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:

Problem types

Recall that the factorial function is defined as follows:

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

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%]

$0 \le a \le b \le 15$ $0 \leq a \leq b \leq 15$

Subtask 2 [30%]

$0 \le a \le b \le 10^5$ $0 \leq a \leq b \leq 10^{5}$

Subtask 3 [50%]

$0 \le a \le b \le 10^9$ $0 \leq a \leq b \leq 10^{9}$

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

Copy

0 2

Sample Output

Copy

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 $a, b \ge 0$ $a, b \geq 0$

1

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

Natural numbers of set $\mathbb N$ $N$ 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 $\mathbb N^*$ $N^{*}$ 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.