Minimum numbers of three lists: Difference between revisions
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Given three lists: |
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<br>Numbers1 = [5,45,23,21,67] |
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*Numbers1 = [5,45,23,21,67] |
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*Numbers2 = [43,22,78,46,38] |
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*Numbers3 = [9,98,12,98,53] |
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then: |
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Revision as of 09:41, 26 October 2021
Minimum numbers of three lists is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- Task
Given three lists:
- Numbers1 = [5,45,23,21,67]
- Numbers2 = [43,22,78,46,38]
- Numbers3 = [9,98,12,98,53]
then:
- Select the minimum of Numbers[n], Numbers2[n] and Numbers3[n], where n <= 5 (one based).
- Add minimum to a new list (Numbers).
- Show Numbers on this page.
C
<lang c>#include <stdio.h>
int min(int a, int b) {
if (a < b) return a; return b;
}
int main() {
int n, m; int numbers1[5] = {5, 45, 23, 21, 67}; int numbers2[5] = {43, 22, 78, 46, 38}; int numbers3[5] = {9, 98, 12, 98, 53}; for (n = 0; n < 5; ++n) { m = min(min(numbers1[n], numbers2[n]), numbers3[n]); printf("%d ", m); } printf("\n");
}</lang>
- Output:
5 22 12 21 38
Go
<lang go>package main
import (
"fmt" "rcu"
)
func main() {
numbers1 := [5]int{5, 45, 23, 21, 67} numbers2 := [5]int{43, 22, 78, 46, 38} numbers3 := [5]int{9, 98, 12, 98, 53} minima := [5]int{} for n := 0; n < 5; n++ { minima[n] = rcu.Min(rcu.Min(numbers1[n], numbers2[n]), numbers3[n]) } fmt.Println(minima)
}</lang>
- Output:
[5 22 12 21 38]
Ring
<lang ring> see "working..." + nl
Numbers = [] Numbers1 = [5,45,23,21,67] Numbers2 = [43,22,78,46,38] Numbers3 = [9,98,12,98,53]
for n = 1 to len(Numbers1)
Temp = [] add(Temp,Numbers1[n]) add(Temp,Numbers2[n]) add(Temp,Numbers3[n]) min = min(Temp) add(Numbers,min)
next
see "The minimum numbers of three lists = " see showArray(Numbers) see nl + "done..." + nl
func showArray(array)
txt = "" see "[" for n = 1 to len(array) txt = txt + array[n] + "," next txt = left(txt,len(txt)-1) txt = txt + "]" see txt
</lang>
- Output:
working... The minimum numbers of three lists = [5,22,12,21,38] done...
Wren
<lang ecmascript>var numbers1 = [ 5, 45, 23, 21, 67] var numbers2 = [43, 22, 78, 46, 38] var numbers3 = [ 9, 98, 12, 98, 53] var minima = List.filled(5, 0) for (n in 0..4) minima[n] = numbers1[n].min(numbers2[n]).min(numbers3[n]) System.print(minima)</lang>
- Output:
[5, 22, 12, 21, 38]