I need to increment number dynamically in Roman
You can increment by int and convert it to number in roman by this library :
numerus
final n = 2;
print(n.toRomanNumeralString());
You could write it yourself as an exercise, or you can use a package like numerus to do this job for you.
From how do I make an integer to roman algorithm in dart?, you can get the following:
const List<int> arabianRomanNumbers = [
1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1
];
final builder = StringBuffer();
for (var a = 0; a < arabianRomanNumbers.length; a++) {
final times = (num / arabianRomanNumbers[a]).truncate(); // equals 1 only when arabianRomanNumbers[a] = num
// executes n times where n is the number of times you have to add
// the current roman number value to reach current num.
builder.write(romanNumbers[a] * times);
num -= times * arabianRomanNumbers[a]; // subtract previous roman number value from num
}
return builder.toString();
Related
I have two different list values i want get the corresponding one list between value to other list between values. Please look at the list below. Here i have 1st list value of 11 and 15 here i need to find the between values of 12, 13 and 15.
11 value =22
15 value =30
List<double> list1 =[11, 15];
List<double> list2 =[22, 30];
If I understand correct you need the interpolation function to generate values between 2 points:
List<double> interpolate(double start, double end, int count) {
if (count < 2) {
throw Exception("interpolate: illegal count!");
}
final array = List.generate(count + 1, (index) => 0.0);
for (int i = 0; i <= count; ++i) {
array[i] = start + i * (end - start) / count;
}
return array;
}
Usage:
void main() {
final input = [1.0, 15.0];
print(interpolate(input.first, input.last, 4));
}
Result:
[1.0, 4.5, 8.0, 11.5, 15.0]
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";
List<AudioWaveBar> bars = [];
var list = List<double>.generate(100, (i) => i as double)..shuffle();
for (var i = 0; i <= list.length; i++) {
bars.add(AudioWaveBar(
heightFactor: list[i],
color: widget.podcast.percentPlayed >= ((i + 1).toDouble() / list.length.toDouble())
? AppTheme.greenStart
: Colors.white10
)); }
I need to have a list of random doubles only (from 0 to 1), but random function doesn't add the numbers to a list and generate function only accepts int numbers that can't be casted. Any help? thanks!
The function Random().nextDouble() generates a random double between 0 and 1.
So just by specifying a max value u can get what u want, like so :
int maxValue = 1000;
var list = List<double>.generate(100, (i) => Random().nextDouble() * maxValue)..shuffle();
I want to write a program that can find the N-th number,which only contains factor 2 , 3 or 5.
def method3(n:Int):Int = {
var q2 = mutable.Queue[Int](2)
var q3 = mutable.Queue[Int](3)
var q5 = mutable.Queue[Int](5)
var count = 1
var x:Int = 0
while(count != n){
val minVal = Seq(q2,q3,q5).map(_.head).min
if(minVal == q2.head){
x = q2.dequeue()
q2.enqueue(2*x)
q3.enqueue(3*x)
q5.enqueue(5*x)
}else if(minVal == q3.head){
x = q3.dequeue()
q3.enqueue(3*x)
q5.enqueue(5*x)
}else{
x = q5.dequeue()
q5.enqueue(5*x)
}
count+=1
}
return x
}
println(method3(1000))
println(method3(10000))
println(method3(100000))
The results
51200000
0
0
When the input number gets larger , I get 0 from the function.
But if I change the function to
def method3(n:Int):Int = {
...
q5.enqueue(5*x)
}
if(x > 1000000000) println(('-',x)) //note here!!!
count+=1
}
return x
}
The results
51200000
(-,1006632960)
(-,1007769600)
(-,1012500000)
(-,1019215872)
(-,1020366720)
(-,1024000000)
(-,1025156250)
(-,1033121304)
(-,1036800000)
(-,1048576000)
(-,1049760000)
(-,1054687500)
(-,1061683200)
(-,1062882000)
(-,1073741824)
0
.....
So I don't know why the result equals to 0 when the input number grows larger.
An Int is only 32 bits (4 bytes). You're hitting the limits of what an Int can hold.
Take that last number you encounter: 1073741824. Multiply that by 2 and the result is negative (-2147483648). Multiply it by 4 and the result is zero.
BTW, if you're working with numbers "which only contains factor 2, 3 or 5", in other words the numbers 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, ... etc., then the 1,000th number in that sequence shouldn't be that big. By my calculations the result should only be 1365.
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.