求1 + 2 + 3...+ n的和

题目：求1 + 2 + 3 + ... + n 的和

要求：不能使用乘除法、for、while、if、else、switch、case等关键字以及条件判断语句(A?B:C)

答：

#include "stdafx.h"
#include <iostream>

using namespace std;

//1、函数查找表法
typedef int (*Func)(int n);
int Sum1(int n)
{
    return 0;
}

int Sum2(int n)
{
    Func sum[2] = {Sum1, Sum2};
    return sum[n > 0](n - 1) + n;
}

//2、&&短路特性
int CountSum(int n, int &total)
{
    n && CountSum(n - 1, total);
    return total += n;
}

//3、构造函数法
class SumClass
{
public:
    SumClass()
    {
        n++;
        sum += n;
    }
    static int GetSum()
    {
        return sum;
    }
private:
    static int n;
    static int sum;
};

int SumClass::n = 0;
int SumClass::sum = 0;

int _tmain(int argc, _TCHAR* argv[])
{
    cout<<"1、函数查找表：1+2+3+...+100 = "<<Sum2(100)<<endl;
    int total = 0;
    CountSum(100, total);
    cout<<"2、&&短路特性：1+2+3+...+100 = "<<total<<endl;
    SumClass *p = new SumClass[100];
    delete []p;
    cout<<"3、构造函数法：1+2+3+...+100 = "<<SumClass::GetSum()<<endl;
    return 0;
}

运行界面如下：