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<p>一个厨师收集了他&nbsp;<code>n</code>&nbsp;道菜的满意程度&nbsp;<code>satisfaction</code>&nbsp;,这个厨师做出每道菜的时间都是 1 单位时间。</p>
<p>一道菜的 「&nbsp;<strong>like-time 系数&nbsp;</strong>」定义为烹饪这道菜结束的时间(包含之前每道菜所花费的时间)乘以这道菜的满意程度,也就是&nbsp;<code>time[i]</code>*<code>satisfaction[i]</code>&nbsp;</p>
<p>返回厨师在准备了一定数量的菜肴后可以获得的最大 <strong>like-time 系数</strong> 总和。</p>
<p>你可以按&nbsp;<strong>任意</strong>&nbsp;顺序安排做菜的顺序,你也可以选择放弃做某些菜来获得更大的总和。</p>
<p>&nbsp;</p>
<p><strong>示例 1</strong></p>
<pre>
<strong>输入:</strong>satisfaction = [-1,-8,0,5,-9]
<strong>输出:</strong>14
<strong>解释:</strong>去掉第二道和最后一道菜,最大的 like-time 系数和为 (-1*1 + 0*2 + 5*3 = 14) 。每道菜都需要花费 1 单位时间完成。</pre>
<p><strong>示例 2</strong></p>
<pre>
<strong>输入:</strong>satisfaction = [4,3,2]
<strong>输出:</strong>20
<strong>解释:可以</strong>按照任意顺序做菜 (2*1 + 3*2 + 4*3 = 20)
</pre>
<p><strong>示例 3</strong></p>
<pre>
<strong>输入:</strong>satisfaction = [-1,-4,-5]
<strong>输出:</strong>0
<strong>解释:</strong>大家都不喜欢这些菜,所以不做任何菜就可以获得最大的 like-time 系数。
</pre>
<p>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>n == satisfaction.length</code></li>
<li><code>1 &lt;= n &lt;= 500</code></li>
<li><code>-1000 &lt;= satisfaction[i] &lt;= 1000</code></li>
</ul>