Positive decimal integers with the digit 1 occurring exactly twice
- Task
- Find numbers n in which number 1 occur twice, where n < 1,000
ALGOL 68
Generates the numbers. In order to print them in order, a table of double 1 numbers yes/no is generated. <lang algol68>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</lang>
- 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
<lang go>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.")
}</lang>
- 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.
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
Raku
<lang perl6>say display 10, '%3d', ^1000 .grep: { .comb.Bag{'1'} == 2 };
sub display {
cache $^c;
"{+$c} matching:\n" ~ $c.batch($^a)».fmt($^b).join: "\n"
}</lang>
- 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
Ring
<lang 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 = row + 1
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 = sum + 1
cstring = substr(cstring,substr(cstring,dstring)+len(string(sum)))
end
return sum
</lang>
- 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...
Wren
<lang ecmascript>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.")</lang>
- 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
<lang 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. "); ]</lang>
- 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.