面试题64. 求1+2+…+n

Author: lzl124631x

leetcode-cn/面试题64. 求1+2+…+n/README.md

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+# [面试题64. 求1+2+…+n](https://leetcode-cn.com/problems/qiu-12n-lcof/)
+
+<div class="content__1Y2H"><div class="notranslate"><p>求 <code>1+2+...+n</code> ，要求不能使用乘除法、for、while、if、else、switch、case等关键字及条件判断语句（A?B:C）。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入:</strong> n = 3
+<strong>输出:&nbsp;</strong>6
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入:</strong> n = 9
+<strong>输出:&nbsp;</strong>45
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>限制：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n&nbsp;&lt;= 10000</code></li>
+</ul>
+</div></div>
+
+## Solution 1.
+
+```cpp
+// OJ: https://leetcode-cn.com/problems/qiu-12n-lcof/
+// Author: github.com/lzl124631x
+// Time: O(1)
+// Space: O(1)
+class Solution {
+    int d(unsigned n, int i) {
+        return -((n >> i) & 1) & (n << i);
+    }
+public:
+    int sumNums(int n) {
+        return (d(n, 0) + d(n, 1) + d(n, 2) + d(n, 3) + d(n, 4) + d(n, 5) + d(n, 6)
+                + d(n, 7) + d(n, 8) + d(n, 9) + d(n, 10) + d(n, 11) + d(n, 12) + d(n, 13) + n) >> 1;
+    }
+}
+```