import kotlin.math.pow

/**
 * Distinct powers
 *
 * Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
 *
 * 2^2=4, 2^3=8, 2^4=16, 2^5=32
 * 3^2=9, 3^3=27, 3^4=81, 3^5=243
 * 4^2=16, 4^3=64, 4^4=256, 4^5=1024
 * 5^2=25, 5^3=125, 5^4=625, 55^=3125
 * If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct
 * terms:
 *
 * 4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
 *
 * How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
 *
 * https://projecteuler.net/problem=29
 */
fun main() {
    println(distinctPowers(limit = 5))
    println(distinctPowers(limit = 100))
}

private fun distinctPowers(limit: Int): Int = sequence {
    for (a in 2..limit) {
        for (b in 2..limit) {
            yield(a.toDouble().pow(b))
        }
    }
}.distinct().count()

