面試OR筆試26——求1到n之和

1 題目及要求

1.1 題目描述

求1到n的和。要求不能使用乘除法，for、while、if、else、switch、case等關鍵字及條件判斷語句(A?B:C)。

 

2 解答

2.1 代碼

#include <iostream>
using namespace std;

// 方案1
class 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2
class 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3
typedef 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;
}