gcd-of-odd-and-even-sums

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Algorithms

You are given an integer <code>n</code>. Your task is to compute the GCD (greatest common divisor) of two values:

<ul>
	<li>
	<code>sumOdd</code>: the sum of the smallest&nbsp;<code>n</code>&nbsp;positive odd numbers.
	</li>
	<li>
	<code>sumEven</code>: the sum of the smallest&nbsp;<code>n</code>&nbsp;positive even numbers.
	</li>
</ul>

Return the GCD of <code>sumOdd</code> and <code>sumEven</code>.

&nbsp;
Example 1:

<div class="example-block">
Input: n = 4

Output: 4

Explanation:

<ul>
	<li>Sum of the first 4 odd numbers <code>sumOdd = 1 + 3 + 5 + 7 = 16</code></li>
	<li>Sum of the first 4 even numbers <code>sumEven = 2 + 4 + 6 + 8 = 20</code></li>
</ul>

Hence, <code>GCD(sumOdd, sumEven) = GCD(16, 20) = 4</code>.
</div>

Example 2:

<div class="example-block">
Input: n = 5

Output: 5

Explanation:

<ul>
	<li>Sum of the first 5 odd numbers <code>sumOdd = 1 + 3 + 5 + 7 + 9 = 25</code></li>
	<li>Sum of the first 5 even numbers <code>sumEven = 2 + 4 + 6 + 8 + 10 = 30</code></li>
</ul>

Hence, <code>GCD(sumOdd, sumEven) = GCD(25, 30) = 5</code>.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= n &lt;= 10​​​​​​​00</code></li>
</ul>

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gcdOfOddEvenSums

中级工程师

GCD of Odd and Even Sums

Math

Number Theory

给你一个整数 <code>n</code>。请你计算以下两个值的&nbsp;最大公约数（GCD）：

<ul>
	<li>
	<code>sumOdd</code>：最小的&nbsp;<code>n</code> 个正奇数的总和。
	</li>
	<li>
	<code>sumEven</code>：最小的&nbsp;<code>n</code> 个正偶数的总和。
	</li>
</ul>

返回 <code>sumOdd</code> 和 <code>sumEven</code> 的 GCD。

&nbsp;

示例 1：

<div class="example-block">
输入： n = 4

输出： 4

解释：

<ul>
	<li>前 4 个奇数的总和 <code>sumOdd = 1 + 3 + 5 + 7 = 16</code></li>
	<li>前 4 个偶数的总和 <code>sumEven = 2 + 4 + 6 + 8 = 20</code></li>
</ul>

因此，<code>GCD(sumOdd, sumEven) = GCD(16, 20) = 4</code>。
</div>

示例 2：

<div class="example-block">
输入： n = 5

输出： 5

解释：

<ul>
	<li>前 5 个奇数的总和 <code>sumOdd = 1 + 3 + 5 + 7 + 9 = 25</code></li>
	<li>前 5 个偶数的总和 <code>sumEven = 2 + 4 + 6 + 8 + 10 = 30</code></li>
</ul>

因此，<code>GCD(sumOdd, sumEven) = GCD(25, 30) = 5</code>。
</div>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>