Fairshare between two and more: Difference between revisions
Added Wren
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With 50000 people: 1
With 50001 people: Only 50000 have a turn</pre>
=={{header|Wren}}==
{{trans|Go}}
{{libheader|Wren-fmt}}
{{libheader|Wren-sort}}
<lang ecmascript>import "/fmt" for Fmt
import "/sort" for Sort
var fairshare = Fn.new { |n, base|
var res = List.filled(n, 0)
for (i in 0...n) {
var j = i
var sum = 0
while (j > 0) {
sum = sum + (j%base)
j = (j/base).floor
}
res[i] = sum % base
}
return res
}
var turns = Fn.new { |n, fss|
var m = {}
for (fs in fss) {
var k = m[fs]
if (!k) {
m[fs] = 1
} else {
m[fs] = k + 1
}
}
var m2 = {}
for (k in m.keys) {
var v = m[k]
var k2 = m2[v]
if (!k2) {
m2[v] = 1
} else {
m2[v] = k2 + 1
}
}
var res = []
var sum = 0
for (k in m2.keys) {
var v = m2[k]
sum = sum + v
res.add(k)
}
if (sum != n) return "only %(sum) have a turn"
Sort.quick(res)
var res2 = List.filled(res.count, "")
for (i in 0...res.count) res2[i] = res[i].toString
return res2.join(" or ")
}
for (base in [2, 3, 5, 11]) {
Fmt.print("$2d : $2d", base, fairshare.call(25, base))
}
System.print("\nHow many times does each get a turn in 50,000 iterations?")
for (base in [191, 1377, 49999, 50000, 50001]) {
var t = turns.call(base, fairshare.call(50000, base))
Fmt.print(" With $5d people: $s", base, t)
}</lang>
{{out}}
<pre>
2 : 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0
3 : 0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1
5 : 0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3
11 : 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4
How many times does each get a turn in 50,000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: only 50000 have a turn
</pre>
=={{header|zkl}}==
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