How to resolve the algorithm Eban numbers step by step in the Wren programming language

Published on 12 May 2024 09:40 PM
#Wren

How to resolve the algorithm Eban numbers step by step in the Wren programming language

Table of Contents

Problem Statement

An   eban   number is a number that has no letter   e   in it when the number is spelled in English. Or more literally,   spelled numbers that contain the letter   e   are banned.

The American version of spelling numbers will be used here   (as opposed to the British). 2,000,000,000   is two billion,   not   two milliard.

Only numbers less than   one sextillion   (1021)   will be considered in/for this task. This will allow optimizations to be used.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Eban numbers step by step in the Wren programming language

Source code in the wren programming language

var rgs = [
    [2, 1000, true],
    [1000, 4000, true],
    [2, 1e4, false],
    [2, 1e5, false],
    [2, 1e6, false],
    [2, 1e7, false],
    [2, 1e8, false],
    [2, 1e9, false]
]
for (rg in rgs) {
    if (rg[0] == 2) {
        System.print("eban numbers up to and including %(rg[1])")
    } else {
        System.print("eban numbers between %(rg[0]) and %(rg[1]) (inclusive):")
    }
    var count = 0
    var i = rg[0]
    while (i <= rg[1]) {
        var b = (i/1e9).floor
        var r = i % 1e9
        var m = (r/1e6).floor
        r = i % 1e6
        var t = (r/1000).floor
        r = r % 1000
        if (m >= 30 && m <= 66) m = m % 10
        if (t >= 30 && t <= 66) t = t % 10
        if (r >= 30 && r <= 66) r = r % 10
        if (b == 0 || b == 2 || b == 4 || b == 6) {     
            if (m == 0 || m == 2 || m == 4 || m == 6) {
                if (t == 0 || t == 2 || t == 4 || t == 6) {
                    if (r == 0 || r == 2 || r == 4 || r == 6) {
                        if (rg[2]) System.write("%(i) ")
                        count = count + 1
                    }
                }
            }
        }
        i = i + 2
    }
    if (rg[2]) System.print()
    System.print("count = %(count)\n")
}

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