Hackerrank - Area Under Curves and Volume of Revolving a Curve

Hackerrank – Popis problému

Celý popis zadania sa nacháza – Hackerrank.

Riešenie

f je volanie funkcie s x ako je v zadaní
funkcia area vypočíta plochu kruhu (
$\Pi r^2$

) so stredom 0 a polomerom s hodnotou function(x)
funkcia summation sa zavolá raz pre funkciu f a raz pre funkciu area

Vytvoril som riešenie v týchto programovacích jazykoch:

Scala

Všetky riešenia sú dostupné aj na mojom GitHub profile.

Scala

def f(coefficients: List[Int], powers:List[Int], x: Double): Double =  {
  coefficients.indices.map(i => coefficients(i) * Math.pow(x, powers(i))).sum
}

def area(coefficients: List[Int], powers: List[Int], x: Double): Double = {
  scala.math.Pi * scala.math.pow(f(coefficients, powers, x), 2)
}

def summation(func: (List[Int], List[Int], Double) => Double,
              upperLimit: Int, lowerLimit: Int, coefficients: List[Int], powers: List[Int]): Double = {
  val range = lowerLimit.toDouble to upperLimit.toDouble by 0.001
  range.map(func(coefficients, powers, _)).sum * 0.001
}

def f(coefficients: List[Int], powers:List[Int], x: Double): Double = {

coefficients.indices.map(i => coefficients(i) * Math.pow(x, powers(i))).sum

}

def area(coefficients: List[Int], powers: List[Int], x: Double): Double = {

scala.math.Pi * scala.math.pow(f(coefficients, powers, x), 2)

}

def summation(func: (List[Int], List[Int], Double) => Double,

upperLimit: Int, lowerLimit: Int, coefficients: List[Int], powers: List[Int]): Double = {

val range = lowerLimit.toDouble to upperLimit.toDouble by 0.001

range.map(func(coefficients, powers, _)).sum * 0.001

}

Pavol Pidanič

Dokážem na 10 prstoch napočítať do 1023

Hackerrank – Area Under Curves and Volume of Revolving a Curve

Hackerrank – Popis problému

Riešenie

Scala

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