# Pow(x, n)
实现 pow(x, n) ，即计算 x 的 n 次幂函数（即，xn）。
示例 1：
输入：x = 2.00000, n = 10
输出：1024.00000
示例 2：
输入：x = 2.10000, n = 3
输出：9.26100
示例 3：
输入：x = 2.00000, n = -2
输出：0.25000
解释：2-2 = 1/22 = 1/4 = 0.25
提示：
-100.0 < x < 100.0-231 <= n <= 231-1-104 <= xn <= 104
以下错误的选项是？
## aop
### before
```c
#include
using namespace std;
```
### after
```c
int main()
{
Solution sol;
double x = 2.00000;
int n = 10;
double res;
res = sol.myPow(x, n);
cout << fixed << setprecision(5) << res;
return 0;
}
```
## 答案
```c
class Solution
{
public:
double myPow(double x, int n)
{
if (n == 0)
return 1.0;
unsigned long long u = llabs(n);
double ans = 1.0;
while (u)
{
if (u & 1)
ans *= x;
x *= x;
u >>= 1;
}
return n > 0 ? 1.0 / ans : ans;
}
};
```
## 选项
### A
```c
class Solution
{
public:
double myPow(double x, int n)
{
if (n == INT_MIN)
{
double t = dfs(x, -(n / 2));
return 1 / t * 1 / t;
}
else
{
return n < 0 ? 1 / dfs(x, -n) : dfs(x, n);
}
}
private:
double dfs(double x, int n)
{
if (n == 0)
{
return 1;
}
else if (n == 1)
{
return x;
}
else
{
double t = dfs(x, n / 2);
return (n % 2) ? (x * t * t) : (t * t);
}
}
};
```
### B
```c
class Solution
{
public:
double myPow(double x, int n)
{
if (n == 0)
return 1;
if (n % 2 == 1)
{
double temp = myPow(x, n / 2);
return temp * temp * x;
}
else if (n % 2 == -1)
{
double temp = myPow(x, n / 2);
return temp * temp / x;
}
else
{
double temp = myPow(x, n / 2);
return temp * temp;
}
}
};
```
### C
```c
class Solution
{
public:
double helper(double x, int n)
{
if (n == 0)
return 1.0;
double y = helper(x, n / 2);
return n % 2 == 0 ? y * y : y * y * x;
}
double myPow(double x, int n)
{
long long N = static_cast(n);
if (N == 0)
return 1;
return N > 0 ? helper(x, N) : 1. / helper(x, -N);
}
};
```