Nombre de zéros de n! (force brute)

Le cadre

On rappelle que, pour $n$ un entier naturel, la factorielle de $n$ se note $n!$ et se définit comme le produit des entiers de $1$ à $n$ compris.

$0! = 1$, par définition (c'est un produit vide).
$1! = 1$
$2! = 1×2 = 2$
$3! = 1×2×3 = 6$
$11! = 1×2×3×4×5×6×7×8×9×10×11 = 39916800$
$42! = 1405006117752879898543142606244511569936384000000000$

On constate que

$3!$ se termine par aucun zéro.
$11!$ se termine par 2 zéros.
$42!$ se termine par 9 zéros.

On pourra vérifier que 1000! se termine par 249 zéros.

Exercice

Coder une fonction nb_zéros qui prend un entier naturel non nul, et qui renvoie le nombre de zéros consécutifs à la droite de son écriture décimale.

Coder une fonction nb_zéros_factorielles qui prend un entier naturel n en paramètre et qui renvoie un tableau de longueur n + 1 qui indique pour chaque indice $i$ le nombre de zéros consécutifs à la droite de l'écriture décimale de $i!$ (la factorielle de $i$).

On garantit que n < 1050, ce qui autorise un algorithme par force brute de finir en temps raisonnable.

Évaluations restantes : ∞/∞

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