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};
};
原文地址:https://www.cnblogs.com/wxdjss/p/5611387.html