求1+2+...+n

题目：求1+2+...+n,要求不能使用乘除法，for,while,if,else,switch,case等关键字及条件判断语句(A?B:C).

  解法一：利用构造函数求解：

  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();

}

 

解法二：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：利用类模板求解

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};

};