AtCoder Beginner Contest 136
E
Max GCD
AtCoder
⌀ 1670
原题
← DF →
翻译中…

Problem Statement

翻译中…

We have a sequence of $N$ integers: $A_1, A_2, \cdots, A_N$.

翻译中…

You can perform the following operation between $0$ and $K$ times (inclusive):

翻译中…
  • Choose two integers $i$ and $j$ such that $i \neq j$, each between $1$ and $N$ (inclusive). Add $1$ to $A_i$ and $-1$ to $A_j$, possibly producing a negative element.
翻译中…

Compute the maximum possible positive integer that divides every element of $A$ after the operations. Here a positive integer $x$ divides an integer $y$ if and only if there exists an integer $z$ such that $y = xz$.

翻译中…

Constraints

翻译中…
  • $2 \leq N \leq 500$
  • $1 \leq A_i \leq 10^6$
  • $0 \leq K \leq 10^9$
  • 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 K
A_1 A_2 \cdots A_{N-1} A_{N}

翻译中…

Output

翻译中…

Print the maximum possible positive integer that divides every element of $A$ after the operations.

翻译中…

样例

Input #1
2 3
8 20
Output #1
7
$7$ will divide every element of $A$ if, for example, we perform the following operation: - Choose $i = 2, j = 1$. $A$ becomes $(7, 21)$. We cannot reach the situation where $8$ or greater integer divides every element of $A$.
Input #2
2 10
3 5
Output #2
8
Consider performing the following five operations: - Choose $i = 2, j = 1$. $A$ becomes $(2, 6)$. - Choose $i = 2, j = 1$. $A$ becomes $(1, 7)$. - Choose $i = 2, j = 1$. $A$ becomes $(0, 8)$. - Choose $i = 2, j = 1$. $A$ becomes $(-1, 9)$. - Choose $i = 1, j = 2$. $A$ becomes $(0, 8)$. Then, $0 = 8 \times 0$ and $8 = 8 \times 1$, so $8$ divides every element of $A$. We cannot reach the situation where $9$ or greater integer divides every element of $A$.
Input #3
4 5
10 1 2 22
Output #3
7
Input #4
8 7
1 7 5 6 8 2 6 5
Output #4
5
题目信息
题号
abc136_e
时间限制
2 s
内存限制
1024 MB

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。