<p>Given a string <code>date</code> representing a <a href="https://en.wikipedia.org/wiki/Gregorian_calendar" target="_blank">Gregorian calendar</a> date formatted as <code>YYYY-MM-DD</code>, return <em>the day number of the year</em>.</p>

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

<pre>
<strong>Input:</strong> date = &quot;2019-01-09&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong> Given date is the 9th day of the year in 2019.
</pre>

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

<pre>
<strong>Input:</strong> date = &quot;2019-02-10&quot;
<strong>Output:</strong> 41
</pre>

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

<ul>
	<li><code>date.length == 10</code></li>
	<li><code>date[4] == date[7] == &#39;-&#39;</code>, and all other <code>date[i]</code>&#39;s are digits</li>
	<li><code>date</code> represents a calendar date between Jan 1<sup>st</sup>, 1900 and Dec 31<sup>th</sup>, 2019.</li>
</ul>

class Solution {
public:
    int dayOfYear(string date) {
        int y, m, d;
        sscanf(date.c_str(), "%d-%d-%d", &y, &m, &d);
        int v = y % 400 == 0 || (y % 4 == 0 && y % 100) ? 29 : 28;
        int days[12] = {31, v, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        int ans = d;
        for (int i = 0; i < m - 1; i++) {
            ans += days[i];
        }
        return ans;
    }
};
11000