<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 d) 
    {
        //31,28,31,30,31,30,,31,31,30,31,30,31
        int a[12]={31,59,90,120,151,181,212,243,273,304,334,365};

        int m=(d[5]-48)*10+d[6]-48;

        int y=(d[0]-48)*1000+(d[1]-48)*100+(d[2]-48)*10+d[3]-48;

        int day=(d[8]-48)*10+d[9]-48;

        (y!=1900 && y%4==0 && m>=3 )?day++:day;

        if(m==1)
           return day;
           
        return a[m-1-1]+day;
    }
};
10800