DMOPC '16 Contest 2 P4 - Zeros - DMOJ: Modern Online Judge

DMOPC '16 Contest 2 P4 - Zeros

Points: 7 (partial)

Time limit: 1.0s

Memory limit: 64M

Author:

jimgao

Problem types

Recall that the factorial function is defined as follows:

$\displaystyle N! = N \times (N-1) \times \dots \times 2 \times 1$

Given integers $a$ ~a~ and $b$ ~b~, please find the number of natural numbers $N$ ~N~ such that $N!$ ~N!~ has a number of trailing zeros in the range of $[a, b]$ ~[a, b]~.

Constraints

Subtask 1 [20%]

$0 \le a \le b \le 15$ ~0 \le a \le b \le 15~

Subtask 2 [30%]

$0 \le a \le b \le 10^5$ ~0 \le a \le b \le 10^5~

Subtask 3 [50%]

$0 \le a \le b \le 10^9$ ~0 \le a \le b \le 10^9~

Input Specification

The first line of the input contains the two integers $a$ ~a~ and $b$ ~b~.

Output Specification

The number of values of $N$ ~N~ that satisfy the condition.

Sample Input

0 2

Sample Output

Explanation

$1! = 1$ ~1! = 1~ is the first element that satisfies the condition, and $14! = 87\,178\,291\,200$ ~14! = 87\,178\,291\,200~ is the last element. Hence, there are $14$ ~14~ values of $N$ ~N~ that satisfy the condition.

Comments

6

thomas0115 commented on Nov. 8, 2016, 10:47 p.m. ← edit 2 →

In the explanation you imply that natural numbers are 1,2,3, etc. but in the input specification you say $a, b \ge 0$ ~a, b \ge 0~

1

aethiraes commented on Nov. 9, 2016, 2:47 p.m. ← edit 2 →

Natural numbers of set $\mathbb N$ ~\mathbb N~ can be $0,1,2,3,\dots$ ~0,1,2,3,\dots~ or $1,2,3,\dots$ ~1,2,3,\dots~

Wikipedia says there's no agreement on which one is "standard", whether $0$ ~0~ is included or not.

The set $\mathbb N^*$ ~\mathbb N^*~ is used to specify positive numbers only, $1,2,3,\dots$ ~1,2,3,\dots~

6

r3mark commented on Nov. 9, 2016, 8:16 p.m.

Yes, but this problem's usage of natural numbers isn't consistent which led to confusion.