Исходя из вопроса, сколько положительных целых чисел <1 000 000 содержит цифру 2? , Я ищу наиболее креативное решение для подсчета всех целых чисел от Xдо , Yсодержащих целое число Z. Zможет быть от 0 до Y.

Каждое найденное целое число считается только один раз, даже если оно Zпоявляется чаще. Например:

Z = 2
123 counts 1
22222 also counts 1

Я начну с очень простого алгоритма, написанного на Java (потому что он любим всеми):

public class Count {
    public static void main(String[] args) {
        int count = 0;
        for (int i = Integer.parseInt(args[0]); i <= Integer.parseInt(args[1]); i++) {
            if (Integer.toString(i).contains(args[2])) {
                count++;
            }
        }
        System.out.println(count);
    }
}

если вы запустите это с

java -jar Count.jar 0 1000000 2

Вы получите это в результате:

468559

Потому что эту проблему не сложно решить, это всего лишь конкурс популярности . Ответ с наибольшим количеством голосов, опубликованный 28 февраля, побеждает!

Баш (20)

seq $1 $2|grep -c $3

использование

$ bash count.sh 0 1000000 2
468559

Funciton

Как обычно, так как высота строки, добавленная StackExchange, разбивает строки, рассмотрите возможность выполнения $('pre').css('line-height',1) в консоли браузера, чтобы это исправить.

В отличие от моих других ответов Funciton, этот не использует никаких объявлений функций. Это просто программа. Однако в нем используется лямбда-выражение - функция, которую я добавил в Funciton в декабре :)

Ожидает ввод в виде трех десятичных целых чисел (может быть отрицательным), разделенных пробелами (то есть x y z). На самом деле zможет быть любая строка; например, это может быть только знак минус ( −, U + 2212) для подсчета количества отрицательных чисел в интервале :)

           ┌───╖
     ┌───┬─┤ ♯ ╟──────────┐
     │   │ ╘═══╝ ╔════╗ ┌─┴─╖             ┌────╖ ╔═══╗
   ┌─┴─╖ └────┐  ║ 21 ║ │ × ╟─────────────┤ >> ╟─╢   ║
 ┌─┤ ʃ ╟───┐  │  ╚══╤═╝ ╘═╤═╝             ╘═╤══╝ ╚═══╝
 │ ╘═╤═╝   │  └──┐  └─────┘   ┌───────────┐ │
 │ ╔═╧═╗ ┌─┴─╖ ┌─┴─╖ ╔════╗ ┌─┴─╖   ┌───╖ ├─┴────────┐
 │ ║   ╟─┤ · ╟─┤ ʘ ╟─╢ 32 ╟─┤ · ╟───┤ ʘ ╟─┘          │
 │ ╚═══╝ ╘═╤═╝ ╘═══╝ ╚════╝ ╘═╤═╝   ╘═╤═╝ ┌─────┐    │
 │         └───────┐  ╔═══╗ ┌─┴─╖     │ ┌─┴─╖   │    │
 │ ┌───────────┐   └──╢ 0 ╟─┤ ʃ ╟─┐   │ │ ♯ ║   │    │
 │ │   ┌───╖ ┌─┴─╖    ╚═══╝ ╘═╤═╝ │   │ ╘═╤═╝ ┌─┴─╖  │
 │ │ ┌─┤ ♯ ╟─┤   ╟─┬─┐ ╔════╗ │ ┌─┴─╖ │   │ ┌─┤ × ║  │
 │ │ │ ╘═══╝ └─┬─╜ └─┘ ║ −1 ║ └─┤ · ╟─┴───┘ │ ╘═╤═╝  │
 │ │ │    ┌────┴────┐  ╚══╤═╝   ╘═╤═╝       │ ╔═╧══╗ │
 │ │ │    │ ┌───╖ ┌─┴─╖ ┌─┴─╖ ┌───┴─────╖   │ ║ 21 ║ │
 │ │ │    └─┤ ♯ ╟─┤ ? ╟─┤ = ║ │ str→int ║   │ ╚════╝ │
 │ │ │      ╘═══╝ ╘═╤═╝ ╘═╤═╝ ╘═╤═══════╝   │ ┌────╖ │
 │ │ │      ╔═══╗ ┌─┴─╖   └─┐ ┌─┴─╖         └─┤ >> ╟─┘
 │ │ │      ║ 0 ╟─┤ ? ╟─┐   └─┤ · ╟───┐       ╘═╤══╝
 │ │ │      ╚═══╝ ╘═╤═╝ └─┐   ╘═╤═╝   └───┐   ┌─┴─╖
 │ │ │            ┌─┴─╖   └─┐ ┌─┴─╖       └───┤ ʘ ║
 │ │ └────────────┤ · ╟─┐   └─┤ ≤ ║           ╘═╤═╝
 │ │              ╘═╤═╝ │     ╘═╤═╝ ┌─────────╖ │
 │ │        ╔═══╗ ╔═╧═╕ │       └─┬─┤ int→str ╟─┘
 │ │        ║ 0 ╟─╢   ├─┤         │ ╘═════════╝
 │ │        ╚═══╝ ╚═╤═╛ └─────────┘
 │ └────────────────┴─┐              │
 │    ┌─────────╖   ┌─┴─╖ ┌─┐   ┌────┴────╖
 └────┤ str→int ╟───┤   ╟─┴─┘   │ int→str ║
      ╘═════════╝   └─┬─╜       ╘════╤════╝
                      └──────────────┘

C #

public class Program
{
    public static void Main(string[] args)
    {
        Console.WriteLine(Enumerable.Range(Convert.ToInt32(args[0]), (Convert.ToInt32(args[1]) + 1) - Convert.ToInt32(args[0])).Count(x => x.ToString().Contains(args[2])));
    }
}

пример

count.exe 0 1000000 2
468559

APL (29)

{+/∨/¨(⍕⍺)∘⍷¨⍕¨⊃{⍺+0,⍳⍵-⍺}/⍵}

Эта функция принимает Zлевый аргумент, а интервал [X,Y]- правый аргумент:

      2 {+/∨/¨(⍕⍺)∘⍷¨⍕¨⊃{⍺+0,⍳⍵-⍺}/⍵} 0 1e6
468559
      0 {+/∨/¨(⍕⍺)∘⍷¨⍕¨⊃{⍺+0,⍳⍵-⍺}/⍵} 0 1e6
402131
      42 {+/∨/¨(⍕⍺)∘⍷¨⍕¨⊃{⍺+0,⍳⍵-⍺}/⍵} 0 1e6
49401

Python 2.7

Жажда Скорости

объяснение

Реализация

def Count(lo,hi,key):
    if hi == 0: return 0
    # Count(lo,hi,key) = Count(0,hi,key) - Count(0,lo - 1,key)
    if lo != 0: return Count(0, hi, key) - Count(0, lo - 1, key)
    # Calculate no of digits in the number to search
    # LOG10(hi) may be a descent trick but because of float approximation
    # this would not be reliable
    n = len(str(hi)) - 1
    # find the most significant digit
    a_n = hi/10**n
    if a_n < key:
        count = a_n*(10**n - 9**n)
    elif a_n > key:
        count = (a_n - 1)*(10**n - 9**n) + 10**n
    else:
        count = a_n*(10**n - 9**n) + 1
    if hi % 10**n != 0:
        if a_n != key:
            return count + Count(0, hi%10**n, key)
        else:
            return count + hi%10**n
    else:
        return count

демонстрация

In [2]: %timeit Count(0,123456789987654321,2)
100000 loops, best of 3: 13.2 us per loop

сравнение

@Dennis

$ \time -f%e bash count.sh 0 1234567 2
585029
11.45

@arshajii

In [6]: %timeit count(0,1234567,2)
1 loops, best of 3: 550 ms per loop

Python 2.7

Решение с использованием регулярных выражений:

>>> from re import findall as f
>>> count=lambda x,y,z:len(f('\d*%d\d*'%z,str(range(x,y+1))))
>>>
>>> count(0,1000000,2)
468559

Баш - 32 31 17 14 символов + длина X, Y и Z

Спасибо Devnull за предложение seq!

seq [X] [Y]|grep -c [Z]

например, X = 100, Y = 200, Z = 20

$ seq 100 200|grep -c 20
2

например, X = 100, Y = 200, Z = 10

$ seq 100 200|grep -c 10
11

например, X = 0, Y = 1000000, Z = 2

$ seq 0 1000000|grep -c 2
468559

PHP

Nothing original, just celebrating my first post here.

<?php

    $x = $argv[1];
    $y = $argv[2];
    $z = $argv[3];
    $count = 0;

    do
    {
        if (!(strpos($x, $z) === false))
            $count++;
        $x++;
    } while ($x <= $y);

    echo $count;

?>

Input

php script.php 0 1000000 2

Output

468559

Scala:

args(0).toInt to args(1).toInt count (_.toString contains args(2))

Рубин

Это отличный пример использования Reduce!

puts (ARGV[0]..ARGV[1]).reduce(0) { |c, n| n.to_s.include?(ARGV[2].to_s) ? c + 1 : c }

Входные данные:

ruby script.rb 0 1000000 2

Выход:

468559

Питон гольф - 61

f=lambda x,y,z:len([i for i in range(x,y)if str(z)in str(i)])

Питон не гольф

def f(x, y, z):
    c = 0
    for i in range(x, y):
        c += str(z) in str(i)
    return c

Java8

Используя новый материал IntStream, это становится по существу одним вкладышем, если вы игнорируете обязательный материал Java Framework:

import java.util.stream.IntStream;
public class A{
  public static void main(String[] args){
    System.out.println(IntStream.rangeClosed(Integer.parseInt(args[0], Integer.parseInt(args[1])).filter(x -> ((Integer)x).toString().contains(args[2])).count());
  }
}

Его можно запустить здесь , хотя мне пришлось жестко задавать значения.

F#

This solution uses IndexOf to search the string, then a little bit of number fiddling to convert the result to 1 if found, and 0 if not found, then sums the result:

let count x y (z : string) = 
    [ x .. y ] |> Seq.sumBy(fun n -> min 1 (n.ToString().IndexOf z + 1))

And it can be called like this:

count 0 1000000 "2" // 468559

Regular Expression

Following will count 1's digits up to 49.

#!/bin/bash

echo "12313451231241241111111111111111111111111111111111111"  |\  
sed "s/[^1]//g;s/11111/5/g;s/1111/4/g;s/111/3/g;s/11/2/g;s/555555555/45/g;s/55555555/40/g;s/5555555/35/g;s/555555/30/g;s/55555/25/g;s/5555/20/g;s/555/15/g;s/55/10/g;s/54/9/g;s/53/8/g;s/52/7/g;s/51/6/g;s/50/5
/g;s/40/4/g;s/30/3/g;s/20/2/g;s/10/1/g"

R 23 25 27chars

Just get the right tool for the job. Simple use of grep in R, nothing fancy.

This is what it does: grep all instances of 2 in the vector 0 until 10e6 and count the number of results using length.

length(grep(2,0:100000,value=TRUE))

length(grep(2,0:10e6))

Result: [1] 468559

Offcourse you can write a function that takes the numbers as an input, just like it is shown in the example.

count = function(x=0, y=1000000, z=2){
  length(grep(z,x:y))
}

Now you can call count with with x, y and z, if unset (that is by default), the values for x, y and z are 0, 1000000 and 2 respectively. Some examples:

count()
[1] 468559

or

count(20, 222, 2)
[1] 59

or

count(0, 100, 10)
[1] 2

Some here think time is of importance, using this function in R takes around 1 second.

system.time(count())
user  system elapsed 
0.979   0.003   0.981

JavaScript (ES6), 63

f=(i,j,n)=>{for(c=0;i<=j;!~(''+i++).indexOf(n)?0:c++);return c}

Usage:

f(0, 1e6, 2)
> 468559

Un-golfed:

f = (i,j,n) => {
  for(
    // Initialize the counter.
    c=0;
    // Iterate through all integers.
    i<=j;
    // Convert current number into string then increment it.
    // Check if the digit appears into the current number.
    !~(''+i++).indexOf(n)
      // Occurence not found.
      ? 0
      // Occurence found.
      // Add 1 to the counter.
      : c++
  );
  return c
}

Ruby

Basically I took Pablo's answer and semi-golfed (38 chars if you drop unnecessary whitespace) it into a not-so-great example of using select.

It selects every index in the range (x .. y) that contains z. This intermediate result is unfortunately stored in an array, whose size is then returned.

x,y,z = $*
p (x..y).select{ |i| i[z] }.size

It looks pretty neat both syntactically and semantically, although the i[z] part doesn't really seem to make sense.

It works because x and y actually are strings, not numbers! Thus each i is also a string, and i[z] of course checks if the string z is contained in i.

$ ruby count-digits.rb 100 200 20
2
$ ruby count-digits.rb 0 1000000 2
468559

Python 2.7, 70 signs

f = lambda x,y,z: sum(map(lambda x: str(z) in str(x), range(0, y+1)))

>>> f(0, 1000000, 2)
468559

Shorter, 65 signs

g = lambda x, y, z: sum(str(z) in str(i) for i in range(0, y+1))
>>> g(0, 1000000, 2)
468559

Using Ruby's Enumerable#grep:

start, stop, target = $*
p (start..stop).grep(Regexp.new target).size

T-SQL

If I can assume variables @X, @Y, and @Z are available:

With an (arbitrarily large ;) existing numbers table - 65

select count(*)from n where n>=@X and n<=@Y and n like '%'+@Z+'%'

With a recursive CTE - 127

with n(n)as(select @X union all select n+1 from n where n<@Y)select count(*)from n where n like'%'+@Z+'%'option(MAXRECURSION 0)

If the variables need to be defined explicitly:

Add 58 to both answers -- Numbers table: 123, Recursive CTE: 185

declare @X int=0;declare @Y int=100;declare @Z varchar(30)='2';

I have no idea how much memory the recursive CTE can use, but it's certainly not going to win any speed contests. The example of searching for 2 in 0 to 1000000 takes 8 seconds on my system.

Here's a SQL Fiddle if anyone wants to play with it. The 1000000 query takes 30+ seconds to run.

Rebol

; version 1 (simple loop counting)

count: func [x [integer!] y [integer!] z [integer!] /local total] [
    total: 0
    for n x y 1 [if found? find to-string n z [++ total]]
    total
]


; version 2 (build series/list and get length)

count: func [x [integer!] y [integer!] z [integer!]] [
    length? collect [for n x y 1 [if find to-string n z [keep true]]]
]

Usage example in Rebol console (REPL):

>> count 0 1000000 2
== 468559

PowerShell

Two solutions, both 40 37 chars.

For all versions of PowerShell:

$a,$b,$c=$args;($a..$b-match$c).count

PowerShell V3 and up have the sls alias for Select-String. This requires the @ to force an array if only one value makes it through the pipeline.

$a,$b,$c=$args;@($a..$b|sls $c).count

Batch

@setLocal enableDelayedExpansion&@set a=0&@for /L %%a in (%1,1,%2) do @set b=%%a&@if "!b:%3=!" NEQ "!b!" @set/aa+=1
@echo !a!

H:\uprof>count 0 1000000 2
468559

H:\uprof>count 1 2 3
0

A bit more readable -

@setLocal enableDelayedExpansion
@set a=0
@for /L %%a in (%1,1,%2) do (
    @set b=%%a
    @if "!b:%3=!" NEQ "!b!" @set/aa+=1
)
@echo !a!

Nice and simple. Uses string manipulation to check if the variable !b! is the same as itself without the third user input, %3 (!b:%3=!).

Mathematica

First way: strings

x, y, z are converted to strings. If a string-integer is not free of z, it is counted.

f[{x_,y_},z_] :=Length[Select[ToString/@Range[Max[x, z], y], !StringFreeQ[#, ToString@z] &]]

Examples

f[{22, 1000}, 23]
f[{0, 10^6}, 2]

20
468559

Second way: lists of digits

g[{x_,y_},z_]:=(t=Sequence@@ IntegerDigits@z;Length@Cases[IntegerDigits@Range[190], 
{s___,t,e___}])

Examples

g[{22, 1000}, 23]
g[{0, 10^6}, 2]

20
468559

GolfScript

I've been trying to improve my GolfScript skills so I thought I'd give it a shot with this question. Here's what I came up with:

`@@0\{.3$>}{.`4$?-1>@+\(}while@;;\;

This can be broken down like this:

0 1000000 2    # parameters

`@@            # convert Z to string and put at bottom of stack
0\             # init counter and swap
{.3$>}         # loop condition: Y > X
{              # loop body
  .`           # convert to string
  4$?          # search for substring
  -1>@+        # if found add to counter
  \(           # decrement Y
}              # end loop body
while          # perform loop
@;;\;          # cleanup

Even though it's GolfScript, by goal was more to try to make it relatively efficient rather than compact, so I'm sure that someone can point out various ways this can be improved.

Demonstration: Note that I've reduced Y in the demo so that it can complete in < 5 seconds.

PHP - 112

No visible loops, but a bit heavy on memory!

<?=count(array_filter(range($argv[1],$argv[2]),function($i)use($argv){return strpos($i,$argv[3].'')!==false;}));

Usage php script.php 0 1000000 2

ECMAScript 3 to 6

using regex:

function f(x,y,z,r){for(r=0,z=RegExp(z);x<y;r+=+z.test(''+x++));return r}

breakdown:

function f(x,y,z,r){        // note argument `r`, eliminating the need for `var `
  for( r=0, z=RegExp(z)     // omitting `new` since ES will add it if omitted
     ; x<y                  // 
     ; r+=+z.test(''+x++)   // `x++` == post increment
                            // `''+Number` == convert Number to string
                            // `test` gives true | false
                            // `+Boolean` converts boolean to 1 | 0
                            // `r+=Number` incrementing r (were Number is always 1 or 0)
     );                     // no body thus semicolon is mandatory!
  return r;                 // returning r
}

using indexOf:

function f(x,y,z,r){for(r=0;x<y;r+=+!!~(''+x++).indexOf(z));return r}

breakdown:

function f(x,y,z,r){                // note argument `r`, eliminating the need for `var `
  for( r=0                          // omitting `new` since ES will add it if omitted
     ; x<y                          // 
     ; r+=+!!~(''+x++).indexOf(z)   // `x++` == post increment
                                    // `''+Number` == convert Number to string
                                    // `indexOf` returns index or `-1` when not found
                                    // `!!~ indexOf` converts sentinel value to boolean
                                    // `+Boolean` converts boolean to 1 | 0
                                    // `r+=Number` incrementing r (were Number is 1 or 0)
     );                             // no body thus semicolon is mandatory!
  return r;                         // returning r
}

^{this function-body is one char less then florent's, so when using ES6 => function notation the total would be 62 char}

Example call: f(0,1e6,2)
Example use: alert( f(0,1e6,2) );

JSFiddle here

PS: both functions above return their local variable r.
So when leaking the result variable r into the global scope, one can again save 10 characters:

function f(x,y,z){for(r=0;i<=j;r+=+!!~(''+i++).indexOf(z));}

Example use: alert( f(0,1e6,2)||r );

Delphi - 120

Bit to much for my taste, going to see if i can get some off.

var x,y,z,i,c:int16;begin readLn(x,y,z);for i:=x to y do if inttostr(i).contains(inttostr(z))then inc(c);writeln(c);end.

Python 2.7 - 50 chars

Bit of a saving on the existing Python answers.

lambda x,y,z:sum(1for n in range(y-x)if`z+x`in`n`)

Using the following tricks:

• Sum can be applied to a generator, unlike len, so use sum(1...) instead of len([n...])
• Use `` instead of str(), which also allows...
• Kill all spaces - see '1for' and 'ifz+xinn'
• Remove the first range() arg by starting at 0 and testing the offset (actually...saves me nothing but I like the look of it better :) )

In action:

In [694]: (lambda x,y,z:sum(1for n in range(y-x)if`z+x`in`n`))(0,1000000,2)
Out[694]: 468559

k [28 chars]

{+/($x+!y)like"*",$:[z],"*"}

Usage

{+/($x+!y)like"*",$:[z],"*"}[0;1000000;2]
468559