pe24.nim

import algorithm

proc nextPermutation[T](x: var openarray[T]): bool {.discardable.} =
  if x.len < 2:
    return false

  var i = x.high
  while i > 0 and x[i-1] >= x[i]:
    dec i

  if i == 0:
    return false

  var j = x.high
  while j >= i and x[j] <= x[i-1]:
    dec j

  swap x[j], x[i-1]
  x.reverse(i, x.high)

  result = true

var v = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for i in 1 .. 999_999:
  v.nextPermutation()

echo "The answer is ", @v