Sum of Digits in the Minimum Number
<p>Given an integer array <code>nums</code>, return <code>0</code><em> if the sum of the digits of the minimum integer in </em><code>nums</code><em> is odd, or </em><code>1</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [34,23,1,24,75,33,54,8]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The minimal element is 1, and the sum of those digits is 1 which is odd, so the answer is 0.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [99,77,33,66,55]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The minimal element is 33, and the sum of those digits is 3 + 3 = 6 which is even, so the answer is 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>
Array; Math
class Solution {
public:
    int sumOfDigits(vector<int>& nums) {
        int x = *min_element(nums.begin(), nums.end());
        int s = 0;
        for (; x > 0; x /= 10) {
            s += x % 10;
        }
        return s & 1 ^ 1;
    }
};