I am writing code in c++ to detect if an input number is a Palindrome Number, which means its reverse is the same as the origin. I have problems computing the reverse int.
e.g.
121 returns true;
123 returns false;
12321 returns true;
10 returns false;
I input 123 and the sum should be 321. However, my code keeps returning 386. I stepped into the function with xcode. Still, I have no idea why reverse += (3 * 10) + 2 turns to be 35 or why the final reverse number to be 386.
int origin = x;
int reverse = 0;
while (x != 0) {
int digit = x % 10;
reverse += ((reverse * 10) + digit);
x /= 10;
}
why reverse += (3 * 10) + 2 turns to be 35
Because += adds what is on the right to the existing value of what’s on the left. (3 * 10) + 2 is 32, but reverse was already 3 and so you are adding your 32 to the existing 3, which is 35.
You don’t want to add to the value of reverse; you want to replace it.
Change
reverse += ((reverse * 10) + digit)
To
reverse = ((reverse * 10) + digit)
Related
I'm looking to create a function that returns a solve for x math equation that can be preformed in ones head (Clearly thats a bit subjective but I'm not sure how else to phrase it).
Example problem: (x - 15)/10 = 6
Note: Only 1 x in the equation
I want to use the operations +, -, *, /, sqrt (Only applied to X -> sqrt(x))
I know that let mathExpression = NSExpression(format: question) converts strings into math equations but when solving for x I'm not sure how to go about doing this.
I previously asked Generating random doable math problems swift for non solving for x problems but I'm not sure how to convert that answer into solving for x
Edit: Goal is to generate an equation and have the user solve for the variable.
Since all you want is a string representing an equation and a value for x, you don't need to do any solving. Just start with x and transform it until you have a nice equation. Here's a sample: (copy and paste it into a Playground to try it out)
import UIKit
enum Operation: String {
case addition = "+"
case subtraction = "-"
case multiplication = "*"
case division = "/"
static func all() -> [Operation] {
return [.addition, .subtraction, .multiplication, .division]
}
static func random() -> Operation {
let all = Operation.all()
let selection = Int(arc4random_uniform(UInt32(all.count)))
return all[selection]
}
}
func addNewTerm(formula: String, result: Int) -> (formula: String, result: Int) {
// choose a random number and operation
let operation = Operation.random()
let number = chooseRandomNumberFor(operation: operation, on: result)
// apply to the left side
let newFormula = applyTermTo(formula: formula, number: number, operation: operation)
// apply to the right side
let newResult = applyTermTo(result: result, number: number, operation: operation)
return (newFormula, newResult)
}
func applyTermTo(formula: String, number:Int, operation:Operation) -> String {
return "\(formula) \(operation.rawValue) \(number)"
}
func applyTermTo(result: Int, number:Int, operation:Operation) -> Int {
switch(operation) {
case .addition: return result + number
case .subtraction: return result - number
case .multiplication: return result * number
case .division: return result / number
}
}
func chooseRandomNumberFor(operation: Operation, on number: Int) -> Int {
switch(operation) {
case .addition, .subtraction, .multiplication:
return Int(arc4random_uniform(10) + 1)
case .division:
// add code here to find integer factors
return 1
}
}
func generateFormula(_ numTerms:Int = 1) -> (String, Int) {
let x = Int(arc4random_uniform(10))
var leftSide = "x"
var result = x
for i in 1...numTerms {
(leftSide, result) = addNewTerm(formula: leftSide, result: result)
if i < numTerms {
leftSide = "(" + leftSide + ")"
}
}
let formula = "\(leftSide) = \(result)"
return (formula, x)
}
func printFormula(_ numTerms:Int = 1) {
let (formula, x) = generateFormula(numTerms)
print(formula, " x = ", x)
}
for i in 1...30 {
printFormula(Int(arc4random_uniform(3)) + 1)
}
There are some things missing. The sqrt() function will have to be implemented separately. And for division to be useful, you'll have to add in a system to find factors (since you presumably want the results to be integers). Depending on what sort of output you want, there's a lot more work to do, but this should get you started.
Here's sample output:
(x + 10) - 5 = 11 x = 6
((x + 6) + 6) - 1 = 20 x = 9
x - 2 = 5 x = 7
((x + 3) * 5) - 6 = 39 x = 6
(x / 1) + 6 = 11 x = 5
(x * 6) * 3 = 54 x = 3
x * 9 = 54 x = 6
((x / 1) - 6) + 8 = 11 x = 9
Okay, let’s assume from you saying “Note: Only 1 x in the equation” that what you want is a linear equation of the form y = 0 = β1*x + β0, where β0 and β1 are the slope and intercept coefficients, respectively.
The inverse of (or solution to) any linear equation is given by x = -β0/β1. So what you really need to do is generate random integers β0 and β1 to create your equation. But since it should be “solvable” in someone’s head, you probably want β0 to be divisible by β1, and furthermore, for β1 and β0/β1 to be less than or equal to 12, since this is the upper limit of the commonly known multiplication tables. In this case, just generate a random integer β1 ≤ 12, and β0 equal to β1 times some integer n, 0 ≤ n ≤ 12.
If you want to allow simple fractional solutions like 2/3, just multiply the denominator and the numerator into β0 and β1, respectively, taking care to prevent the numerator or denominator from getting too large (12 is again a good limit).
Since you probably want to make y non-zero, just generate a third random integer y between -12 and 12, and change your output equation to y = β1*x + β0 + y.
Since you mentioned √ could occur over the x variable only, that is pretty easy to add; the solution (to 0 = β1*sqrt(x) + β0) is just x = (β0/β1)**2.
Here is some very simple (and very problematic) code for generating random integers to get you started:
import func Glibc.srand
import func Glibc.rand
import func Glibc.time
srand(UInt32(time(nil)))
print(rand() % 12)
There are a great many answers on this website that deal with better ways to generate random integers.
I want to know how am I supposed to count the number of time a loop has repeated itself? More specifically how do I extract and output the number of repeats?
var x = 20
while x < 100 {
x += 10
}
The loop has executed 8 times in order to get x == 100. Is there a way to extract the number '8' so it can be used somewhere else (e.g. to make it a variable elsewhere)?
You said it yourself: you want to count. So count!
var x = 20
var numtimes = 0
while x < 100 {
x += 10
numtimes += 1 // count!
}
numtimes // 8
I'm trying to get the average of an array of Ints using the following code:
let numbers = [1,2,3,4,5]
let avg = numbers.reduce(0) { return $0 + $1 / numbers.count }
print(avg) // 1
Which is obviously incorrect. However, if I remove the division to the outside of the closure:
let numbers = [1,2,3,4,5]
let avg = numbers.reduce(0) { return $0 + $1 } / numbers.count
print(avg) // 3
Bingo! I think I remember reading somewhere (can't recall if it was in relation to Swift, JavaScript or programming math in general) that this has something to do with the fact that dividing the sum by the length yields a float / double e.g. (1 + 2) / 5 = 0.6 which will be rounded down within the sum to 0. However I would expect ((1 + 2) + 3) / 5 = 1.2 to return 1, however it too seems to return 0.
With doubles, the calculation works as expected whichever way it's calculated, as long as I box the count integer to a double:
let numbers = [1.0,2.0,3.0,4.0,5.0]
let avg = numbers.reduce(0) { return $0 + $1 / Double(numbers.count) }
print(avg) // 3
I think I understand the why (maybe not?). But I can't come up with a solid example to prove it.
Any help and / or explanation is very much appreciated. Thanks.
The division does not yield a double; you're doing integer division.
You're not getting ((1 + 2) + 3 etc.) / 5.
In the first case, you're getting (((((0 + (1/5 = 0)) + (2/5 = 0)) + (3/5 = 0)) + (4/5 = 0)) + (5/5 = 1)) = 0 + 0 + 0 + 0 + 0 + 1 = 1.
In the second case, you're getting ((((((0 + 1) + 2) + 3) + 4) + 5) / 5) = 15 / 5 = 3.
In the third case, double precision loss is much smaller than the integer, and you get something like (((((0 + (1/5.0 = 0.2)) + (2/5.0 = 0.4)) + (3/5.0 = 0.6)) + (4/5.0 = 0.8)) + (5/5.0 = 1.0)).
The problem is that what you are attempting with the first piece of code does not make sense mathematically.
The average of a sequence is the sum of the entire sequence divided by the number of elements.
reduce calls the lambda function for every member of the collection it is being called on. Thus you are summing and dividing all the way through.
For people finding it hard to understand the original answer.
Consider.
let x = 4
let y = 3
let answer = x/y
You expect the answer to be a Double, but no, it is an Int. For you to get an answer which is not a rounded down Int. You must explicitly state the values to be Double. See below
let doubleAnswer = Double(x)/Double(y)
Hope this helped.
I've got this question, and I'm a bit confused as to what would be printed, especially for pass-by-reference. What value would be passed to x if there are two parameters? Thanks!
Consider the following program. For each of the following parameter-passing methods, what is printed?
a. Passed by value
b. Passed by reference
c. Passed by value-result
void main()
{
int x = 5;
foo (x,x);
print (x);
}
void foo (int a, int b)
{
a = 2 * b + 1;
b = a - 1;
a = 3 * a - b;
}
The first two should be pretty straightforward, the last one is probably throwing you because it's not really a C++ supported construct. It's something that had been seen in Fortran and Ada some time ago. See this post for more info
As for your results, I think this is what you would get:
1)
5
2)
x = 5,
a = 2 * 5 + 1 = 11
b = 11 - 1 = 10
a = 3 * 10 - 10 = 20; // remember, a and b are the same reference!
x = 20
3) Consider this (in C++ style). We will copy x into a variable, pass that by reference, and then copy the result back to x:
void main()
{
int x = 5;
int copy = x;
foo (copy,copy); // copy is passed by reference here, for sake of argument
x = copy;
print (x);
}
Since nothing in the foo function is doing anything with x directly, your result will be the same as in #2.
Now, if we had something like this for foo
void foo (int a, int b)
{
a = 2 * b + 1;
x = a - 1; // we'll assume x is globally accessible
a = 3 * a - b;
}
Then # 2 would produce the same result, but #3 would come out like so:
a = 2 * 5 + 1 = 11
x = 11 - 1 = 10 // this no longer has any effect on the final result
a = 3 * 11 - 11 = 22
x = 22
This is a simple number series question, I have numbers in series like
2,4,8,16,32,64,128,256 these numbers are formed by 2,2(square),2(cube) and so on.
Now if I add 2+4+8 = 14. 14 will get only by the addition 2,4 and 8.
so i have 14in my hand now, By some logic i need to get the values which are helped to get 14
Example:
2+4+8 = 14
14(some logic) = 2,4,8.
This is an easy one:
2+4+8=14 ... 14+2=16
2+4+8+16=30 ... 30+2=32
2+4+8+16+32=62 ... 62+2=64
So you just need to add 2 to your sum, then calculate ld (binary logarithm), and then subtract 1. This gives you the number of elements of your sequence you need to add up.
e.g. in PHP:
$target=14;
$count=log($target+2)/log(2)-1;
echo $count;
gives 3, so you have to add the first 3 elements of your sequence to get 14.
Check the following C# code:
x = 14; // In your case
indices = new List<int>();
for (var i = 31; i >= i; i--)
{
var pow = Math.Pow(2, i);
if x - pow >= 0)
{
indices.Add(pow);
x -= pow;
}
}
indices.Reverse();
assuming C:
unsigned int a = 14;
while( a>>=1)
{
printf("%d ", a+1);
}
if this is programming, something like this would suffice:
int myval = 14;
int maxval = 256;
string elements = "";
for (int i = 1; i <= maxval; i*=2)
{
if ((myval & i) != 0)
elements += "," + i.ToString();
}
Use congruency module 2-powers: 14 mod 2 = 0, 14 mod 4 = 2, 14 mod 8 = 6, 14 mod 16 = 14, 14 mod 32 = 14...
The differences of this sequence are the numbers you look for 2 - 0 = 2, 6 - 2 = 4, 14 - 6 = 8, 14 - 14 = 0, ...
It's called the p-adic representation and is formally a bit more difficult to explain, but I hope this gives you an idea for an algorithm.