How do I scale a number up to the nearest ten, hundred, thousand, etc...
Ex.
num = 11 round up to 20
num = 15 round up to 20
num = 115 round up to 200
num = 4334 round up to 5000
I guess this formula might work? Unless you have more examples to show.
power = floor(log10(n))
result = (floor(n/(10^power)) + 1) * 10^power
import math
exp = math.log10(num)
exp = math.floor(exp)
out = math.ceil(num/10**exp)
out = out * 10**exp
Convert the number to a decimal (i.e. 11 goes to 1.1, 115 goes to 1.15), then take the ceiling of the number, then multiply it back. Example:
public static int roundByScale(int toRound) {
int scale = (int)Math.pow(10.0, Math.floor(Math.log10(toRound)));
double dec = toRound / scale;
int roundDec = (int)Math.ceil(dec);
return roundDec * scale;
}
In this case, if you input 15, it will be divided by 10 to become 1.5, then rounded up to 2, then the method will return 2 * 10 which is 20.
public static int ceilingHighestPlaceValue(int toCeil)
{
int placeValue = Math.Pow(10,toCeil.ToString().Length()-1);
double temp = toCeil / placeValue;
return= ceil(temp) * placeValue;
}
Related
I'm trying to build some training app for functions for school. The problem is:
Everytime the randomly picked number is lower than 0, my function shows +-, because
I have a fixed format for my function.
EXAMPLE
I tried to use the NumberFormat of the Intl-Package, but then I can't use the int-values correctly. Is there a way to show a plus sign for positive numbers, while they are still usable to work with them?
Code so far:
int randomNumberMinMax(int min, int max){
int randomminmax = min + Random().nextInt(max - min);
if(randomminmax==0){
randomminmax = min + Random().nextInt(max - min);
}
//generate random number within minimum and maximum value
return randomminmax;
}
int a = randomNumberMinMax(-5, 5);
int b = randomNumberMinMax(-10, 10);
int c = randomNumberMinMax(-10, 10);
String task = "f(x) = $a(x+$b)²+ $c";
You could only show the plus when the number is positive like this for example
String task = "f(x) = $a(x${b >= 0 ? "+" : ""}$b)²${c >= 0 ? "+" : ""} $c";
I'm using Swift 2. I have double myMinute and I want to convert it to double hours(myHours). How can I do it? My codes under below
let myMinute : Double = 62.0
let myHours : Double = ?
I want to show with math example: myhours = 1.0
Found this here!
func minutesToHoursMinutes (minutes : Int) -> (hours : Int , leftMinutes : Int {
return (minutes / 60, (minutes % 60))
}
let timeTuple = minutesToHoursMinutes(minutes: 100)
timeTuple.hours /// 1
timeTuple.leftMinutes /// 40
The value should be 1.03 not 1.02 if you use the current value of minute (you can check with calculator) and it is a simple math, i think what you want beside the calculation is how to round the value to 2 decimal point. You should've made it clearer. This will do the work.
myHours = Double(round(myMinute / 60 * 100) / 100)
print(myHours)
Please could somebody help me. I am trying to run a simple compounding calculation in Swift.
Formula I am trying to recreate:
T = P(1+r/n)^(n*t), where
T = Total, P = Starting amount, r = interest rate, n = number of times compounded and t = number of years
My code as follows:
import Darwin
var total: Double
var startingAmount: Double = 5000.00
var interestRate: Double = 0.05
var numberOfTimesCompounded: Double = 4.0
var numberOfYears: Double = 2.0
var totalYear1: Double
var toThePowerOf: Double
totalYear1 = startingAmount * (1 + interestRate / numberOfTimesCompounded)
toThePowerOf = numberOfTimesCompounded * number of years
total = pow(totalYear1,toThePowerOf)
The answer to the formula should be 5,522.43
In my code above, TotalYear1 = 5062.50 (which is correct) and toThePowerOf = 8.0 (which is correct) However, total shows = 4314398832739892000000.00 which clearly isn't right. Could anyone tell me what I am doing wrong with my calculation of total?
Many thanks
You've actually implemented T = (P(1+r/n))^(n*t), which doesn't even make dimensional sense.
startingAmount needs to be multiplied at the end, it can't be part of the pow:
totalYear1 = (1 + interestRate / numberOfTimesCompounded)
toThePowerOf = numberOfTimesCompounded * number of years // [sic]
total = startingAmount * pow(totalYear1,toThePowerOf)
guys, I'm making simple graph drawer and want to find beautiful values for horizontal lines.
For example, if I have value equals to 72089.601562, beautiful is 70000, or 75000. So, I think that beautifulNumber%5 = 0.
Have you any ideas?
How about this?
#import <math.h>
#import <stdio.h>
#define ROUNDING 5000
int beautify(float input)
{
// Cast to int, losing the decimal value.
int value = (int)input;
value = (value / ROUNDING) * ROUNDING;
if ((int)input % ROUNDING > ROUNDING / 2 )
{
value += ROUNDING;
}
return value;
}
int main()
{
printf("%d\n", beautify(70000.601562)); // 70000
printf("%d\n", beautify(72089.601562)); // 70000
printf("%d\n", beautify(76089.601562)); // 75000
printf("%d\n", beautify(79089.601562)); // 80000
printf("%d\n", beautify(70000.601562)); // 70000
return 0;
}
It depends whether you want a floor value, a ceiling value or just to round to the nearest 5000.
For a floor value:
int beautiful = (int)(floor(ugly / 5000.0) * 5000.0);
For a ceiling value:
int beautiful = (int)(ceil(ugly / 5000.0) * 5000.0);
For rounding:
int beautiful = (int)(round(ugly / 5000.0) * 5000.0);
For making graph lines, I'd probably find the minimum and maximum values you have to graph, start with a floor value for the minimum value and then add your desired interval until you have surpassed your maximum value.
For instance:
float minValue = 2.34;
float maxValue = 7.72;
int interval = 1;
NSMutableArray *horizLines = [NSMutableArray array];
int line = (int)(floor(minValue / interval) * interval);
[horizLines addObject:[NSNumber numberWithInt:line]];
do {
line = (int)(ceil(minValue / interval) * interval);
[horizLines addObject:[NSNumber numberWithInt:line]];
if (minValue >= maxValue) break;
minValue = minValue + interval;
}
Use as needed!
Well, it seems like you'd want it to scale based on the size of the number. If the range only goes to 10, then obviously rounding to the nearest 5,000 doesn't make sense. There's probably a really elegant way to code it using bit shifting but just something like this will do the trick:
float value = 72089.601562
int beautiful = 0;
// EDIT to support returning a float for small numbers:
if (value < 0.2) beautiful = int(value*100)/100.;
else if (value < 2.) beautiful = int(value*10)/10.;
// Anything bigger is easy:
else if (value < 20) beautiful = (int)value;
else if (value < 200) beautiful = (int)value/10;
else if (value < 2000) beautiful = (int)value/100;
else if (value < 20000) beautiful = (int)value/1000;
// etc
Sounds like what you want to do is round to 1 or perhaps 2 significant digits. Rounding to n significant digits is pretty easy:
double roundToNDigits(double x, int n) {
double basis = pow(10.0, floor(log10(x)) - (n-1));
return basis * round(x / basis);
}
This will give you roundToNDigits(74518.7, 1) == 70000.0 and roundToNDigits(7628.54, 1) == 8000.00
If you want to round to 1 or 2 digits (only 2 where the second digit is 5), you want something like:
double roundSpecial(double x) {
double basis = pow(10.0, floor(log10(x))) / 2.0;
return basis * round(x / basis);
}
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.