Positive decimal integers with the digit 1 occurring exactly twice: Difference between revisions

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Task

Find positive decimal integers n in which the digit 1 occurs exactly twice, where n < 1,000.

11l

L(n) 1..999
   I String(n).count(‘1’) == 2
      print(n, end' ‘ ’)

Output:

11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911

8080 Assembly

        org     100h
loop:   lda     tho             ; Are we there yet?
        cpi     '0'
        rnz
        call    incr            ; If not, increment
        lxi     b,0203h         ; Need two ones, 3 steps
        mvi     a,'1'
        lxi     h,acc
ones:   cmp     m               ; Count the ones
        jnz     $+4
        dcr     b
        dcr     c
        inx     h
        jnz     ones
        xra     a               ; If two ones, print
        ora     b
        cz      prn
        jmp     loop

incr:   lxi     h,acc           ; Increment accumulator
        mvi     a,'9'+1
incrl:  inr     m
        cmp     m
        rnz
        mvi     m,'0'
        inx     h
        jmp     incrl

prn:    lxi     h,acc+4         ; Print accumulator w/o leading zero
        mvi     a,'0'
skip:   dcx     h
        cmp     m
        jz      skip
prl:    push    h
        mvi     c,2             ; CP/M print character
        mov     e,m
        call    5
        pop     h
        dcx     h
        xra     a
        cmp     m
        jnz     prl
        mvi     c,9             ; CP/M print newline
        lxi     d,nl
        jmp     5

acc:    db      '000'           ; Accumulator (stored backwards)
tho:    db      '0'             ; Thousands digit (stop if not 0 anymore)
nl:     db      13,10,'$'       ; Newline

Output:

Action!

BYTE FUNC OnesCount(INT x)
  BYTE c,d

  c=0
  WHILE x#0
  DO
    d=x MOD 10
    IF d=1 THEN
      c==+1
    FI
    x==/10
  OD
RETURN (c)

PROC Main()
  INT i

  FOR i=0 TO 999
  DO
    IF OnesCount(i)=2 THEN
      PrintI(i) Put(32)
    FI
  OD
RETURN

Output:

Screenshot from Atari 8-bit computer

11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911

Ada

with Ada.Text_IO; use Ada.Text_IO;

procedure Positives_With_1_Twice is

   function One_Twice (N : Positive) return Boolean is
      NN        : Natural := N;
      One_Count : Natural := 0;
   begin
      while NN > 0 loop
         if NN mod 10 = 1 then
            One_Count := One_Count + 1;
         end if;
         NN := NN / 10;
      end loop;
      return One_Count = 2;
   end One_Twice;

begin
   for N in 1 .. 999 loop
      if One_Twice (N) then
         Put (N'Image); Put (" ");
      end if;
   end loop;
   New_Line;
end Positives_With_1_Twice;

Output:

 11  101  110  112  113  114  115  116  117  118  119  121  131  141  151  161  171  181  191  211  311  411  511  611  711  811  911

ALGOL 68

Generates the numbers. In order to print them in order, a table of double 1 numbers yes/no is generated.

BEGIN # find numbers where the digit 1 occurs twice, up to 999 #
    [ 1 : 999 ]BOOL double 1; FOR i TO UPB double 1 DO double 1[ i ] := FALSE OD;
    # generte the numbers                                      #
    FOR i FROM 0 TO 9 DO
        IF i /= 1 THEN
            double 1[ 110 + i ] := TRUE;
            double 1[ 101 + ( i * 10 ) ] := TRUE;
            double 1[ ( i * 100 ) + 11 ] := TRUE
        FI
    OD;
    # print the numbers in order                               #
    INT double 1 count := 0;
    FOR i TO UPB double 1 DO
        IF double 1[ i ] THEN
            print( ( " ", whole( i, -3 ) ) );
            IF ( double 1 count +:= 1 ) MOD 10 = 0 THEN print( ( newline ) ) FI
        FI
    OD
END

Output:

  11 101 110 112 113 114 115 116 117 118
 119 121 131 141 151 161 171 181 191 211
 311 411 511 611 711 811 911

ALGOL W

Generates the numbers and sorts them into order using Sorting algorithms/Quicksort#ALGOL W.

begin % find numbers where the digit 1 occurs twice, up to 999 %
    integer double1Count;
    integer array double1 ( 1 :: 100 ); % assume there will be at most 100 numbers %
    % Quicksorts in-place the array of integers v, from lb to ub - external %
    procedure quicksort ( integer array v( * )
                        ; integer value lb, ub
                        ) ; algol "sortingAlgorithms_Quicksort" ;
    % increments n by 1 and returns its new value %
    integer procedure inc ( integer value result n ) ; begin n := n + 1; n end inc ;
    % generate the numbers %
    double1Count := 0;
    for i := 0 until 9 do begin
        if i not = 1 then begin
            double1( inc( double1Count ) ) := 110 + i;
            double1( inc( double1Count ) ) := 101 + ( i * 10 );
            double1( inc( double1Count ) ) := ( i * 100 ) + 11
        end if_i_ne_1
    end for_i ;
    % sort the numbers %
    quickSort( double1, 1, double1Count );
    % print the numbers %
    for i := 1 until double1Count do writeon( i_w := 1, s_w := 1, double1( i ) );
    write();
    write( i_w := 1, s_w := 0, "Found ", double1Count, " numbers" )
end.

Output:

11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911

Found 27 numbers

APL

(⊢(/⍨)(2='1'+.=⍕)¨) ⍳1000

Output:

11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911

Arturo

prints [11 101 110]
loop 2..9 'd -> prints ~"|d|11 1|d|1 11|d| "

Output:

11 101 110 211 121 112 311 131 113 411 141 114 511 151 115 611 161 116 711 171 117 811 181 118 911 191 119

AWK

# syntax: GAWK -f NUMBERS_N_IN_WHICH_NUMBER_1_OCCUR_TWICE.AWK
BEGIN {
    start = 1
    stop = 999
    for (i=start; i<=stop; i++) {
      if (gsub(/1/,"&",i) == 2) {
        printf("%4d%1s",i,++count%10?"":"\n")
      }
    }
    printf("\nNumber 1 occurs twice %d-%d: %d\n",start,stop,count)
    exit(0)
}

Output:

  11  101  110  112  113  114  115  116  117  118
 119  121  131  141  151  161  171  181  191  211
 311  411  511  611  711  811  911
Number 1 occurs twice 1-999: 27

COBOL

       IDENTIFICATION DIVISION.
       PROGRAM-ID. TWO-ONES.
       
       DATA DIVISION.
       WORKING-STORAGE SECTION.
       77 NUM       PIC 9999.
       77 FMT       PIC ZZ9.
       77 ONES      PIC 9.
       
       PROCEDURE DIVISION.
       BEGIN.
           PERFORM COUNT-ONES
           VARYING NUM FROM 1 BY 1 UNTIL NUM IS EQUAL TO 1000. 
           STOP RUN.
           
       COUNT-ONES.
           MOVE ZERO TO ONES.
           INSPECT NUM TALLYING ONES FOR ALL '1'.
           IF ONES IS EQUAL TO 2,
               MOVE NUM TO FMT,
               DISPLAY FMT.

Output:

Euler

As with the Arturo, F# etc. samples, generates the sequence but doesn't sort it into order.

begin      new dPos; new non1digits; label digitLoop;
           non1digits <- ( 0, 2, 3, 4, 5, 6, 7, 8, 9 );
           dPos <- 0;
digitLoop: if [ dPos <- dPos + 1 ] <= length non1digits then begin
              new d;
              d <- non1digits[ dPos ];
              out [ d * 100 ] + 11; out [ d * 10 ] + 101; out 110 + d;
              goto digitLoop
           end else 0
end $

Output:

    NUMBER                  11
    NUMBER                 101
    NUMBER                 110
    NUMBER                 211
    NUMBER                 121
    NUMBER                 112
    NUMBER                 311
    NUMBER                 131
    NUMBER                 113
    NUMBER                 411
    NUMBER                 141
    NUMBER                 114
    NUMBER                 511
    NUMBER                 151
    NUMBER                 115
    NUMBER                 611
    NUMBER                 161
    NUMBER                 116
    NUMBER                 711
    NUMBER                 171
    NUMBER                 117
    NUMBER                 811
    NUMBER                 181
    NUMBER                 118
    NUMBER                 911
    NUMBER                 191
    NUMBER                 119

F#

// 3 digit numbers with 2 ones. Nigel Galloway: July 6th., 2021
[0;2;3;4;5;6;7;8;9]|>List.collect(fun g->[[g;1;1];[1;g;1];[1;1;g]])|>List.iter(fun(n::g::l::_)->printf "%d " (n*100+g*10+l)); printfn ""

Output:

11 101 110 211 121 112 311 131 113 411 141 114 511 151 115 611 161 116 711 171 117 811 181 118 911 191 119

Factor

Translation of: F#

Works with: Factor version 0.99 2021-06-02

USING: io math math.functions prettyprint sequences
sequences.extras ;

{ 0 2 3 4 5 6 7 8 9 }
[| n | { { n 1 1 } { 1 n 1 } { 1 1 n } } ] map-concat
[ <reversed> 0 [ 10^ * + ] reduce-index pprint bl ] each nl

Output:

11 101 110 211 121 112 311 131 113 411 141 114 511 151 115 611 161 116 711 171 117 811 181 118 911 191 119

FreeBASIC

function numdig(byval n as integer, d as const integer) as uinteger
    'counts the number of occurrences of digit d in the number n
    dim as uinteger m = 0
    while n
        if n mod 10 = d then m+=1
        n\=10
    wend
    return m
end function

for i as uinteger = 1 to 999
    if numdig(i, 1) = 2 then print i;"  ";
next i
print

Output:

11  101  110  112  113  114  115  116  117  118  119  121  131  141  151  161  171  181  191  211  311  411  511  611  711  811  911

Go

Translation of: Wren

Library: Go-rcu

package main

import (
    "fmt"
    "rcu"
)

func main() {
    fmt.Println("Decimal numbers under 1,000 whose digits include two 1's:")
    var results []int
    for i := 11; i <= 911; i++ {
        digits := rcu.Digits(i, 10)
        count := 0
        for _, d := range digits {
            if d == 1 {
                count++
            }
        }
        if count == 2 {
            results = append(results, i)
        }
    }
    for i, n := range results {
        fmt.Printf("%5d", n)
        if (i+1)%7 == 0 {
            fmt.Println()
        }
    }
    fmt.Println("\n\nFound", len(results), "such numbers.")
}

Output:

Decimal numbers under 1,000 whose digits include two 1's:
   11  101  110  112  113  114  115
  116  117  118  119  121  131  141
  151  161  171  181  191  211  311
  411  511  611  711  811  911

Found 27 such numbers.

J

Generating the numbers as the cartesian product of the eligible digits (and sorting the result, for "prettiness"):

   ~./:~10 #. >,{~.(2 1#1;i.10){~(! A.&i. ])3
11 101 110 111 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911

jq

Works with: jq

Works with gojq, the Go implementation of jq

This entry contains two solutions: the first is fast for this problem, but algorithmically inefficient; the second uses `countEquals` so candidates can be discarded quickly. The output is the same in both cases and so is shown only once.

Using `count`

def count(s): reduce s as $x (0; .+1);

range(1;1000)
| select( tostring | explode | 2 == count( select(.[] == 49))) # "1"

Using `countEquals`

def countEquals($n; s):
  label $out
  | foreach (s, null) as $x (-1;
      . + 1;
      if $x == null then . == $n
      elif . > $n then false, break $out
      else empty
      end);
      
range(1;1000)
| select( tostring
          | countEquals(2; select(explode[] == 49))) # "1"

Output:

Julia

totalddigisn(x, d = 1, n = 2; base=10) = count(j -> j == d, digits(x; base)) == n

println(filter(totalddigisn, 1:1000))

Output:

[11, 101, 110, 112, 113, 114, 115, 116, 117, 118, 119, 121, 131, 141, 151, 161, 171, 181, 191, 211, 311, 411, 511, 611, 711, 811, 911]

Mathematica / Wolfram Language

sol = Cases[Range[999], _?(Count[IntegerDigits@#, 1] == 2 &)];
Partition[sol, Max[FactorInteger[Length@sol][[All, 1]]]] // TableForm

Output:

11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911

Ksh

#!/bin/ksh

# Positive decimal integers with the digit 1 occurring twice

#	# Variables:
#
integer MAX=999

#	# Functions:
#


 ######
# main #
 ######

for ((i=10; i<MAX; i++)); do
	[[ ${i} == *{2}(1)* ]] && printf "%d " ${i}
done
echo

Output:

11 110 111 112 113 114 115 116 117 118 119 211 311 411 511 611 711 811 911

MiniZinc

%Permutations with some identical elements. Nigel Galloway: July 6th., 2021
include "count.mzn";
array [1..3] of var 0..9: N; constraint count(N,1,2);
output [show((N[1]*100)+(N[2]*10)+N[3])]

Output:

110
----------
101
----------
11
----------
211
----------
311
----------
411
----------
511
----------
611
----------
711
----------
811
----------
911
----------
121
----------
131
----------
141
----------
151
----------
161
----------
171
----------
181
----------
191
----------
112
----------
113
----------
114
----------
115
----------
116
----------
117
----------
118
----------
119
----------
==========
Finished in 206msec

Nim

import sugar, strutils

let result = collect(newSeq):
               for n in 1..<1000:
                 if count($n, '1') == 2: n
echo "Found ", result.len, " numbers:"
echo result.join(" ")

Output:

Found 27 numbers:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911

Pascal

program positiveDecimalIntegersWithTheDigit1occurringExactlyTwice(output);
var
	n: integer;
begin
	for n := 1 to 999 do
	begin
		if ord(n mod 10 = 1) + ord(n mod 100 div 10 = 1) + ord(n div 100 = 1) = 2 then
		begin
			writeLn(n)
		end
	end
end.

Perl

#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Numbers_n_in_which_number_1_occur_twice
use warnings;

my @twoones = grep tr/1// =~ 2, 1 .. 1000;
print "@twoones\n" =~ s/.{60}\K /\n/gr;

Output:

11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161
171 181 191 211 311 411 511 611 711 811 911

Phix

function two_ones(string s) return length(find_all('1',s))=2 end function
sequence res = filter(apply(tagset(999),sprint),two_ones)
printf(1,"%d found:\n  %s\n",{length(res),join_by(res,1,9," ","\n ")})

Output:

27 found:
  11 101 110 112 113 114 115 116 117
 118 119 121 131 141 151 161 171 181
 191 211 311 411 511 611 711 811 911

Pike

int main() {
	int limit = 1000;

	for(int i = 0; i < limit + 1; i++) {
			if(String.count((string)i, "1") == 2) {
				write((string)i + " ");
			}
		}
	write("\n");

	return 0;
}

Output:

11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911

PL/M

Works with: 8080 PL/M Compiler

... under CP/M (or an emulator)

100H: /* FIND NUMBERS BELOW 1000 WHERE 1 OCCURS TWICE IN THIER DECIMAL      */
      /* REPRESENTATION                                                     */

   /* CP/M SYSTEM CALL AND I/O ROUTINES                                     */
   BDOS:      PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
   PR$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C );  END;
   PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S );  END;
   PR$NL:     PROCEDURE;   CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
   PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
      DECLARE N ADDRESS;
      DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
      V = N;
      W = LAST( N$STR );
      N$STR( W ) = '$';
      N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      DO WHILE( ( V := V / 10 ) > 0 );
         N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      END;
      CALL PR$STRING( .N$STR( W ) );
   END PR$NUMBER;

   /* TASK                                                                  */
   DECLARE ( I, V, COUNT, ONE$COUNT ) ADDRESS;
   COUNT = 0;
   DO I = 10 TO 999;
      V = I;
      ONE$COUNT = 0;
      DO WHILE V > 0;
         IF V MOD 10 = 1 THEN DO;
            ONE$COUNT = ONE$COUNT + 1;
         END;
         V = V / 10;
      END;
      IF ONE$COUNT = 2 THEN DO;
         CALL PR$CHAR( ' ' );
         IF I < 100 THEN CALL PR$CHAR( ' ' );
         CALL PR$NUMBER( I );
         IF ( COUNT := COUNT + 1 ) MOD 10 = 0 THEN CALL PR$NL;
      END;
   END;

EOF

Output:

  11 101 110 112 113 114 115 116 117 118
 119 121 131 141 151 161 171 181 191 211
 311 411 511 611 711 811 911

PROMAL

;;; find numbers below 1000 where 1 occurs twice in their decimal representation
PROGRAM oneTwice 
INCLUDE LIBRARY

WORD i
WORD v
WORD count
WORD oneCount

BEGIN
count = 0
FOR i = 10 TO 999
  v = i
  oneCount = 0
  WHILE v > 0
    IF v % 10 = 1
      oneCount = oneCount + 1
    v = v / 10
  IF oneCount = 2
    OUTPUT " #3I", i
    count = count + 1
    IF count % 10 = 0
      OUTPUT "#C"
END

Output:

  11 101 110 112 113 114 115 116 117 118
 119 121 131 141 151 161 171 181 191 211
 311 411 511 611 711 811 911

Python

#Aamrun, 5th October 2021

from itertools import permutations

for i in range(0,10):
    if i!=1:
        baseList = [1,1]
        baseList.append(i)
        [print(int(''.join(map(str,j)))) for j in sorted(set(permutations(baseList)))]

Output:

Quackery

  [ 0 swap
    [ dup 0 != while
      10 /mod 1 = if
        [ dip 1+ ]
      again ] 
    drop 2 = ]        is two-1s ( n --> b )

  [] 1000 times
    [ i^ two-1s if
        [ i^ join ] ] 
  echo

Output:

[ 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 ]

Raku

say display 10, '%3d', ^1000 .grep: { .comb.Bag{'1'} == 2 };

sub display {
    cache $^c;
    "{+$c} matching:\n" ~ $c.batch($^a)».fmt($^b).join: "\n"
}

Yields:

27 matching:
 11 101 110 112 113 114 115 116 117 118
119 121 131 141 151 161 171 181 191 211
311 411 511 611 711 811 911

REXX

version 1

Programming note: the three filters (marked below) could've been incorporated into one REXX statement if code golf is the desired goal.

/*REXX program finds  positive decimal integers  which contain exactly two  ones  (1s). */
parse arg hi cols .                              /*obtain optional argument from the CL.*/
if   hi=='' |   hi==","  then   hi= 1000         /*Not specified?  Then use the default.*/
if cols=='' | cols==","  then cols=   10         /* "      "         "   "   "     "    */
w= 10                                            /*width of a number in any column.     */
title= ' positive decimal integers which contain exactly two ones (1s)  which are  <'  hi
say ' index │'center(title,  1 + cols*(w+1)     )
say '───────┼'center(""   ,  1 + cols*(w+1), '─')
found= 0;                    idx= 1              /*initialize # integers and the index. */
$=                                               /*a list of integers found  (so far).  */
     do j=1  for  hi-1                           /*find positive integers within range. */
     p= pos(1, j);       if p==0  then iterate   /*integer doesn't have a one (1)? Skip.*/       /* ◄■■■■■■■ a filter.*/
     p= pos(1, j, p+1);  if p==0  then iterate   /*   "       "      "  a 2nd one?   "  */       /* ◄■■■■■■■ a filter.*/
     p= pos(1, j, p+1);  if p>0   then iterate   /*   "      does    "  a 3rd one?   "  */       /* ◄■■■■■■■ a filter.*/
     found= found + 1                            /*bump the number of integers found.   */
     $= $ right(j, w)                            /*add an integer to the ──►  $  list.  */
     if found//cols\==0           then iterate   /*have we populated a line of output?  */
     say center(idx, 7)'│'  substr($, 2);   $=   /*display what we have so far  (cols). */
     idx= idx + cols                             /*bump the  index  count for the output*/
     end   /*j*/
                                                 /*stick a fork in it,  we're all done. */
if $\==''  then say center(idx, 7)"│"  substr($, 2)  /*possible display residual output.*/
say '───────┴'center(""   ,  1 + cols*(w+1), '─')
say
say 'Found '       found      title

output when using the default inputs:

 index │                positive decimal integers which contain exactly two ones (1s)  which are  < 1000
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
   1   │         11        101        110        112        113        114        115        116        117        118
  11   │        119        121        131        141        151        161        171        181        191        211
  21   │        311        411        511        611        711        811        911
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────

Found  27  positive decimal integers which contain exactly two ones (1s)  which are  < 1000

version 2

Programming note: not all REXXes have the countstr BIF.

/*REXX program finds  positive decimal integers  which contain exactly two  ones  (1s). */
parse arg hi cols .                              /*obtain optional argument from the CL.*/
if   hi=='' |   hi==","  then   hi= 1000         /*Not specified?  Then use the default.*/
if cols=='' | cols==","  then cols=   10         /* "      "         "   "   "     "    */
w= 10                                            /*width of a number in any column.     */
title= ' positive decimal integers which contain exactly two ones (1s)  which are  <'  hi
say ' index │'center(title,  1 + cols*(w+1)     )
say '───────┼'center(""   ,  1 + cols*(w+1), '─')
found= 0;                    idx= 1              /*initialize # integers and the index. */
$=                                               /*a list of integers found  (so far).  */
     do j=1  for  hi-1                           /*find positive integers within range. */
     if countstr(1, j)\==2  then iterate         /*Doesn't have exactly 2 one's?  Skip. */       /* ◄■■■■■■■ a filter.*/
     found= found + 1                            /*bump the number of integers found.   */
     $= $ right(j, w)                            /*add an integer to the ──►  $  list.  */
     if found//cols\==0     then iterate         /*have we populated a line of output?  */
     say center(idx, 7)'│'  substr($, 2);   $=   /*display what we have so far  (cols). */
     idx= idx + cols                             /*bump the  index  count for the output*/
     end   /*j*/
                                                 /*stick a fork in it,  we're all done. */
if $\==''  then say center(idx, 7)"│"  substr($, 2)  /*possible display residual output.*/
say '───────┴'center(""   ,  1 + cols*(w+1), '─')
say
say 'Found '       found      title

output is identical to the 1^st REXX version.

Ring

load "stdlib.ring"
see "working..." + nl
see "Numbers n in which number 1 occur twice:" + nl

row = 0
sum = 0
limit = 1000

for n = 1 to limit
    strn = string(n)
    ind = count(strn,"1")
    if ind = 2
       see "" + n + " " 
       row++
       if row%5 = 0
          see nl
       ok
    ok
next

see nl + "Found " + row + " numbers" + nl
see "done..." + nl

func count(cstring,dstring)
     sum = 0
     while substr(cstring,dstring) > 0
           sum++
           cstring = substr(cstring,substr(cstring,dstring)+len(string(sum)))
     end
     return sum

Output:

working...
Numbers n in which number 1 occur twice:
11 101 110 112 113 
114 115 116 117 118 
119 121 131 141 151 
161 171 181 191 211 
311 411 511 611 711 
811 911 
Found 27 numbers
done...

RPL

≪ →STR 0
     1 3 PICK SIZE FOR j
        OVER j DUP SUB "1" == + NEXT 
     SWAP DROP 2 ==
≫ 'ONLY2?' STO   

≪ { } 
   1 1000 FOR j
     IF j ONLY2? THEN j + END NEXT
≫ EVAL

Output:

1: { 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 }

Ruby

p (1..1000).select{|n| n.digits.count(1) == 2}

Output:

[11, 101, 110, 112, 113, 114, 115, 116, 117, 118, 119, 121, 131, 141, 151, 161, 171, 181, 191, 211, 311, 411, 511, 611, 711, 811, 911]

SETL

program two_ones;
    print([n : n in [1..999] | 2 = #[d : d in str n | val d = 1]]);
end program;

Output:

[11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911]

Sidef

say (1..1000 -> grep { .digits.count { _ == 1 } == 2 })

Output:

[11, 101, 110, 112, 113, 114, 115, 116, 117, 118, 119, 121, 131, 141, 151, 161, 171, 181, 191, 211, 311, 411, 511, 611, 711, 811, 911]

Wren

Library: Wren-math

Library: Wren-fmt

import "./math" for Int
import "./fmt" for Fmt

System.print("Decimal numbers under 1,000 whose digits include two 1's:")
var results = (11..911).where { |i| Int.digits(i).count { |d| d == 1 } == 2 }.toList
Fmt.tprint("$5d", results, 7)
System.print("\nFound %(results.count) such numbers.")

Output:

Decimal numbers under 1,000 whose digits include two 1's:
   11   101   110   112   113   114   115
  116   117   118   119   121   131   141
  151   161   171   181   191   211   311
  411   511   611   711   811   911

Found 27 such numbers.

XPL0

func Ones(N);           \Return count of 1's in N
int N, Count;
[Count:= 0;
repeat  N:= N/10;
        if rem(0) = 1 then Count:= Count+1;
until   N = 0;
return Count;
];

int  N, Count;
[for N:= 1 to 1000-1 do
    if Ones(N) = 2 then
        [IntOut(0, N);
        Count:= Count+1;
        if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\);
        ];
CrLf(0);
IntOut(0, Count);
Text(0, " such numbers found below 1000.
");
]

Output:

11      101     110     112     113     114     115     116     117     118
119     121     131     141     151     161     171     181     191     211
311     411     511     611     711     811     911     
27 such numbers found below 1000.