AtCoder Beginner Contest 182
翻译中…
Problem Statement
翻译中…
Given is an integer sequence $A$: $A_1, A_2, A_3, \dots, A_N$.
Let the GCD-ness of a positive integer $k$ be the number of elements among $A_1, A_2, A_3, \dots, A_N$ that are divisible by $k$.
Among the integers greater than or equal to $2$, find the integer with the greatest GCD-ness. If there are multiple such integers, you may print any of them.
翻译中…
Constraints
翻译中…
$1 \le N \le 100$$2 \le A_i \le 1000$- All values in input are integers.
翻译中…
Input
翻译中…
Input is given from Standard Input in the following format:
翻译中…
N
A_1 \hspace{7pt} A_2 \hspace{7pt} A_3 \hspace{5pt} \dots \hspace{5pt} A_N翻译中…
Output
翻译中…
Print an integer with the greatest GCD-ness among the integers greater than or equal to $2$. If there are multiple such integers, any of them will be accepted.
翻译中…
样例
Input #1
3
3 12 7
Output #1
3
Among $3$, $12$, and $7$, two of them - $3$ and $12$ - are divisible by $3$, so the GCD-ness of $3$ is $2$.
No integer greater than or equal to $2$ has greater GCD-ness, so $3$ is a correct answer.
Input #2
5
8 9 18 90 72
Output #2
9
In this case, the GCD-ness of $9$ is $4$.
$2$ and $3$ also have the GCD-ness of $4$, so you may also print $2$ or $3$.
Input #3
5
1000 1000 1000 1000 1000
Output #3
1000
题目信息
题面与样例版权归 AtCoder 及出题人所有；中文翻译由 AI 生成（CI 管线），如有出入请以原文为准。