double Pow(int m, int n)
{
if(m == 0 && n < 0)
{
return double.PositiveInfinity;
}
if(n > 0)
{
double result = 1;
for(int i = 1; i <= n; i++)
{
result = result * m;
}
return result;
}
else
{
double result = 1;
for (int i = 1; i <= -n; i++)
{
result = result / m;
}
return result;
}
}
Console.WriteLine(Math.Pow(2,5));
Console.WriteLine(Pow(2, 5));
C#实现Pow函数
最新推荐文章于 2025-04-13 12:38:51 发布