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<>小明问了我一个问题：
<>如何用python判断一个数是否为2的n次方呢？
<>Sample Input：
2
<>Sample Output
True
<>我的代码如下：
n = int(input()) if n & (n - 1) == 0: print('True') else: print('False')
<>我的思路
*
看到这个问题首先想到的就是循环来做，应该能做出来，不过要是数字较大估计内存会溢出。
*
转换为二进制，转换几个看看规律
for i in range(0, 10): print(bin(2 ** i)) 结果： 0b1 0b10 0b100 0b1000 0b10000
0b100000 0b1000000 0b10000000 0b100000000 0b1000000000
可以很明显的观察出来，如果一个数是2的n次方，那么将这个数转换为二进制再转换为字符串其中只有一个1。于是我找到了了这样的方法:
n = str(bin(n)) if n.count('1') == 1: print('这个数是2的n次方')
但是，如果只是这样子的话从程序设计的角度来讲健壮性太差
我们可以在深入的观察一下：
for i in range(0, 10): print(bin(2 ** i - 1)) 结果： 0b0 0b1 0b11 0b111 0b1111
0b11111 0b111111 0b1111111 0b11111111 0b111111111
得到规律：如果一个数字i为2的n次方，则 i&(i-1)=0。
注：& ：按位与运算符：参与运算的两个值,如果两个相应位都为1,则该位的结果为1,否则为0。
当然还需要做一些判断和异常处理，不过这里记录一下方法，也就不用关心那些细节了。
<>学习体会
综合了运算符和判断的一个有趣的例子，回顾了一下二进制的计算。解决问题之后的喜悦让人感觉不错，继续努力，总会有成长的不是么？
<>加油！！！
<>你可以的！你总是这样相信着自己！
