Positive decimal integers with the digit 1 occurring exactly twice

Positive decimal integers with the digit 1 occurring exactly twice 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.

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:
```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```

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:
```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
```

```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:
``` 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```

See Pascal

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`

See Pascal

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:
```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
```

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 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;

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

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 1st 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-seq
Library: Wren-fmt
```import "/math" for Int
import "/seq" for Lst
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
for (chunk in Lst.chunks(results, 7)) Fmt.print("\$5d", chunk)
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.
```