题目:求1+2+…+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)
//1、递归
int add(int n, int &sum)
{
//n为0的时候即退出
n && add(n - 1, sum);
return (sum += n);
}
void test1()
{
int sum = 0;
int n = 10;
int ret=add(n,sum);
cout << "1+2+...+10=" << ret << "\n";
}
//2、利用构造函数求解
//先定义一个类型,接着创建n个该类型的实例,那么这个类的构造函数将确定会被调用n次,故可以把累加的代码放在构造函数里。
class temp
{
public:
temp()
{
n++;
sum += n;
}
static void reset()
{
n = 0;
sum = 0;
}
static unsigned int Getsum()
{
return sum;
}
private:
static unsigned int n;
static unsigned int sum;
};
unsigned int temp::n = 0;
unsigned int temp::sum = 0;
unsigned int Sum(unsigned int n)
{
temp::reset();
temp* a = new temp[n];
delete []a;
a = NULL;
return temp::Getsum();
}
void test2()
{
unsigned int n = 10;
unsigned int ret = Sum(n);
cout << "1+2+3+...+10=" << ret << "\n";
}
//3、利用虚函数求解
//一个函数充当递归函数的角色,一个充当终止递归函数的角色,可以用bool变量来控制选择哪个函数,故要把数值n转换为布尔值,可以对n进行两次反运算,即!!n
class A;
A* array[2];
class A
{
public:
virtual unsigned int sum(unsigned int n)
{
return 0;
}
};
class B :public A
{
virtual unsigned int sum(unsigned int n)
{
return array[!!n]->sum(n - 1) + n;
}
};
unsigned int Sum(unsigned int n)
{
A a;
B b;
array[0] = &a;
array[1] = &b;
unsigned int r = array[1]->sum(n);
return r;
}
void test3()
{
unsigned int n = 10;
unsigned int ret = Sum(n);
cout << "1+2+3...+10=" << ret << "\n";
}
//4、利用函数指针求解
//用函数指针模拟虚函数
typedef unsigned int(*fun)(unsigned int);
unsigned int sum(unsigned int n)
{
return 0;
}
unsigned int sum1(unsigned int n)
{
static fun f[2] = { sum, sum1 };
return f[!!n](n - 1) + n;
}
void test4()
{
unsigned int ret = sum1(10);
cout << "1+2+3...+10=" << ret << "\n";
}