《剑指offer》第六十四题：求1+2+…+n

// 面试题64：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）。

#include <cstdio>

// ====================方法一====================
// 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:
    //静态成员, 所有实例共享
    static unsigned int N;  //当前数
    static unsigned int Sum; //累加和 
};

unsigned int Temp::N = 0;
unsigned int Temp::Sum = 0;

unsigned int Sum_Solution1(unsigned int n)
{
    Temp::Reset();  //初始化

    Temp* a = new Temp[n];  //创建n个实例, 则构造函数会被调用n次
    delete[] a;
    a = nullptr;

    return Temp::GetSum();
}

// ====================方法二====================
// 使用虚函数求解
class A;
A* Array[2];

class A
{
public:
    virtual unsigned int Sum(unsigned int n) //处理函数中止
    {
        return 0;
    }
};

class B : public A
{
public:
    virtual unsigned int Sum(unsigned int n)  //充当递归函数
    {
        // !!n 非零值为true 零值为false
        return Array[!!n]->Sum(n - 1) + n;
    }
};

int Sum_Solution2(int n)
{
    A a;
    B b;
    Array[0] = &a;
    Array[1] = &b;

    //当n大于0时, 总是执行B中sum, 直到n==0
    int value = Array[1]->Sum(n);

    return value;
}

// ====================方法三====================
// 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)
{
    //静态成员, 实现方法类似于方法二
    static fun f[2] = { Solution3_Teminator, Sum_Solution3 }; //两个函数

    //当n大于0时调用函数Sum_Solution3
    //当n==0时调用函数Solution3_Teminator
    return n + f[!!n](n - 1); 
}

// ====================方法四====================
// 利用模板类型求解
template <unsigned int n> struct Sum_Solution4
{
    //实现方法类似
    enum Value { N = Sum_Solution4<n - 1>::N + n };
};

template <> struct Sum_Solution4<1>
{
    enum Value { N = 1 };
};

template <> struct Sum_Solution4<0>
{
    enum Value { N = 0 };
};

// ====================测试代码====================
void Test(int n, int expected)
{
    printf("Test for %d begins:
", n);

    if (Sum_Solution1(n) == expected)
        printf("Solution1 passed.
");
    else
        printf("Solution1 failed.
");

    if (Sum_Solution2(n) == expected)
        printf("Solution2 passed.
");
    else
        printf("Solution2 failed.
");

    if (Sum_Solution3(n) == expected)
        printf("Solution3 passed.
");
    else
        printf("Solution3 failed.
");
}

void Test1()
{
    const unsigned int number = 1;
    int expected = 1;
    Test(number, expected);
    if (Sum_Solution4<number>::N == expected)
        printf("Solution4 passed.
");
    else
        printf("Solution4 failed.
");
}

void Test2()
{
    const unsigned int number = 5;
    int expected = 15;
    Test(number, expected);
    if (Sum_Solution4<number>::N == expected)
        printf("Solution4 passed.
");
    else
        printf("Solution4 failed.
");
}

void Test3()
{
    const unsigned int number = 10;
    int expected = 55;
    Test(number, expected);
    if (Sum_Solution4<number>::N == expected)
        printf("Solution4 passed.
");
    else
        printf("Solution4 failed.
");
}

void Test4()
{
    const unsigned int number = 0;
    int expected = 0;
    Test(number, expected);
    if (Sum_Solution4<number>::N == expected)
        printf("Solution4 passed.
");
    else
        printf("Solution4 failed.
");
}

int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();

    return 0;
}

分析：通过各种方法实现递归操作。

typedef int (*fun)( int );
class Solution {
public:
    
    static int Solution3_Teminator(int n) { return 0; }
    
    static int Sum_Solution(int n) {
        
        static fun f[2] = {Solution3_Teminator, Sum_Solution};
        return n + f[!!n](n - 1);
    }
};