Maximum Ascending Subarray Sum
<p>Given an array of positive integers <code>nums</code>, return the <strong>maximum</strong> possible sum of an <span data-keyword="strictly-increasing-array">strictly increasing subarray</span> in<em> </em><code>nums</code>.</p>

<p>A subarray is defined as a contiguous sequence of numbers in an array.</p>

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

<pre>
<strong>Input:</strong> nums = [10,20,30,5,10,50]
<strong>Output:</strong> 65
<strong>Explanation: </strong>[5,10,50] is the ascending subarray with the maximum sum of 65.
</pre>

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

<pre>
<strong>Input:</strong> nums = [10,20,30,40,50]
<strong>Output:</strong> 150
<strong>Explanation: </strong>[10,20,30,40,50] is the ascending subarray with the maximum sum of 150.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [12,17,15,13,10,11,12]
<strong>Output:</strong> 33
<strong>Explanation: </strong>[10,11,12] is the ascending subarray with the maximum sum of 33.
</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
class Solution {
public:
    int maxAscendingSum(vector<int>& nums) {
        int ans = 0, t = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (i == 0 || nums[i] > nums[i - 1]) {
                t += nums[i];
                ans = max(ans, t);
            } else {
                t = nums[i];
            }
        }
        return ans;
    }
};