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求解1+2+3+...+n,要求不能使用乘除法、while、for、if 、else、switch、case等关键字

2017-07-14 09:52 573 查看
方法一:利用递归以及&&的短路原理

int Sum(int num, int& sum)
{
num&&Sum(num-1,sum);//递归跳出条件,当num为假时,无需判断后序语句,返回sum+=n
return sum +=num;
}
int main()
{
int num= 0;
int sum = 0;
cin>>num;
cout << Sum(num, sum) << endl;
system("pause");
return 0;
}


方法二:利用构造函数求解

class Sum
{
public:
Sum()
{
++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 Sum::n = 0;
unsigned int Sum::sum = 0;

unsigned int Sum_Solution(unsigned int n)
{
Sum::Reset();
Sum* a = new Sum
;
delete[]a;
a = NULL;
return Sum::GetSum();
}
int main()
{
int n = 3;
cout << Sum_Solution(n) << endl;
system("pause");
return 0;
}


方法三: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_Solution(int n)
{
A a;
B b;
Array[0] = &a;
Array[1] = &b;
int value = Array[1]->Sum(n);
return value;
}


方法四:利用函数指针求解

typedef unsigned int(*fun)(unsigned int);
unsigned int Solution(unsigned int n)
{
return 0;
}
unsigned int Sum_Solution(unsigned int n)
{
static fun f[2] = { Solution, Sum_Solution };
return n + f[!!n](n - 1);
}


方法五:利用模板类型求解

template <unsigned int n>
struct Sum_Solution
{
enum Value
{
N = Sum_Solution<n-1>::N+n;

};

};
template<>
struct Sum_Solution<1>
{
enum Value
{
N = 1
};

};
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标签:  1+2+3+-+n
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