How to resolve the algorithm Kaprekar numbers step by step in the V (Vlang) programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Kaprekar numbers step by step in the V (Vlang) programming language
Table of Contents
Problem Statement
A positive integer is a Kaprekar number if: Note that a split resulting in a part consisting purely of 0s is not valid, as 0 is not considered positive.
10000 (1002) splitting from left to right:
Generate and show all Kaprekar numbers less than 10,000.
Optionally, count (and report the count of) how many Kaprekar numbers are less than 1,000,000.
The concept of Kaprekar numbers is not limited to base 10 (i.e. decimal numbers); if you can, show that Kaprekar numbers exist in other bases too.
For this purpose, do the following:
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Kaprekar numbers step by step in the V (Vlang) programming language
Source code in the v programming language
import strconv
fn kaprekar(n u64, base u64) (bool, int) {
mut order := 0
if n == 1 {
return true, -1
}
nn, mut power := n*n, u64(1)
for power <= nn {
power *= base
order++
}
power /= base
order--
for ; power > 1; power /= base {
q, r := nn/power, nn%power
if q >= n {
return false, -1
}
if q+r == n {
return true, order
}
order--
}
return false, -1
}
fn main() {
mut max := u64(10000)
println("Kaprekar numbers < $max:")
for m := u64(0); m < max; m++ {
isk, _ := kaprekar(m, 10)
if isk {
println(" $m")
}
}
// extra credit
max = u64(1e6)
mut count := 0
for m := u64(0); m < max; m++ {
isk, _ := kaprekar(m, 10)
if isk {
count++
}
}
println("\nThere are $count Kaprekar numbers < ${max}.")
// extra extra credit
base := 17
max_b := "1000000"
println("\nKaprekar numbers between 1 and ${max_b}(base ${base}):")
max, _ = strconv.common_parse_uint2(max_b, base, 64)
println("\n Base 10 Base ${base} Square Split")
for m := u64(2); m < max; m++ {
isk, pos := kaprekar(m, u64(base))
if !isk {
continue
}
sq := strconv.format_uint(m*m, base)
str := strconv.format_uint(m, base)
split := sq.len-pos
println("${m:8} ${str:7} ${sq:12} ${sq[..split]:6} + ${sq[split..]}") // optional extra extra credit
}
}
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