面试OR笔试26——求1到n之和

1 题目及要求

1.1 题目描述

求1到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)。

 

2 解答

2.1 代码

#include <iostream>using namespace std;// 方案1class Temp{public:Temp(){++k;s+=k;}static void reset(){k=s=0;}static unsigned getSum(){return s;}private:static unsigned k,s;};unsigned Temp::k = 0;unsigned Temp::s = 0;unsigned limitSum1(unsigned n){Temp::reset();Temp *a = new Temp[n];delete[] a;a = nullptr;return Temp::getSum();}// 方案2class A{public:virtual unsigned sum(unsigned n){return 0;}};A *a[2];class B:public A{public:virtual unsigned sum(unsigned n){return a[!!n]->sum(n-1)+n;}};unsigned limitSum2(unsigned n){A a1; B b1;a[0]=&a1;a[1]=&b1;return a[1]->sum(n);}// 方案3typedef unsigned (*fun)(unsigned);unsigned limitSum31(unsigned n){return 0;}unsigned limitSum3(unsigned n){static fun a[2]={limitSum31,limitSum3};return a[!!n](n-1)+n;}int main(){cout << limitSum1(7) << endl;cout << limitSum2(7) << endl;cout << limitSum3(7) << endl;    return 0;}