B)) - Math.ceil(Math.sqrt(A)) + 1
Thanks! :)
Instead of going through each number in the range and figuring out if its a perfect square, I would suggest the below
Find a square root of the start number and find the integer part of it.
Lets say start number is 5. So integer part of the square root will be 2.
Now do the same for the range end number
Lets say end range was 1000, so the integer part of its square root would be 31. Now iterate from 2+1 to 31 and keep printing its square. That would give you the perfect squares between the given range.
Instead of the if(sqrt * sqrt == number) you could also check whether the double returned by Math.srt(number) is a integer. The algorithm would than become as follows:
for(int number =A ; number<= B; number++){
if((Math.sqrt(number) % 1) == 0) {
cnt++;
}
}
Note: Haven't tried the code myself so might not work as I expect.
Regarding the question on how you can improve the performance. The checking on whether the number is perfect could be done in parallel by executing per number a task. The access to the counter has to be synchronized than, (to be on the safe side).
This program is essentially a game where the user must enter numbers to see which numbers are good: numbers with an even number of even digits, and an odd number of odd digits.
So first of all, the program ends when I enter a one digit number, which is not intentional. I assume that has something to do with the while being while (n > 0). There also is likely an issue with the if (numEven % 2 == 0......) because the print results seem almost random, with a number being good and the same number not being good sometimes.
Honestly, I am lost at this point. Thank you so much in advance for any help.
UPDATE: This code is working how I want it to, I just wanted to thank everybody who helped out! It's my first semester of computer science class, so I'm still rather new at this...excuse my mistakes that were likely pretty stupid :)
package quackygame;
import java.util.Scanner;
public class QuackyGame
{
public static void main(String[] args)
{
System.out.println("Welcome to the Number Game!"
+ " Try to figure out the pattern "
+ "in the numbers that Wallace likes!");
Scanner scan = new Scanner (System.in);
int n;
int numEven = 0;
int numOdd = 0;
boolean isEven;
do
{
System.out.print("Enter a number > 0: ");
n = scan.nextInt();
while (n > 0)
{
if (n % 2 == 0)
{
//n is even
isEven = true;
numEven++;
}
else
{
//n is odd
isEven = false;
numOdd++;
}
n /= 10;
}
//if numEven is even and numOdd is odd
if (numEven % 2 == 0 && numOdd % 2 == 1)
System.out.println("Wallace liked your number!");
else
{
System.out.println("Wallace didn't like your number.");
}
numEven = 0;
numOdd = 0;
}
while (n >= 0);
}
}
There are a few core issues in the code based on the desired results that you described. The most glaring issue I see is that you intend for the game to essentially "start from scratch" at the beginning of each round, but you never actually reset the numEven and numOdd variables. This is the source of your print results seeming random. For example, if you started a game and input the number:
34567
The game would process the number and say that it is a favorable number because it is odd, has an odd number of odd digits (3), and has an even number of even digits (2). However, upon playing the game again, it would execute the same code without setting the variables back to 0, which means that upon entering:
34567
The game would process this number as a bad number because the accumulated value of odd digits would be 6 instead of 3 (since 3 the first time + 3 the second time results in 6), and 6 is even. So what we want to do is this:
...
int n;
do
{
int numEven = 0;
int numOdd = 0;
System.out.print("Enter a number: ");
n = scan.nextInt();
...
By placing the numEven and numOdd declarations inside of the "do" block, they are local variables which only exist for the duration of the do block. We could also do something as simple as this:
...
else
{
System.out.println("Wallace didn't like your number.");
}
numEven = 0;
numOdd = 0;
}
while (n > 0);
...
Just resetting the values will help us to keep track of the actual intended values of numOdd and numEven more consistently.
With regard to the program closing when you input a single digit number, I'm not sure. That doesn't make sense because since it is a do-while loop it should at least execute once, and issue one of the print statements. I'm loading this code into my IDE right now to give it a run through. I'll update my answer if I find something.
-EDIT-: Upon reading your question again, it seems that you may not be suggesting that the program closes before actually completing any of its functions, but simply that it closes at all. The reason for the closing of the program is that you are performing an integer division arithmetic function where you probably want to be using a different type of number. Let me explain:
In normal human counting, we have our natural set of numbers which have no decimal points. They usually start like this:
1, 2, 3, 4, 5 ...
Then we have a separate set of numbers for math where we operate with more precision:
0.5, 1.4232, 3.142 ...
When we are talking about numbers with normal human language, we assume that dividing 1 by 2 results in 0.5. However, computers do not implicitly know this. In order for a computer to reach the conclusion "0.5" from the division of 1 by 2, you need to explicitly tell it that it should use a certain type of number to produce that output.
The "normal" numbers I referenced earlier are most loosely related to the integer in programming. It's basically a number without a decimal point. What that means is that whenever you divide two integers together, you always get another integer as the result. So if you were to divide 1 by 2, the computer would not interpret the result as 0.5 because that number has a decimal. Instead, it would round it down to the nearest integer, which in this case is 0.
So for a more specific example referencing the actual question at hand, let's say we input the number 5 into our program. It goes through all of the calculations for odds and evens, but eventually gets to this line:
n /= 10
This is where things get funky. We are dividing two integers, but their result does not come out as a perfect integer. In this case, the result of 5 / 10 is again 0.5. But for the computer, since we are dividing two integers, the result 0.5 just won't do, so after rounding down to the nearest integer we get 0. At this point, there is one fatal mistake:
(while n > 0);
When we perform this check, we get false and the while loop ends. Why? Because after performing n /= 10, n becomes 0. And 0 is not greater than 0.
How can we fix this? The best thing to do is probably just use a floating point number to perform the calculations. In Java, this is pretty easy. All we really have to do is:
n /= 10.0
When Java sees that we are dividing by 10.0, which is not an integer, it automatically converts "n" to a floating point number to divide by 10.0. In this case then, if n is 5, our result in dividing 5 by 10.0 will be 0.5. Then, when we run:
(while n > 0);
This becomes true! And the loop does not break.
I am going to put all of these changes into my IDE just to confirm that everything is working as intended for me. I would suggest you give it a try too to see if it fixes your problems.
Hope this helps.
You are increasing numEven or numOdd count each time you input a number, and then you use if (numEven % 2 == 0 && numOdd % 2 == 1) , it is random because if you put number 33 for the first time => numOdd = 1; => true => "Wallace likes" , but next time you put 33 for the second time => numOdd = 2; => false => "Wallace doesnt like".
Edit* Maybe you wanted something like this?
public static void main(String[] args)
{
System.out.println("Welcome to the Number Game!"
+ " Try to figure out the pattern "
+ "in the numbers that Wallace likes!");
Scanner scan = new Scanner (System.in);
int n;
boolean isEven;
do
{
System.out.print("Enter a number: ");
n = scan.nextInt();
//if 0, you leave the loop
if(n==0) {
System.out.println("You pressed 0, have a nice day");
break;
}
if (n % 2 == 0)
{
//it is even
isEven = true;
}
else
{
//it is not even
isEven = false;
}
//if even then he likes it, otherwise he does not
if (isEven)
System.out.println("Wallace liked your number!");
else
{
System.out.println("Wallace didn't like your number.");
}
}
//put any contition here, lets say if you press 0 , you leave the loop
while (n != 0);
}
All the solutions online I can find use BigInteger but I have to solve this using arrays.
I'm only a beginner and I even took this to my Computer Science club and even couldn't figure it out.
Every time I enter a number greater than 31, the output is always zero.
Also, when I enter a number greater than 12, the output is always incorrect.
E.g. fact(13) returns 1932053504 when it should return 6227020800
Here's what I have so far:
import java.util.Scanner;
class Fact
{
public static void main(String[] args)
{
Scanner kb = new Scanner(System.in);
System.out.println("Enter the number you wish to factorial");
int x = kb.nextInt();
System.out.println(fact(x));
}
public static int fact(int x)
{
int[] a = new int[x];
int product = 1;
for(int i = 0; i < a.length; i++)
{
a[i] = x;
x--;
}
for(int i = 0; i < a.length; i++)
{
product = product * a[i];
}
return product;
}
}
Maximum Values Make Large Numbers Terrible
Sadly, because of the maximum values of integers and longs, you are unable to go any larger than
For Longs:
2^63 - 1
9223372036854775807
9 quintillion 223 quadrillion 372 trillion 36 billion 854 million 775 thousand 807
and For Ints:
2^31 - 1
2147483647
2 billion 147 million 483 thousand 647
(I put the written names in to show the size)
At any point in time during the calculation you go over these, "maximum values," you will overflow the variable causing it to behave differently than you would expect, sometimes causing weird zeros to form.
Even BigIntegers have problems with this although it can go up to numbers way higher than just longs and ints which is why they is used with methods that generate massive numbers like factorials.
You seem to want to avoid using BigInteger and only use primatives so long will be the largest data type that you can use.
Even if you convert everything over to long (except the array iterators of course), you will only be able to calculate the factorial up to 20 accurately. Anything over that would overflow the variable. This is because 21! goes over the "maximum value" for longs.
In short, you would need to either use BigInteger or create your own class to calculate the factorials for numbers greater than 20.
Edit: This problem is solved. If you would like to help on another problem, please visit Java Biasing Random Numbers in a Triangular Array.
I'm doing a multiplication game, so I pick 2 numbers between 0 and 12 inclusive. If I do that like this:
int num1 = (int)(Math.random() * 13);
int num2 = (int)(Math.random() * 13);
the squares (0x0,1x1,2x2,etc) are picked half the time (because 1x2 is the same as 2x1). How can I make all combinations picked at the same frequency? There are 91 possible combinations (n(n+1)/2). If it helps, here is a 13 by 13 triangular array:
{{0},
{0,0},
{0,0,0},
{0,0,0,0},
{0,0,0,0,0},
{0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0}};
I've tried picking the first number and giving the second number a 50% chance of being the first one. This did not work. I tried giving the second number a 1/91 chance of being the first one. This resulted in the smaller numbers being picked a far greater number of times (around 7/91 of the time; it is a smooth, curved increase). I thought about having a single random number: int roll = random.next(91) and then splitting it into 2 entries (like a coordinate (x,y)) but I could not figure out how to split it.
The int roll = random.next(91) strategy will work just fine. You get guaranteed, worry-free uniform distribution, and better performance to boot, since you're only picking 1 random number. You simply need to find a formula that identifies where one "row" ends and another begins. Look for the pattern:
0, 1, 3, 6, 10, 15, ...
There's a reason they're called "triangular numbers..."
Let's flesh this out a bit more. You want to actually find the nearest triangle number smaller than the random roll that you picked: that gets you to the right row, and the difference of that triangle number and roll gets you the offset into that row.
Given that the nth triangle number is given by n*(n+1)/2, how do you find the largest one smaller than roll? Given the small size of the array, a naïve implementation should be plenty fast:
int largestTriangleNumberSmallerThan(int x) {
int i = 0;
int last = 0;
while (true) {
int triangle = i*(i+1)/2;
if (triangle > x) return last;
last = triangle;
i++;
}
}
http://ideone.com/vzQEBz
Of course, that's boring and didn't take any thought. We can do better! We can do it in constant* time, regardless of how big the input is! Start by inverting the function (we only care about the positive root, of course):
n = (Math.sqrt(8y + 1) - 1)/2
Then truncate the decimal part, and run it back through:
int largestTriangleNumberSmallerThan(int x) {
int n = (int) (Math.sqrt(8*x + 1) - 1)/2;
return n*(n+1)/2;
}
http://ideone.com/1qBHfX
To put it all together:
int roll = random.nextInt(91);
int num1 = (int) (Math.sqrt(8*roll + 1) - 1)/2;
int num2 = roll - num1*(num1+1)/2;
That's it!
*assuming that the native StrictMath#sqrt(double) function is constant time - I'm actually not sure about this.