Chrome 11+（可能还有IE）在设置值后不会触发OnChange(Chrome 11+ (and possibly IE) not triggering OnChange after value set)

 分析聚合中的代码片段的运行时可能是最容易的。 在外循环的第一次迭代中，内循环将运行1次。 在外循环的第二次迭代中，内循环将运行2次。 在外循环的第三次迭代中，内循环将运行4次。 更一般地说，在外循环的第k次迭代中，内循环将运行2k次。 一旦i变得大于N，外循环就会停止，这发生在log 2 N次迭代之后。 
 
 如果我们总结完成的工作总量，我们会看到它 
 
 
 
 1 + 2 + 4 + 8 + ... + 2 log 2 N = 2 log 2 N + 1 - 1 = 2n - 1 
 
 
 （这使用1 + 2 + 4 + 8 + ... + 2 k = 2 k + 1 - 1的事实）。 因此，对整个代码（即包括两个循环）所做的总工作是O（n）。 

It might be easiest to analyze the runtime of the piece of code in the aggregate. On the first iteration of the outer loop, the inner loop will run 1 time. On the second iteration of the outer loop, the inner loop will run 2 times. On the third iteration of the outer loop, the inner loop will run 4 times. More generally, on the kth iteration of the outer loop, the inner loop will run 2k times. The outer loop stops as soon as i becomes greater than N, which happens after log2 N iterations.
 
If we sum up the total work done, we'll see that it's
 
 
 
1 + 2 + 4 + 8 + ... + 2log2 N = 2log2 N + 1 - 1 = 2n - 1
 
 
(This uses the fact that 1 + 2 + 4 + 8 + ... + 2k = 2k+1 - 1). Therefore, the total work done for the entire piece of code (that is, including both loops) is O(n).