求1+2+3+...+n，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

来源：互联网发布：linux ulimit 编辑：程序博客网时间：2024/05/29 16:56

方法一：利用构造函数和静态数据成员#include <iostream>using namespace std;class Temp{public:Temp(){++N;Sum+=N;}static void Reset(){N=0;Sum=0;}static int GetSum(){return Sum;}private:static int N;static int Sum;};int Temp::N=0;int Temp::Sum=0;int solution_Sum(int n){Temp::Reset();Temp *a=new Temp[n];delete []a;a=0;return Temp::GetSum();}int main(){cout<<solution_Sum(100)<<endl;return 0;}方法二：利用虚函数#include <iostream>using namespace std;class A;A* Array[2];class A{public:virtual int Sum(int n){return 0;}};class B:public A{public:virtual int Sum(int n){return Array[!!n]->Sum(n-1)+n;}};int solution2_Sum(int n){A a;B b;Array[0]=&a;Array[1]=&b;int value=Array[1]->Sum(n);return value;}int main(){cout<<solution2_Sum(100)<<endl;return 0;}利用函数指针#include <iostream>using namespace std;typedef int (*fun)(int);int solution_f1(int i){return 0;}int solution_f2(int i){fun f[2]={solution_f1, solution_f2};return i+f[!!i](i-1);}void main(){cout<<solution_f2(100)<<endl;} 三。利用&&的短路特性#include <stdio.h>#include <stdlib.h>#include <string.h>int add_fun(int n, int &sum){n && add_fun(n-1, sum);return (sum+=n);}int main(){int sum=0;int n=100;printf("1+2+3+...+n=%d\n",add_fun(n, sum));return 0;}

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