프로그래머스 / 기사단원의 무기

https://school.programmers.co.kr/learn/courses/30/lessons/136798

파이썬

약수는 제곱근(루트)를 이용해서 계산을 줄일 수 있다. 

def solution(number, limit, power):
    answer = 0
    for i in range(1, number + 1):
        sqrt_i = i ** .5
        count = sum(1 for j in range(1, int(sqrt_i) + 1) if i % j == 0) * 2 - (0 if sqrt_i % 1 else 1)
        answer += power if limit < count else count
    return answer

코틀린

class Solution {
    fun solution(number: Int, limit: Int, power: Int): Int {
        var answer = 0
        for (each in 1..number) {
            var temp = 0
            for (num in 1..each)
                if (each % num == 0)
                    temp++
            answer += if (temp > limit) power else temp
        }
        return answer
    }
}

import kotlin.math.sqrt

class Solution {
    fun solution(number: Int, limit: Int, power: Int): Int {
        var answer = 0
        for (each in 1..number) {
            var temp = 0
            val sqrtEach = sqrt(each.toDouble())
            for (num in 1 until sqrtEach.toInt() + 1)
                if (each % num == 0)
                    temp++
            temp = temp * 2 - if (sqrtEach % 1 > 0) 0 else 1
            answer += if (temp > limit) power else temp
        }
        return answer
    }
}

class Solution {
    fun solution(number: Int, limit: Int, power: Int) = (1..number)
        .map { (1..it).count { each -> it % each == 0 } }
        .sumOf { if (it > limit) power else it }
}

import kotlin.math.sqrt

class Solution {
    fun solution(number: Int, limit: Int, power: Int) = (1..number)
        .sumOf {
            val sqrtEach = sqrt(it.toDouble())
            val temp = (
                    (1 until sqrtEach.toInt() + 1).count { each -> it % each == 0 }
                    ) * 2 - if (sqrtEach % 1 > 0) 0 else 1
            if (temp > limit) power else temp
        }
}