Largest five adjacent number
- Task
Generate random 1000-digit number.
Find the five adjacent digits in the 1000-digit number that form the largest 5-digit number.
- Extra credit
Find the five adjacent digits in the 1000-digit number that form the smallest 5-digit number.
11l
V largeNum = [random:(1..9)] [+] (0.<999).map(i -> random:(0..9))
V (maxNum, minNum) = (0, 99999)
L(i) 996
V num = Int(largeNum[i.+5].join(‘’))
I num > maxNum
maxNum = num
E I num < minNum
minNum = num
print(‘Largest 5-digit number extracted from random 1000-digit number: ’maxNum)
print(‘Smallest 5-digit number extracted from random 1000-digit number: #05’.format(minNum))
- Output:
Largest 5-digit number extracted from random 1000-digit number: 99902 Smallest 5-digit number extracted from random 1000-digit number: 00043
Ada
with Ada.Text_Io;
with Ada.Numerics.Discrete_Random;
procedure Adjacent_Numbers is
Adjacent_Length : constant := 5;
subtype Digit is Character range '0' .. '9';
package Random_Digits
is new Ada.Numerics.Discrete_Random (Digit);
Gen : Random_Digits.Generator;
Line : String (1 .. 1000);
Large : Natural := Natural'First;
Small : Natural := Natural'Last;
begin
Random_Digits.Reset (Gen);
Line := (others => Random_Digits.Random (Gen));
for I in Line'First .. Line'Last - Adjacent_Length + 1 loop
declare
Window : String renames Line (I .. I + Adjacent_Length - 1);
begin
Large := Natural'Max (Large, Natural'Value (Window));
Small := Natural'Min (Small, Natural'Value (Window));
end;
end loop;
Ada.Text_Io.Put_Line ("The largest number : " & Natural'Image (Large));
Ada.Text_Io.Put_Line ("The smallest number: " & Natural'Image (Small));
end Adjacent_Numbers;
- Output:
The largest number : 99625 The smallest number: 102
ALGOL 68
Adding the minimum number for good measure...
BEGIN # generate 1000 random digits and find the largest/smallest numbers formed from 5 consecutive digits #
[ 1 : 1000 ]CHAR digits;
FOR i TO UPB digits DO digits[ i ] := REPR ( ENTIER ( next random * 10 ) + ABS "0" ) OD;
STRING max number := digits[ 1 : 5 ];
STRING min number := digits[ 1 : 5 ];
FOR i FROM 2 TO UPB digits - 4 DO
STRING next number = digits[ i : i + 4 ];
IF next number > max number
THEN
# found a new higher number #
max number := next number
FI;
IF next number < min number
THEN
# found a new lower number #
min number := next number
FI
OD;
print( ( "Largest 5 consecutive digits from 1000 random digits: ", max number, newline ) );
print( ( "Smallest 5 consecutive digits from 1000 random digits: ", min number, newline ) )
END
- Output:
Largest 5 consecutive digits from 1000 random digits: 99987 Smallest 5 consecutive digits from 1000 random digits: 00119
APL
⌈/((⊣+10×⊢)/(⌽↓))⌺5⊢(-⎕IO)+?1000/10
- Output:
(example)
99994
Arturo
N: join to [:string] map 1..100 'x -> random 1000000000 9999999999
i: 0
maxNum: 0
minNum: ∞
while [i < (size N)-5][
num: to :integer join @[N\[i] N\[i+1] N\[i+2] N\[i+3] N\[i+4]]
if num > maxNum -> maxNum: num
if num < minNum -> minNum: num
i: i + 1
]
print "Our random 1000-digit number is:"
print N
print ""
print ["Max 5-adjacent number found:" maxNum]
print ["Min 5-adjacent number found:" (repeat "0" 5-(size to :string minNum)) ++ (to :string minNum)]
- Output:
Our random 1000-digit number is: 2540956677308157418624519953263471599696918276171651168484519407031160813613006352660058588944602704848634276542837184618726044674117357036813240557325769932073351534364289297094415941273117151277729576200542643185699525405079189015204192029912043004161916366921458912887890652627268028071729897387395041640352395354106991129061548748712499227024213135531365974620993813773921850969630855401781344832397898392812417729744785629765286216304456806870691502938136795922685099816652448188701308354551593078486609811394420601431484916913833634669083737749230355341380266781803894385432741405633278873213701238310761908151961510643290964548205746238459266137202173265468217401777681775761126374654289733873900330799576500024068191362342162163615972164105625935627483920193464168192083262176697432155066174175594837721476581087940310642712981291006889657297350894628612724944063786324456854104801432247483498384207351647946918119868105898645178074174003550762101547842674605061792172905254724197215648686667 Max 5-adjacent number found: 99816 Min 5-adjacent number found: 00024
AutoHotkey
maxNum := 0, str := ""
loop, 1000
{
Random, rnd, 0, 9
str .= rnd
output .= rnd . (Mod(A_Index, 148) ? "" : "`n")
if A_Index < 5
continue
num := SubStr(str, A_Index-4, 5)
maxNum := maxNum > num ? maxNum : num
minNum := A_Index = 5 ? num : minNum < num ? minNum : num
}
MsgBox % result := output "`n`nLargest five adjacent digits = " maxNum
. "`n`nSmallest five adjacent digits = " minNum
- Output:
3893212622395522104846091986776081862634026945871752892124324578621089065097043281907406149009719673003318226562809101957181871693776164191416491334 2509291361707848297387923254298547833186351133036771338719578505791529263806019240009497155124458943732581184022226943392528107498748575424217651885 3083736872582691290721469942482918430078673685947447234032602113276631102983248999047362916320523840282929255314468323644427797630259187509914424396 1523615571637081320270791095221484894567420630155741441396012393172769867922862248399483054652921274863786220527596050784952102267710198517665662903 6335615800351254988779849447078262460051794249274045128158246939351902901862546960248213286880570086476859341012102414828750098051948784732121573660 9618754338433412518619240496583375235634416473003920360759949694724646721954909867058588446320222792637823988375313876167705092153587245148819122980 2777308429997046827297505483667631338885207838402941712216614732232703459440770039141898763110002290662921501156 Largest five adjacent digits = 99970 Smallest five adjacent digits = 00022
AWK
# syntax: GAWK -f LARGEST_FIVE_ADJACENT_NUMBER.AWK
BEGIN {
limit = 1000
width = 5
max_n = 0
for (i=1; i<=width; i++) {
min_n = min_n "9"
}
srand()
for (i=1; i<=limit; i++) {
digits = digits int(rand() * 10)
}
for (i=1; i<=limit-width+1; i++) {
n = substr(digits,i,width)
if (n > max_n) {
max_n = n
max_pos = i
}
if (n < min_n) {
min_n = n
min_pos = i
}
}
printf("look for %d digit number using %d digits\n",width,limit)
printf("largest %0*d in positions %d-%d\n",width,max_n,max_pos,max_pos+width-1)
printf("smallest %0*d in positions %d-%d\n",width,min_n,min_pos,min_pos+width-1)
exit(0)
}
- Output:
look for 5 digit number using 1000 digits largest 99873 in positions 300-304 smallest 00099 in positions 697-701
BASIC
BASIC256
dim number(1000)
highest = 0
lowest = 100000
for i = 0 to 999
number[i] = int(rand*10)
if i >= 4 then
tmp = number[i] + 10*number[i-1] + 100*number[i-2] + 1000*number[i-3] + 10000*number[i-4]
if tmp < lowest then lowest = tmp
if tmp > highest then highest = tmp
end if
next i
print highest, lowest
Chipmunk Basic
100 randomize timer
110 dim number(999)
120 highest = 0
130 lowest = 100000
140 for i = 0 to 999
150 number(i) = int(rnd(10))
160 if i >= 4 then
170 tmp = number(i)+10*number(i-1)+100*number(i-2)+1000*number(i-3)+10000*number(i-4)
180 if tmp < lowest then lowest = tmp
190 if tmp > highest then highest = tmp
200 endif
210 next i
220 print highest,lowest
230 end
Gambas
Public number[1000] As Byte
Public Sub Main()
Randomize
Dim tmp As Integer, highest As Integer = 0, lowest As Integer = 100000
For i As Integer = 0 To 999
number[i] = Int(Rnd(10))
If i >= 4 Then
tmp = number[i] + 10 * number[i - 1] + 100 * number[i - 2] + 1000 * number[i - 3] + 10000 * number[i - 4]
If tmp < lowest Then lowest = tmp
If tmp > highest Then highest = tmp
End If
Next
Print highest, lowest
End
PureBasic
OpenConsole()
Dim number.i(999)
highest.i = 0
lowest.i = 100000
For i.i = 0 To 999
number(i) = Random(10)
If i >= 4:
tmp = number(i) + 10*number(i-1) + 100*number(i-2) + 1000*number(i-3) + 10000*number(i-4)
If tmp < lowest: lowest = tmp: EndIf
If tmp > highest: highest = tmp: EndIf
EndIf
Next i
PrintN(Str(highest) + #TAB$ + Str(lowest))
QBasic
RANDOMIZE TIMER
DIM number(0 TO 999)
highest = 0
lowest = 100000
FOR i = 0 TO 999
number(i) = INT(RND * 10)
IF i >= 4 THEN
tmp = number(i) + 10 * number(i - 1) + 100 * number(i - 2) + 1000 * number(i - 3) + 10000 * number(i - 4)
IF tmp < lowest THEN lowest = tmp
IF tmp > highest THEN highest = tmp
END IF
NEXT i
PRINT highest, lowest
END
True BASIC
RANDOMIZE
DIM number(0 TO 999)
LET highest = 0
LET lowest = 100000
FOR i = 0 TO 999
LET number(i) = INT(RND*10)
IF i >= 4 THEN
LET tmp = number(i)+10*number(i-1)+100*number(i-2)+1000*number(i-3)+10000*number(i-4)
IF tmp < lowest THEN LET lowest = tmp
IF tmp > highest THEN LET highest = tmp
END IF
NEXT i
PRINT highest, lowest
END
Yabasic
dim number(999)
highest = 0
lowest = 100000
for i = 0 to 999
number(i) = int(ran(10))
if i >= 4 then
tmp = number(i) + 10*number(i-1) + 100*number(i-2) + 1000*number(i-3) + 10000*number(i-4)
if tmp < lowest lowest = tmp
if tmp > highest highest = tmp
fi
next i
print highest, lowest
BQN
⌈´(⊣+10×⊢)˝⌽⍉5↕1000 •rand.Range 10
- Output:
(example)
99991
C
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <time.h>
#define DIGITS 1000
#define NUMSIZE 5
uint8_t randomDigit() {
uint8_t d;
do {d = rand() & 0xF;} while (d >= 10);
return d;
}
int numberAt(uint8_t *d, int size) {
int acc = 0;
while (size--) acc = 10*acc + *d++;
return acc;
}
int main() {
uint8_t digits[DIGITS];
int i, largest = 0;
srand(time(NULL));
for (i=0; i<DIGITS; i++) digits[i] = randomDigit();
for (i=0; i<DIGITS-NUMSIZE; i++) {
int here = numberAt(&digits[i], NUMSIZE);
if (here > largest) largest = here;
}
printf("%d\n", largest);
return 0;
}
- Output:
(example)
99931
CLU
% Generate a number of N random digits
random_digits = proc (n: int) returns (sequence[int])
digits: array[int] := array[int]$predict(1,n)
% A number never starts with a zero
array[int]$addh(digits, 1+random$next(9))
for i: int in int$from_to(1,n-1) do
array[int]$addh(digits, random$next(10))
end
return(sequence[int]$a2s(digits))
end random_digits
% Find the largest and smallest N-adjacent number in the digits
find_min_max = proc (n: int, digits: sequence[int]) returns (int,int)
min: int := 10**n % Guaranteed to be bigger than any N-adjacent number
max: int := 0
for i: int in int$from_to(1, sequence[int]$size(digits)-n) do
cur: int := 0
for j: int in int$from_to(0, n-1) do
cur := 10*cur + digits[i+j]
end
if cur<min then min:=cur end
if cur>max then max:=cur end
end
return(min, max)
end find_min_max
start_up = proc ()
% Seed the RNG with the current time
d: date := now()
random$seed(d.second + 60*(d.minute + 60*d.hour))
% Find the minimum and maximum 5-adjacent numbers in a 1000-digit number
min, max: int := find_min_max(5, random_digits(1000))
po: stream := stream$primary_output()
stream$putl(po, "Smallest: " || int$unparse(min))
stream$putl(po, "Largest: " || int$unparse(max))
end start_up
- Output:
Smallest: 144 Largest: 99951
Delphi
function Get5DigitNumber(S: string; Off: integer): integer;
{Extract 5 digit number from string at Off}
var I: integer;
var NS: string;
begin
NS:=Copy(S,Off,5);
Result:=StrToIntDef(NS,-1);
end;
function BreakupString(S: string): string;
{Breakup thousand digit number for easy display}
var I: integer;
begin
for I:=1 to Length(S) do
begin
Result:=Result+S[I];
if (I mod 55)=0 then Result:=Result+#$0D#$0A;
end;
end;
procedure FiveDigitNumber(Memo: TMemo);
{Find the largest and small 5 digit sequence}
{in 1000 digit number}
var S: string;
var N,I: integer;
var Largest,Smallest: integer;
begin
Smallest:=High(Integer);
Largest:=0;
for I:=1 to 1000 do
S:=S+Char(Random(10)+$30);
for I:=1 to Length(S)-5 do
begin
N:=Get5DigitNumber(S,I);
if N>Largest then Largest:=N;
if N<Smallest then Smallest:=N;
end;
Memo.Lines.Add(BreakupString(S));
Memo.Lines.Add('Largest: '+IntToStr(Largest));;
Memo.Lines.Add('Smallest: '+IntToStr(Smallest));;
end;
- Output:
0082263134040802937368731342824182794880115050767752659 6926207485596307977119758620628125911215421677000178364 7438810001625238336693427757455861441056098692774612931 9301856160395349334087194184285169534216966507128749101 0333045468523586265833674268791722749102838792380205401 7335212073765802860114410575280403628540910018912794058 9569778977033072890894634763659190635686944921467068416 0978402580498879216810854417805724457730620420683349740 8203884243646784563247619038458645194136841413688117232 0960606571886477139587251334596793042923055521495533796 5592094928040937883628134090110628164451939278452734493 5741344340195488542852682604882967292438604245256357719 4755578568409079269700382959730067457921191314413220282 3502307407547002586284406642530858066838890257743184196 5040611036453640792847940715686736822030381083124941163 3588177613294220880152655471721880286144478485085399563 1095924640071825166992021998152653370680394470682198029 3879102724160697653653330275506532525946257246355415772 4978409544 Largest: 99815 Smallest: 16
F#
// Largest five adjacent number. Nigel Galloway: September 28th., 2021
let n()=let n()=System.Random().Next(10) in Seq.unfold(fun g->Some(g,(g%10000)*10+n()))(n()*10000+n()*1000+n()*100+n()*10+n())
printfn $"Largest 5 adjacent digits are %d{(n()|>Seq.take 995|>Seq.max)}"
- Output:
Largest 5 adjacent digits are 99914
EasyLang
for i to 1000
n$ &= random 10 - 1
.
min = 99999
for i = 1 to 995
n = number substr n$ i 5
min = lower min n
max = higher max n
.
print min & " " & max
- Output:
21 99768
Factor
USING: grouping io math.functions prettyprint random sequences ;
1000 10^ random unparse 5 <clumps> supremum print
- Output:
99987
FreeBASIC
Generate the number digit by digit, and test as we go. If the task didn't specifically ask to generate the whole 1,000 digit number I wouldn't bother storing more than five of its digits at a time.
randomize timer
dim as ubyte number(0 to 999)
dim as uinteger seg, highest = 0, lowest = 100000
for i as uinteger = 0 to 999
number(i) = int(rnd*10)
if i >= 4 then
seg = number(i) + 10*number(i-1) + 100*number(i-2) +_
1000*number(i-3) + 10000*number(i-4)
if seg < lowest then lowest = seg
if seg > highest then highest = seg
end if
next i
print highest, lowest
- Output:
99748 31
Go
package main
import (
"fmt"
"math/rand"
"rcu"
"strings"
"time"
)
func main() {
rand.Seed(time.Now().UnixNano())
var sb strings.Builder
for i := 0; i < 1000; i++ {
sb.WriteByte(byte(rand.Intn(10) + 48))
}
number := sb.String()
for i := 99999; i >= 0; i-- {
quintet := fmt.Sprintf("%05d", i)
if strings.Contains(number, quintet) {
ci := rcu.Commatize(i)
fmt.Printf("The largest number formed from 5 adjacent digits (%s) is: %6s\n", quintet, ci)
break
}
}
for i := 0; i <= 99999; i++ {
quintet := fmt.Sprintf("%05d", i)
if strings.Contains(number, quintet) {
ci := rcu.Commatize(i)
fmt.Printf("The smallest number formed from 5 adjacent digits (%s) is: %6s\n", quintet, ci)
return
}
}
}
- Output:
Sample run:
The largest number formed from 5 adjacent digits (99928) is: 99,928 The smallest number formed from 5 adjacent digits (00120) is: 120
J
>./5([+10*])/@|:\?1000#10
- Output:
(example)
99929
jq
Also works with gojq, the Go implementation of jq.
First, a direct solution using only jq's standard library and a line for generating the PRN:
< /dev/random tr -cd '0-9' | head -c 1000 | jq -R '
length as $n
| . as $s
| ($s[0:5] | tonumber) as $m
| reduce range(1; $n - 5) as $i ( {min: $m, max: $m};
($s[$i: $i+5] | tonumber) as $x
| if $x < .min then .min = $x
elif $x > .max then .max = $x
else . end)
'
- Output:
{ "min": 224, "max": 99772 }
Next, a "one-line solution" apart from generic helper functions and the line for generating the PRN:
< /dev/random tr -cd '0-9' | head -c 1000 | jq -R '
# Input: an array
# Output: a stream of the width-long subarrays
def windows(width):
range(0; 1 + length - width) as $i | .[$i:$i+width];
def minmax(s):
reduce s as $x ( {};
if .min == null then {min: $x, max: $x}
elif $x < .min then .min = $x
elif $x > .max then .max = $x else . end);
explode | minmax(windows(5) | implode | tonumber)
Julia
dig = rand(0:9, 1000)
@show maximum(evalpoly(10, dig[i:i+4]) for i in 1:length(dig)-4)
- Output:
maximum((evalpoly(10, dig[i:i + 4]) for i = 1:length(dig) - 4)) = 99993
Or, using strings, and see Nigel's comment in the discussion:
julia> setprecision(3324) 3324 julia> s = string(BigFloat(pi))[3:end] "141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365" ⋯ 180 bytes ⋯ "66940513200056812714526356082778577134275778960917363717872146844090122495343014654958537105079227968925892354201995611212902196086403441815981362977477130996051870721134999999837297804995105973173281609631859502445945534690830264252230825334468503526193118817101000313783875288658753320838142061717766914730359825349042875546873115956286388235378759375195778185778053217122680661300192787661119590921642019898" julia> m, pos = maximum((s[i:i+4], i) for i in 1:length(s)-4) ("99999", 763) julia> println("Maximum is $m at position $pos.") Maximum is 99999 at position 763.
Mathematica / Wolfram Language
MinMax[FromDigits /@ Partition[RandomInteger[{0, 9}, 1000], 5, 1]]
- Output:
{104,99984}
Nim
import random, strutils
randomize()
const N = 1000
type Digit = 0..9
# Build a 1000-digit number.
var number: array[1..N, Digit]
number[1] = rand(1..9) # Make sure the first digit is not 0.
for i in 1..N: number[i] = rand(9)
func `>`(s1, s2: seq[Digit]): bool =
## Compare two digit sequences.
## Defining `<` rather than `>` would work too.
assert s1.len == s2.len
for i in 0..s1.high:
let comp = cmp(s1[i], s2[i])
if comp != 0: return comp == 1
result = false
var max = @[Digit 0, 0, 0, 0, 0]
for i in 5..N:
let n = number[i-4..i]
if n > max: max = n
echo "Largest 5-digit number extracted from random 1000-digit number: ", max.join()
- Output:
Largest 5-digit number extracted from random 1000-digit number: 99855
Pascal
inspired by Wren
Assumes that there at least is a "1" 4 digits before end of all digits.Else I have to include sysutils and s := Format('%.5d',[i]); for leading zeros.
var
digits,
s : AnsiString;
i : LongInt;
begin
randomize;
setlength(digits,1000);
for i := 1 to 1000 do
digits[i] := chr(random(10)+ord('0'));
for i := 99999 downto 0 do
begin
str(i:5,s);
if Pos(s,digits) > 0 then
break;
end;
writeln(s, ' found as largest 5 digit number ')
end.
- Output:
99889 found as largest 5 digit number
Perl
#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Largest_five_adjacent_number
use warnings;
$_ = join '', map int rand 10, 1 .. 1e3;
my @n;
@n[ /(?=(\d{5}))/g ] = ();
print "$#n\n";
- Output:
99899
Phix
with javascript_semantics procedure shlong(string s) string hi5 = s[1..5], lo5 = hi5 for i=2 to length(s)-4 do string s5 = s[i..i+4] hi5 = max(hi5,s5) lo5 = min(lo5,s5) end for printf(1,"String %s: hi5:%s, lo5:%s\n",{shorten(s),hi5,lo5}) end procedure string s = repeat(' ',1000) for i=1 to length(s) do s[i] = rand(10)-1+'0' end for shlong(s) include mpfr.e mpfr pi = mpfr_init(0,-1001) -- (set precision to 1,000 dp, plus the "3.") mpfr_const_pi(pi) s = mpfr_get_fixed(pi,1000) s = s[3..$] shlong(s)
- Output:
String 35369847249221789712...55814915156742014134 (1,000 digits): hi5:99969, lo5:00013 String 14159265358979323846...66111959092164201989 (1,000 digits): hi5:99999, lo5:00031
Python
Seeding the random number generator directly with the datetime stamp produces a warning that it will be deprecated in Python 3.9, hence the "hack" of creating a string out of the timestamp and then seeding with it.
#Aamrun, 5th October 2021
from random import seed,randint
from datetime import datetime
seed(str(datetime.now()))
largeNum = [randint(1,9)]
for i in range(1,1000):
largeNum.append(randint(0,9))
maxNum,minNum = 0,99999
for i in range(0,994):
num = int("".join(map(str,largeNum[i:i+5])))
if num > maxNum:
maxNum = num
elif num < minNum:
minNum = num
print("Largest 5-adjacent number found ", maxNum)
print("Smallest 5-adjacent number found ", minNum)
Results from multiple runs :
- Output:
Largest 5-adjacent number found 99743 Smallest 5-adjacent number found 102 Largest 5-adjacent number found 99965 Smallest 5-adjacent number found 84 Largest 5-adjacent number found 99808 Smallest 5-adjacent number found 58 Largest 5-adjacent number found 99938 Smallest 5-adjacent number found 10 Largest 5-adjacent number found 99957 Smallest 5-adjacent number found 35
Quackery
9 random 1+
999 times [ 10 * 10 random + ]
dup 10 995 ** / swap
say "1000 digit number"
cr cr
dup echo
cr cr
[] swap 1000 times
[ 10 /mod
rot join swap ]
drop dup
0 0 rot witheach
[ swap 10 *
100000 mod +
tuck max swap ]
drop
say "largest 5 adjacent digits " echo
cr cr
5 split nip
dip dup witheach
[ swap 10 *
100000 mod +
tuck min swap ]
drop
say "smallest 5 adjacent digits "
number$
char 0 over size 5 swap - of
swap join echo$
- Output:
1000 digit number 6840907174710710253578773992410923828010161316527489025709598588564725782830158923520744533291356763925463645174705745049218864529135157750600471363289558510223445011025163844199130052941524405130793922050669143532883592357897096269697903780509770222546659289832999639637730759831717125055857319129937934353617386529810429642261048827016148476352187592939822910964334104828550764225596939965675519243696921514153715258715961987394884393797714002723369560598384723111928648279375269590756880538160907807290640466592734345970439851284217252141914792365031610947925633607292897379320456985054219371373707477609843617810620097343420379245258762479642377134776965386535533204636182773979582543243782455626021964121509778973939346873293400502531060571761381532229278485105166678017234489439222625767334040651185482277484204647473910364297105035077787620562600454016296114868335345408156093266755340971022669397814048919735693462065796634326535292979494128432997646841467835174156471055078228524511787150409 largest 5 adjacent digits 99963 smallest 5 adjacent digits 00272
Raku
Show minimum too because... why not?
Use some Tamil Unicode numbers for brevity, and for amusement purposes.
௰ - Tamil number ten ௲ - Tamil number one thousand
Do it 5 times for variety, it's random after all.
(^௰).roll(௲).rotor(5 => -4)».join.minmax.bounds.put xx 5
- Sample output:
00371 99975 00012 99982 00008 99995 00012 99945 00127 99972
Ring
digit = ""
max = 0
min = 99999
limit = 1000
for n = 1 to limit
rand = random(9)
randStr = string(rand)
digit += randStr
next
for n = 1 to len(digit)-5
res = substr(digit,n,5)
resNum = number(res)
if resNum > max
max = resNum
ok
if resNum < min
min = res
ok
next
see "The largest number is:" + nl
see max + nl
see "The smallest number is:" + nl
see min + nl
- Output:
The largest number is: 99638 The smallest number is: 00118
RPL
RPL can not handle 1000-digit numbers, so we use a 1000-digit string.
≪ ""
1 1000 START RAND 9 * 0 RND →STR + NEXT
{ -99999 0 }
1 3 PICK SIZE 4 - FOR j
OVER j DUP 4 + SUB
STR→ NEG LASTARG 2 →LIST MAX
NEXT ABS
≫ 'TASK' STO
- Output:
2: "46725324552811522… 1: { 198 99886 }
Ruby
digits = %w(0 1 2 3 4 5 6 7 8 9)
arr = Array.new(1000){ digits.sample }
puts "minimum sequence %s, maximum sequence %s." % arr.each_cons(5).minmax_by{|slice| slice.join.to_i}.map(&:join)
- Output:
minimum sequence 00096, maximum sequence 99508.
Sidef
var k = 5
var n = 1e1000.irand
say "length(n) = #{n.len}"
var c = n.digits.cons(k)
say ("Min #{k}-digit sub-number: ", c.min_by { .digits2num }.flip.join)
say ("Max #{k}-digit sub-number: ", c.max_by { .digits2num }.flip.join)
- Output:
length(n) = 1000 Min 5-digit sub-number: 00072 Max 5-digit sub-number: 99861
V (Vlang)
import rand
import rand.seed
import strings
fn main() {
rand.seed(seed.time_seed_array(2))
mut sb := strings.new_builder(128)
for _ in 0..1000 {
sb.write_byte(u8(rand.intn(10) or {0} + 48))
}
number := sb.str()
println('>> $number')
for i := 99999; i >= 0; i-- {
quintet := "${i:05}"
if number.contains(quintet) {
println("The largest number formed from 5 adjacent digits ($quintet) is: ${i:6}")
break
}
}
for i := 0; i <= 99999; i++ {
quintet := "${i:05}"
if number.contains(quintet) {
println("The smallest number formed from 5 adjacent digits ($quintet) is: ${i:6}")
return
}
}
}
- Output:
The largest number formed from 5 adjacent digits (99928) is: 99,928 The smallest number formed from 5 adjacent digits (00120) is: 120
Wren
Very simple approach as there's little need for speed here.
import "random" for Random
import "./fmt" for Fmt
var rand = Random.new()
var digits = List.filled(1000, 0)
for (i in 0...999) digits[i] = rand.int(10)
var number = digits.join()
for (r in [99999..0, 0..99999]) {
var target = (r.from == 0) ? "smallest" : "largest "
for (i in r) {
var quintet = Fmt.swrite("$05d", i)
if (number.contains(quintet)) {
Fmt.print("The $s number formed from 5 adjacent digits ($s) is: $,6d", target, quintet, i)
break
}
}
}
- Output:
Sample output:
The largest number formed from 5 adjacent digits (99830) is: 99,830 The smallest number formed from 5 adjacent digits (00154) is: 154
XPL0
char Number(1000);
int Num, Max, I, J;
[for I:= 0 to 1000-1 do \generate 1000-digit number
Number(I):= Ran(10);
Max:= 0; \find its largest 5-digit number
for I:= 0 to 1000-5 do
[Num:= 0;
for J:= 0 to 5-1 do
Num:= Num*10 + Number(I+J);
if Num > Max then
Max:= Num;
];
IntOut(0, Max);
]
- Output:
99930
- Draft Programming Tasks
- 11l
- Ada
- ALGOL 68
- APL
- Arturo
- AutoHotkey
- AWK
- BASIC
- BASIC256
- Chipmunk Basic
- Gambas
- PureBasic
- QBasic
- True BASIC
- Yabasic
- BQN
- C
- CLU
- Delphi
- SysUtils,StdCtrls
- F Sharp
- EasyLang
- Factor
- FreeBASIC
- Go
- Go-rcu
- J
- Jq
- Julia
- Mathematica
- Wolfram Language
- Nim
- Pascal
- Perl
- Phix
- Python
- Quackery
- Raku
- Ring
- RPL
- Ruby
- Sidef
- V (Vlang)
- Wren
- Wren-fmt
- XPL0