AtCoder Beginner Contest 136
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Problem Statement
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We have a sequence of $N$ integers: $A_1, A_2, \cdots, A_N$.
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You can perform the following operation between $0$ and $K$ times (inclusive):
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- 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.
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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 $y = xz$.
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Constraints
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$2 \leq N \leq 500$$1 \leq A_i \leq 10^6$$0 \leq K \leq 10^9$- All values in input are integers.
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Input
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Input is given from Standard Input in the following format:
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N K
A_1 A_2 \cdots A_{N-1} A_{N}
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Output
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Print the maximum possible positive integer that divides every element of $A$ after the operations.
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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。