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：