日期:2014-05-17 浏览次数:20908 次
public class MainClass
{
public static void Main()
{
Console.WriteLine(Foo(30));
}
public static int Foo(int i)
{
if (i <= 0)
return 0;
else if(i > 0 && i <= 2)
return 1;
else return Foo(i -1) + Foo(i - 2);
}
}
------解决方案--------------------
static void Main(string[] args)
{
Console.WriteLine(fab(3));//n表示第几个斐波那契数
}
static int fab(int n)
{
if (n == 1 || n == 2)
return 1;
else
return fab(n - 1) + fab(n - 2);
}
------解决方案--------------------
public class MainClass
{
public static void Main()
{
Console.WriteLine(Foo(10));
}
public static int Foo(int i)
{
if (i <= 0)
return 0;
else if(i > 0 && i <= 2)
Console.WriteLine("1 ");
else
Console.WriteLine((Foo(i -1) + Foo(i - 2)).ToString());
}
}
------解决方案--------------------
int a = 1, b = 1;
for (int i = 0; i < 10; i++)
{
int temp = a;
if (i < 2)
Console.WriteLine(temp);
else
{
Console.WriteLine(a+b);
a = b;
b = temp + b;
}
}