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Pow(x, n)
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Implement pow(x, n).
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显然是用利用快速幂，不过要注意n<0的情况。快速幂的话，可以参考我的另一篇博客 http://blog.csdn.net/asdfghjkl1993/article/details/16967897
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class Solution {
public:
double pow(double x, int n)
{
double res=1;
if(n<0)
{
x=1/x;
n=-n;
}
while(n>0)
{
if(n&1)
res=res*x;
x=x*x;
n>>=1;
}
return res;
}
};
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