发散思维能力 求1+2+3+...+n

       发散思维的特点是思维活动的多向性和交通性，也就是我们在思考问题时注重运用多思路、多方案、多途径地解决问题。对于同一个问题，我们可以从不同的方向、侧面和层次，采用探索、转换、迁移、组合和分解等方法，提出多种创新的解法。

        问题:求1+2+3+...+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）通常有三种解法:公式n(n+1)/2,递归和循环

// ====================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:    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();}// ====================方法二: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;}// ====================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){    static fun f[2] = {Solution3_Teminator, Sum_Solution3};     return n + f[!!n](n - 1);}// ====================方法四:利用模板类型求解====================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};};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", n);    if(Sum_Solution1(n) == expected)        printf("Solution1 passed.\n");    else        printf("Solution1 failed.\n");    if(Sum_Solution2(n) == expected)        printf("Solution2 passed.\n");    else        printf("Solution2 failed.\n");    if(Sum_Solution3(n) == expected)        printf("Solution3 passed.\n");    else        printf("Solution3 failed.\n");}void Test1(){    const unsigned int number = 1;    int expected = 1;    Test(number, expected);    if(Sum_Solution4<number>::N == expected)        printf("Solution4 passed.\n");    else        printf("Solution4 failed.\n");}void Test2(){    const unsigned int number = 5;    int expected = 15;    Test(number, expected);    if(Sum_Solution4<number>::N == expected)        printf("Solution4 passed.\n");    else        printf("Solution4 failed.\n");}void Test3(){    const unsigned int number = 10;    int expected = 55;    Test(number, expected);    if(Sum_Solution4<number>::N == expected)        printf("Solution4 passed.\n");    else        printf("Solution4 failed.\n");}void Test4(){    const unsigned int number = 0;    int expected = 0;    Test(number, expected);    if(Sum_Solution4<number>::N == expected)        printf("Solution4 passed.\n");    else        printf("Solution4 failed.\n");}int _tmain(int argc, _TCHAR* argv[]){    Test1();    Test2();    Test3();    Test4();    return 0;}