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Positive decimal integers with the digit 1 occurring exactly twice

From Rosetta Code
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.
Task

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

Action![edit]

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

ALGOL 68[edit]

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[edit]

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

AWK[edit]

 
# 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

F#[edit]

 
// 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

</pre>

Factor[edit]

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 

Go[edit]

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.

jq[edit]

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[edit]

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]

Ksh[edit]

 
#!/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[edit]

 
%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[edit]

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[edit]

Free Pascal[edit]

 
program OneOne;
const
SUBSTITUTES ='023456789';
ONES = '111';
var
col,idx: integer;
new : string[3];
begin
For col := 1 to 3 do
begin
new := ONES;
For idx := 1 to length(SUBSTITUTES) do
begin
new[col] := SUBSTITUTES[idx];
write(new:4);
end;
writeln;
end;
end.
Output:
 011 211 311 411 511 611 711 811 911
 101 121 131 141 151 161 171 181 191
 110 112 113 114 115 116 117 118 119

Perl[edit]

#!/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[edit]

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

Python[edit]

 
#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

Raku[edit]

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[edit]

version 1[edit]

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[edit]

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[edit]

 
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...

Sidef[edit]

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[edit]

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[edit]

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.