面试题：求1+2+....+n的和

题目：求1+2+…+n的和，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）

//1、递归
int add(int n, int &sum)
{
//n为0的时候即退出
    n && add(n - 1, sum);
    return (sum += n);
}
void test1()
{
    int sum = 0;
    int n = 10;
    int ret=add(n,sum);
    cout << "1+2+...+10=" << ret << "\n";
}

//2、利用构造函数求解
//先定义一个类型，接着创建n个该类型的实例，那么这个类的构造函数将确定会被调用n次，故可以把累加的代码放在构造函数里。
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(unsigned int n)
{
    temp::reset();
    temp* a = new temp[n];
    delete []a;
    a = NULL;
    return temp::Getsum();
}
void test2()
{
    unsigned int n = 10;
    unsigned int ret = Sum(n);
    cout << "1+2+3+...+10=" << ret << "\n";
}

//3、利用虚函数求解
//一个函数充当递归函数的角色，一个充当终止递归函数的角色，可以用bool变量来控制选择哪个函数，故要把数值n转换为布尔值，可以对n进行两次反运算，即！！n
class A;
A* array[2];
class A
{
public:
    virtual unsigned int sum(unsigned int n)
    {
        return 0;
    }
};
class B :public A
{
    virtual unsigned int sum(unsigned int n)
    {
        return array[!!n]->sum(n - 1) + n;
    }
};
unsigned int Sum(unsigned int n)
{
    A a;
    B b;
    array[0] = &a;
    array[1] = &b;

    unsigned int r = array[1]->sum(n);
    return r;
}

void test3()
{
    unsigned int n = 10;
    unsigned int ret = Sum(n);
    cout << "1+2+3...+10=" << ret << "\n";
}

//4、利用函数指针求解
//用函数指针模拟虚函数
typedef unsigned int(*fun)(unsigned int);
unsigned int sum(unsigned int n)
{
    return 0;
}
unsigned int sum1(unsigned int n)
{
    static fun f[2] = { sum, sum1 };
    return f[!!n](n - 1) + n;
}
void test4()
{
    unsigned int ret = sum1(10);
    cout << "1+2+3...+10=" << ret << "\n";
}