347 votes

Le moyen le plus rapide de lister tous les nombres premiers inférieurs à N

C'est le meilleur algorithme que j'ai pu trouver.

def get_primes(n):
    numbers = set(range(n, 1, -1))
    primes = []
    while numbers:
        p = numbers.pop()
        primes.append(p)
        numbers.difference_update(set(range(p*2, n+1, p)))
    return primes

>>> timeit.Timer(stmt='get_primes.get_primes(1000000)', setup='import   get_primes').timeit(1)
1.1499958793645562

Peut-on le rendre encore plus rapide ?

Ce code a une faille : puisque numbers est un ensemble non ordonné, il n'y a aucune garantie que numbers.pop() supprimera le nombre le plus bas de l'ensemble. Néanmoins, cela fonctionne (du moins pour moi) pour certains nombres d'entrée :

>>> sum(get_primes(2000000))
142913828922L
#That's the correct sum of all numbers below 2 million
>>> 529 in get_primes(1000)
False
>>> 529 in get_primes(530)
True

357voto

unutbu Points 222216

Attention : timeit les résultats peuvent varier en raison des différences de matériel ou de version de Python.

Vous trouverez ci-dessous un script qui compare un certain nombre d'implémentations :

Un grand merci à stephan pour avoir attiré l'attention sur la roue à tamis_30. Le mérite revient à Robert William Hanks pour les nombres premiers de 2 à, les nombres premiers de 3 à, rwh_primes, rwh_primes1 et rwh_primes2.

Parmi les méthodes Python simples testées, avec psyco pour n=1000000, rwh_primes1 était le plus rapide des tests.

+---------------------+-------+
| Method              | ms    |
+---------------------+-------+
| rwh_primes1         | 43.0  |
| sieveOfAtkin        | 46.4  |
| rwh_primes          | 57.4  |
| sieve_wheel_30      | 63.0  |
| rwh_primes2         | 67.8  |    
| sieveOfEratosthenes | 147.0 |
| ambi_sieve_plain    | 152.0 |
| sundaram3           | 194.0 |
+---------------------+-------+

Parmi les méthodes Python simples testées, sans psyco pour n=1000000, rwh_primes2 était le plus rapide.

+---------------------+-------+
| Method              | ms    |
+---------------------+-------+
| rwh_primes2         | 68.1  |
| rwh_primes1         | 93.7  |
| rwh_primes          | 94.6  |
| sieve_wheel_30      | 97.4  |
| sieveOfEratosthenes | 178.0 |
| ambi_sieve_plain    | 286.0 |
| sieveOfAtkin        | 314.0 |
| sundaram3           | 416.0 |
+---------------------+-------+

De toutes les méthodes testées, autoriser numpy pour n=1000000, nombres premiers de 2 à était le plus rapide des tests.

+---------------------+-------+
| Method              | ms    |
+---------------------+-------+
| primesfrom2to       | 15.9  |
| primesfrom3to       | 18.4  |
| ambi_sieve          | 29.3  |
+---------------------+-------+

Les temps ont été mesurés en utilisant la commande :

python -mtimeit -s"import primes" "primes.{method}(1000000)"

avec {method} remplacé par chacun des noms de méthode.

primes.py :

#!/usr/bin/env python
import psyco; psyco.full()
from math import sqrt, ceil
import numpy as np

def rwh_primes(n):
    # https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    """ Returns  a list of primes < n """
    sieve = [True] * n
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
    return [2] + [i for i in xrange(3,n,2) if sieve[i]]

def rwh_primes1(n):
    # https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    """ Returns  a list of primes < n """
    sieve = [True] * (n/2)
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i/2]:
            sieve[i*i/2::i] = [False] * ((n-i*i-1)/(2*i)+1)
    return [2] + [2*i+1 for i in xrange(1,n/2) if sieve[i]]

def rwh_primes2(n):
    # https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    """ Input n>=6, Returns a list of primes, 2 <= p < n """
    correction = (n%6>1)
    n = {0:n,1:n-1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6]
    sieve = [True] * (n/3)
    sieve[0] = False
    for i in xrange(int(n**0.5)/3+1):
      if sieve[i]:
        k=3*i+1|1
        sieve[      ((k*k)/3)      ::2*k]=[False]*((n/6-(k*k)/6-1)/k+1)
        sieve[(k*k+4*k-2*k*(i&1))/3::2*k]=[False]*((n/6-(k*k+4*k-2*k*(i&1))/6-1)/k+1)
    return [2,3] + [3*i+1|1 for i in xrange(1,n/3-correction) if sieve[i]]

def sieve_wheel_30(N):
    # http://zerovolt.com/?p=88
    ''' Returns a list of primes <= N using wheel criterion 2*3*5 = 30

Copyright 2009 by zerovolt.com
This code is free for non-commercial purposes, in which case you can just leave this comment as a credit for my work.
If you need this code for commercial purposes, please contact me by sending an email to: info [at] zerovolt [dot] com.'''
    __smallp = ( 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
    61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139,
    149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227,
    229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311,
    313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401,
    409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491,
    499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599,
    601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683,
    691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797,
    809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887,
    907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997)

    wheel = (2, 3, 5)
    const = 30
    if N < 2:
        return []
    if N <= const:
        pos = 0
        while __smallp[pos] <= N:
            pos += 1
        return list(__smallp[:pos])
    # make the offsets list
    offsets = (7, 11, 13, 17, 19, 23, 29, 1)
    # prepare the list
    p = [2, 3, 5]
    dim = 2 + N // const
    tk1  = [True] * dim
    tk7  = [True] * dim
    tk11 = [True] * dim
    tk13 = [True] * dim
    tk17 = [True] * dim
    tk19 = [True] * dim
    tk23 = [True] * dim
    tk29 = [True] * dim
    tk1[0] = False
    # help dictionary d
    # d[a , b] = c  ==> if I want to find the smallest useful multiple of (30*pos)+a
    # on tkc, then I need the index given by the product of [(30*pos)+a][(30*pos)+b]
    # in general. If b < a, I need [(30*pos)+a][(30*(pos+1))+b]
    d = {}
    for x in offsets:
        for y in offsets:
            res = (x*y) % const
            if res in offsets:
                d[(x, res)] = y
    # another help dictionary: gives tkx calling tmptk[x]
    tmptk = {1:tk1, 7:tk7, 11:tk11, 13:tk13, 17:tk17, 19:tk19, 23:tk23, 29:tk29}
    pos, prime, lastadded, stop = 0, 0, 0, int(ceil(sqrt(N)))
    # inner functions definition
    def del_mult(tk, start, step):
        for k in xrange(start, len(tk), step):
            tk[k] = False
    # end of inner functions definition
    cpos = const * pos
    while prime < stop:
        # 30k + 7
        if tk7[pos]:
            prime = cpos + 7
            p.append(prime)
            lastadded = 7
            for off in offsets:
                tmp = d[(7, off)]
                start = (pos + prime) if off == 7 else (prime * (const * (pos + 1 if tmp < 7 else 0) + tmp) )//const
                del_mult(tmptk[off], start, prime)
        # 30k + 11
        if tk11[pos]:
            prime = cpos + 11
            p.append(prime)
            lastadded = 11
            for off in offsets:
                tmp = d[(11, off)]
                start = (pos + prime) if off == 11 else (prime * (const * (pos + 1 if tmp < 11 else 0) + tmp) )//const
                del_mult(tmptk[off], start, prime)
        # 30k + 13
        if tk13[pos]:
            prime = cpos + 13
            p.append(prime)
            lastadded = 13
            for off in offsets:
                tmp = d[(13, off)]
                start = (pos + prime) if off == 13 else (prime * (const * (pos + 1 if tmp < 13 else 0) + tmp) )//const
                del_mult(tmptk[off], start, prime)
        # 30k + 17
        if tk17[pos]:
            prime = cpos + 17
            p.append(prime)
            lastadded = 17
            for off in offsets:
                tmp = d[(17, off)]
                start = (pos + prime) if off == 17 else (prime * (const * (pos + 1 if tmp < 17 else 0) + tmp) )//const
                del_mult(tmptk[off], start, prime)
        # 30k + 19
        if tk19[pos]:
            prime = cpos + 19
            p.append(prime)
            lastadded = 19
            for off in offsets:
                tmp = d[(19, off)]
                start = (pos + prime) if off == 19 else (prime * (const * (pos + 1 if tmp < 19 else 0) + tmp) )//const
                del_mult(tmptk[off], start, prime)
        # 30k + 23
        if tk23[pos]:
            prime = cpos + 23
            p.append(prime)
            lastadded = 23
            for off in offsets:
                tmp = d[(23, off)]
                start = (pos + prime) if off == 23 else (prime * (const * (pos + 1 if tmp < 23 else 0) + tmp) )//const
                del_mult(tmptk[off], start, prime)
        # 30k + 29
        if tk29[pos]:
            prime = cpos + 29
            p.append(prime)
            lastadded = 29
            for off in offsets:
                tmp = d[(29, off)]
                start = (pos + prime) if off == 29 else (prime * (const * (pos + 1 if tmp < 29 else 0) + tmp) )//const
                del_mult(tmptk[off], start, prime)
        # now we go back to top tk1, so we need to increase pos by 1
        pos += 1
        cpos = const * pos
        # 30k + 1
        if tk1[pos]:
            prime = cpos + 1
            p.append(prime)
            lastadded = 1
            for off in offsets:
                tmp = d[(1, off)]
                start = (pos + prime) if off == 1 else (prime * (const * pos + tmp) )//const
                del_mult(tmptk[off], start, prime)
    # time to add remaining primes
    # if lastadded == 1, remove last element and start adding them from tk1
    # this way we don't need an "if" within the last while
    if lastadded == 1:
        p.pop()
    # now complete for every other possible prime
    while pos < len(tk1):
        cpos = const * pos
        if tk1[pos]: p.append(cpos + 1)
        if tk7[pos]: p.append(cpos + 7)
        if tk11[pos]: p.append(cpos + 11)
        if tk13[pos]: p.append(cpos + 13)
        if tk17[pos]: p.append(cpos + 17)
        if tk19[pos]: p.append(cpos + 19)
        if tk23[pos]: p.append(cpos + 23)
        if tk29[pos]: p.append(cpos + 29)
        pos += 1
    # remove exceeding if present
    pos = len(p) - 1
    while p[pos] > N:
        pos -= 1
    if pos < len(p) - 1:
        del p[pos+1:]
    # return p list
    return p

def sieveOfEratosthenes(n):
    """sieveOfEratosthenes(n): return the list of the primes < n."""
    # Code from: <dickinsm@gmail.com>, Nov 30 2006
    # http://groups.google.com/group/comp.lang.python/msg/f1f10ced88c68c2d
    if n <= 2:
        return []
    sieve = range(3, n, 2)
    top = len(sieve)
    for si in sieve:
        if si:
            bottom = (si*si - 3) // 2
            if bottom >= top:
                break
            sieve[bottom::si] = [0] * -((bottom - top) // si)
    return [2] + [el for el in sieve if el]

def sieveOfAtkin(end):
    """sieveOfAtkin(end): return a list of all the prime numbers <end
    using the Sieve of Atkin."""
    # Code by Steve Krenzel, <Sgk284@gmail.com>, improved
    # Code: https://web.archive.org/web/20080324064651/http://krenzel.info/?p=83
    # Info: http://en.wikipedia.org/wiki/Sieve_of_Atkin
    assert end > 0
    lng = ((end-1) // 2)
    sieve = [False] * (lng + 1)

    x_max, x2, xd = int(sqrt((end-1)/4.0)), 0, 4
    for xd in xrange(4, 8*x_max + 2, 8):
        x2 += xd
        y_max = int(sqrt(end-x2))
        n, n_diff = x2 + y_max*y_max, (y_max << 1) - 1
        if not (n & 1):
            n -= n_diff
            n_diff -= 2
        for d in xrange((n_diff - 1) << 1, -1, -8):
            m = n % 12
            if m == 1 or m == 5:
                m = n >> 1
                sieve[m] = not sieve[m]
            n -= d

    x_max, x2, xd = int(sqrt((end-1) / 3.0)), 0, 3
    for xd in xrange(3, 6 * x_max + 2, 6):
        x2 += xd
        y_max = int(sqrt(end-x2))
        n, n_diff = x2 + y_max*y_max, (y_max << 1) - 1
        if not(n & 1):
            n -= n_diff
            n_diff -= 2
        for d in xrange((n_diff - 1) << 1, -1, -8):
            if n % 12 == 7:
                m = n >> 1
                sieve[m] = not sieve[m]
            n -= d

    x_max, y_min, x2, xd = int((2 + sqrt(4-8*(1-end)))/4), -1, 0, 3
    for x in xrange(1, x_max + 1):
        x2 += xd
        xd += 6
        if x2 >= end: y_min = (((int(ceil(sqrt(x2 - end))) - 1) << 1) - 2) << 1
        n, n_diff = ((x*x + x) << 1) - 1, (((x-1) << 1) - 2) << 1
        for d in xrange(n_diff, y_min, -8):
            if n % 12 == 11:
                m = n >> 1
                sieve[m] = not sieve[m]
            n += d

    primes = [2, 3]
    if end <= 3:
        return primes[:max(0,end-2)]

    for n in xrange(5 >> 1, (int(sqrt(end))+1) >> 1):
        if sieve[n]:
            primes.append((n << 1) + 1)
            aux = (n << 1) + 1
            aux *= aux
            for k in xrange(aux, end, 2 * aux):
                sieve[k >> 1] = False

    s  = int(sqrt(end)) + 1
    if s  % 2 == 0:
        s += 1
    primes.extend([i for i in xrange(s, end, 2) if sieve[i >> 1]])

    return primes

def ambi_sieve_plain(n):
    s = range(3, n, 2)
    for m in xrange(3, int(n**0.5)+1, 2): 
        if s[(m-3)/2]: 
            for t in xrange((m*m-3)/2,(n>>1)-1,m):
                s[t]=0
    return [2]+[t for t in s if t>0]

def sundaram3(max_n):
    # https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/2073279#2073279
    numbers = range(3, max_n+1, 2)
    half = (max_n)//2
    initial = 4

    for step in xrange(3, max_n+1, 2):
        for i in xrange(initial, half, step):
            numbers[i-1] = 0
        initial += 2*(step+1)

        if initial > half:
            return [2] + filter(None, numbers)

################################################################################
# Using Numpy:
def ambi_sieve(n):
    # http://tommih.blogspot.com/2009/04/fast-prime-number-generator.html
    s = np.arange(3, n, 2)
    for m in xrange(3, int(n ** 0.5)+1, 2): 
        if s[(m-3)/2]: 
            s[(m*m-3)/2::m]=0
    return np.r_[2, s[s>0]]

def primesfrom3to(n):
    # https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    """ Returns a array of primes, p < n """
    assert n>=2
    sieve = np.ones(n/2, dtype=np.bool)
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i/2]:
            sieve[i*i/2::i] = False
    return np.r_[2, 2*np.nonzero(sieve)[0][1::]+1]    

def primesfrom2to(n):
    # https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    """ Input n>=6, Returns a array of primes, 2 <= p < n """
    sieve = np.ones(n/3 + (n%6==2), dtype=np.bool)
    sieve[0] = False
    for i in xrange(int(n**0.5)/3+1):
        if sieve[i]:
            k=3*i+1|1
            sieve[      ((k*k)/3)      ::2*k] = False
            sieve[(k*k+4*k-2*k*(i&1))/3::2*k] = False
    return np.r_[2,3,((3*np.nonzero(sieve)[0]+1)|1)]

if __name__=='__main__':
    import itertools
    import sys

    def test(f1,f2,num):
        print('Testing {f1} and {f2} return same results'.format(
            f1=f1.func_name,
            f2=f2.func_name))
        if not all([a==b for a,b in itertools.izip_longest(f1(num),f2(num))]):
            sys.exit("Error: %s(%s) != %s(%s)"%(f1.func_name,num,f2.func_name,num))

    n=1000000
    test(sieveOfAtkin,sieveOfEratosthenes,n)
    test(sieveOfAtkin,ambi_sieve,n)
    test(sieveOfAtkin,ambi_sieve_plain,n) 
    test(sieveOfAtkin,sundaram3,n)
    test(sieveOfAtkin,sieve_wheel_30,n)
    test(sieveOfAtkin,primesfrom3to,n)
    test(sieveOfAtkin,primesfrom2to,n)
    test(sieveOfAtkin,rwh_primes,n)
    test(sieveOfAtkin,rwh_primes1,n)         
    test(sieveOfAtkin,rwh_primes2,n)

L'exécution du script teste que toutes les implémentations donnent le même résultat.

132voto

Un code Python pur plus rapide et plus économe en mémoire :

def primes(n):
    """ Returns  a list of primes < n """
    sieve = [True] * n
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
    return [2] + [i for i in range(3,n,2) if sieve[i]]

ou en commençant par un demi-tamis

def primes1(n):
    """ Returns  a list of primes < n """
    sieve = [True] * (n//2)
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i//2]:
            sieve[i*i//2::i] = [False] * ((n-i*i-1)//(2*i)+1)
    return [2] + [2*i+1 for i in range(1,n//2) if sieve[i]]

Code numpy plus rapide et plus économe en mémoire :

import numpy
def primesfrom3to(n):
    """ Returns a array of primes, 3 <= p < n """
    sieve = numpy.ones(n//2, dtype=bool)
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i//2]:
            sieve[i*i//2::i] = False
    return 2*numpy.nonzero(sieve)[0][1::]+1

une variation plus rapide en commençant par un tiers de tamis :

import numpy
def primesfrom2to(n):
    """ Input n>=6, Returns a array of primes, 2 <= p < n """
    sieve = numpy.ones(n//3 + (n%6==2), dtype=bool)
    for i in range(1,int(n**0.5)//3+1):
        if sieve[i]:
            k=3*i+1|1
            sieve[       k*k//3     ::2*k] = False
            sieve[k*(k-2*(i&1)+4)//3::2*k] = False
    return numpy.r_[2,3,((3*numpy.nonzero(sieve)[0][1:]+1)|1)]

Une version (difficile à coder) en pur-python du code ci-dessus serait :

def primes2(n):
    """ Input n>=6, Returns a list of primes, 2 <= p < n """
    n, correction = n-n%6+6, 2-(n%6>1)
    sieve = [True] * (n//3)
    for i in range(1,int(n**0.5)//3+1):
      if sieve[i]:
        k=3*i+1|1
        sieve[      k*k//3      ::2*k] = [False] * ((n//6-k*k//6-1)//k+1)
        sieve[k*(k-2*(i&1)+4)//3::2*k] = [False] * ((n//6-k*(k-2*(i&1)+4)//6-1)//k+1)
    return [2,3] + [3*i+1|1 for i in range(1,n//3-correction) if sieve[i]]

Malheureusement, pure-python n'adopte pas la méthode numpy, plus simple et plus rapide, pour effectuer des affectations, et l'appel à len() à l'intérieur de la boucle comme dans [False]*len(sieve[((k*k)//3)::2*k]) est trop lent. J'ai donc dû improviser pour corriger l'entrée (et éviter plus de maths) et faire de la math-magie extrême (et douloureuse).

Personnellement, je trouve dommage que numpy (qui est si largement utilisé) ne fasse pas partie de la bibliothèque standard de Python, et que les améliorations en termes de syntaxe et de vitesse semblent être complètement négligées par les développeurs Python.

42voto

Alex Martelli Points 330805

Il y a un exemple très intéressant dans le livre de cuisine Python. aquí -- la version la plus rapide proposée sur cette URL est :

import itertools
def erat2( ):
    D = {  }
    yield 2
    for q in itertools.islice(itertools.count(3), 0, None, 2):
        p = D.pop(q, None)
        if p is None:
            D[q*q] = q
            yield q
        else:
            x = p + q
            while x in D or not (x&1):
                x += p
            D[x] = p

donc cela donnerait

def get_primes_erat(n):
  return list(itertools.takewhile(lambda p: p<n, erat2()))

En mesurant à l'invite du shell (comme je préfère le faire) avec ce code dans pri.py, j'observe :

$ python2.5 -mtimeit -s'import pri' 'pri.get_primes(1000000)'
10 loops, best of 3: 1.69 sec per loop
$ python2.5 -mtimeit -s'import pri' 'pri.get_primes_erat(1000000)'
10 loops, best of 3: 673 msec per loop

donc il semble que la solution Cookbook soit deux fois plus rapide.

28voto

jbochi Points 12280

Utilisation de Le tamis de Sundaram je pense avoir battu le record de pure-Python :

def sundaram3(max_n):
    numbers = range(3, max_n+1, 2)
    half = (max_n)//2
    initial = 4

    for step in xrange(3, max_n+1, 2):
        for i in xrange(initial, half, step):
            numbers[i-1] = 0
        initial += 2*(step+1)

        if initial > half:
            return [2] + filter(None, numbers)

Comparaison :

C:\USERS>python -m timeit -n10 -s "import get_primes" "get_primes.get_primes_erat(1000000)"
10 loops, best of 3: 710 msec per loop

C:\USERS>python -m timeit -n10 -s "import get_primes" "get_primes.daniel_sieve_2(1000000)"
10 loops, best of 3: 435 msec per loop

C:\USERS>python -m timeit -n10 -s "import get_primes" "get_primes.sundaram3(1000000)"
10 loops, best of 3: 327 msec per loop

18voto

sth Points 91594

L'algorithme est rapide, mais il présente un grave défaut :

>>> sorted(get_primes(530))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73,
79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163,
167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251,
257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349,
353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443,
449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 527, 529]
>>> 17*31
527
>>> 23*23
529

Vous supposez que numbers.pop() renverrait le plus petit nombre de l'ensemble, mais ce n'est pas du tout garanti. Les ensembles ne sont pas ordonnés et pop() supprime et renvoie un arbitraire Il ne peut donc pas être utilisé pour sélectionner le prochain nombre premier parmi les nombres restants.

Prograide.com

Prograide est une communauté de développeurs qui cherche à élargir la connaissance de la programmation au-delà de l'anglais.
Pour cela nous avons les plus grands doutes résolus en français et vous pouvez aussi poser vos propres questions ou résoudre celles des autres.

Powered by:

X