题目:求1+2+3+...+n。要求不能使用乘除法,for、while、if、else、switch、case等关键字以及条件判断语句(A?B:C).
方法一:利用构造函数
class Temp
{
public:
Temp(){++N;Sum+=N;}
static void Reset(){N=0;Sum=0;}
static unsigned int GetSum(){return Sum;}
private:
static unsigned int N;
static unsigned int Sum;
};
unsigned int Temp::N=0;
unsigned int Temp::Sum=0;
unsigned int Sum_Solution1(unsigned int n)
{
Temp::Reset();
Temp *a=new Temp[n];
delete []a;
a=NULL;
return Temp::GetSum();
}方法二:利用虚函数求解:
class A;
A* Array[2];
class A
{
public:
virtual unsigned int Sum(unsigned int n)
{
return 0;
}
};
class B:public A
{
public:
virtual unsigned int Sum(unsigned int n)
{
return Array[!!n]->Sum(n-1)+n;
}
};
int Sum_Solution2(int n)
{
A a;
B b;
Array[0]=&a;
Array[1]=&b;
int value=Array[1]->Sum(n);
return value;
}解法三:利用函数指针求解
typedef unsigned int (*fun)(unsigned int);
unsigned int Solution3_Teminator(unsigned int n)
{
return 0;
}
unsigned int Sum_Solution3(unsigned int n)
{
static fun f[2]=(Solution3_Teminator,Sum-Solution3);
return n+f[!!n](n-1);
}解法四:利用模板类型求解
template <unsigned int n> struct Sum_Solution4
{
enum Value{N=Sum_Solution4<n-1>::N+n};
};
template <> struct Sum_Solution4<1>
{
enum Value{N=1};
};
















