add leetcode problem-cn part1

算法题（国内版）/problem (English)/K 次串联后最大子数组之和(English) [k-concatenation-maximum-sum].html

@@ -0,0 +1,38 @@
+<p>Given an integer array <code>arr</code> and an integer <code>k</code>, modify the array by repeating it <code>k</code> times.</p>
+
+<p>For example, if <code>arr = [1, 2]</code> and <code>k = 3 </code>then the modified array will be <code>[1, 2, 1, 2, 1, 2]</code>.</p>
+
+<p>Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be <code>0</code> and its sum in that case is <code>0</code>.</p>
+
+<p>As the answer can be very large, return the answer <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr = [1,2], k = 3
+<strong>Output:</strong> 9
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr = [1,-2,1], k = 5
+<strong>Output:</strong> 2
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr = [-1,-2], k = 7
+<strong>Output:</strong> 0
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= arr[i] &lt;= 10<sup>4</sup></code></li>
+</ul>