another solution for "Bulb Switcher"

Author: haoel

algorithms/cpp/bulbSwitcher/bulbSwitcher.cpp

@@ -1,5 +1,5 @@
 // Source : https://leetcode.com/problems/bulb-switcher/
-// Author : Calinescu Valentin
+// Author : Calinescu Valentin, Hao Chen
 // Date   : 2015-12-28
 
 /*************************************************************************************** 
@@ -21,6 +21,42 @@
  * So you should return 1, because there is only one bulb is on.
  * 
  ***************************************************************************************/
+
+ /* Solution
+  * --------
+  *
+  * We know, 
+  *   - if a bulb can be switched to ON eventually, it must be switched by ODD times
+  *   - Otherwise, if a bulb has been switched by EVEN times, it will be OFF eventually.
+  * So, 
+  *   - If bulb `i` ends up ON if and only if `i` has an ODD numbers of divisors.
+  * And we know, 
+  *   - the divisors come in pairs. for example: 
+  *     12 - [1,12] [2,6] [3,4] [6,2] [12,1] (the 12th bulb would be switched by 1,2,3,4,6,12)
+  *   - the pairs means almost all of the numbers are switched by EVEN times.
+  *
+  * But we have a special case - square numbers
+  *   - A square number must have a divisors pair with same number. such as 4 - [2,2], 9 - [3,3]
+  *   - So, a square number has a ODD numbers of divisors.
+  *
+  * At last, we figure out the solution is: 
+  *    
+  *    Count the number of the squre numbers!! 
+  */
+
+class Solution {
+public:
+    int bulbSwitch(int n) {
+        int cnt = 0;
+        for (int i=1; i*i<=n; i++) { 
+            cnt++;
+        }
+        return cnt;
+    }
+};
+
+
+
  /* 
  * Solution 1 - O(1)
  * =========