用C++编写输入两个数,如,X,Y,求X的Y次方.

[yjip267]

这是我编的,Y是正数能实现,Y是负数就不对了.不知道怎么才能完善这个程序请大家指点一下.谢谢!我在VC++里面编的.
#include <iostream>
using namespace std;
long fun(int x,int y);
int main()
{ int number,power;
long answer;
cout<<"输入一个数";
cin>>number;
cout<<"输入次方数:";
cin>>power;
answer = fun(number,power);
cout<<number<<"to the"<<power<<"th power is"<<answer<<endl;
return 0;
}
long fun(int x,int y)
{ if (y==1)
return x;
else
return (x*fun(x,y-1));
}

[小猪瑞比]

汗 这么小都递归？？？？

X的3次方，不就是X * X * X 嘛

进来判断一个，如果是负数，则先取绝对值，最后加上‘-’就行了呗。

[ouyh12345]

long fun(int x,int y)
{
if (x==0 || x==1)
return x;

if (y==1)
return x;

if (y == 0)
return 1;

if (y > 0)
return (x*fun(x,y-1));

return fun(x, y+1) / x;
}

[ouyh12345]

如果y是负数的话，返回值为long就不对了，得返回double

[Lynn_Ran]

#include  <iostream> 
using namespace std;  

long fun(int iNumber,int iPower); 

int main() 
{ 
	int iNumber, iPower; 
	long iAnswer; 
	cout << "输入一个数"; 
	cin >> iNumber; 
	cout <<"输入次方数:"; 
	cin >> iPower; 
	iAnswer = fun(iNumber, iPower); 
	if ( iPower >= 0 )
		cout << iNumber << " to the " << iPower << "th power is " << iAnswer << endl; 
	else
		cout << iNumber << " to the " << iPower << "th power is 1/" << iAnswer << endl;
	return 0; 
} 

long fun(int iNumber,int iPower) 
{ 
	long iAnswer = 1;

	if ( iNumber==0 || iNumber==1 || iPower==1 )
		return iNumber;
	else if ( iPower == 0 )
		return 1;
	else if ( iPower < 0 )
		iPower = -iPower;
	for ( int i=1; i<=iPower; i++)
		iAnswer *= iNumber;
	return iAnswer;
}

这么简单的题别用递归了吧……

[paidfighting]

爱用递归的人不少啊

[WZL19870523]

用递归，很好

[hasen2008]

#include <iostream>
#include <cmath>
using namespace std;

template <typename _T>
_T Square(int x, int y)
{
if ((0 == x) || (1 == x))
{
return x;
}
if (0 == y)
{
return 1;
}
if (1 == y)
{
return x;
}

if (0 < y)
{
return x * Square<_T>(x, y - 1);
}
else
{
return 1 / (x * Square<_T>(x, abs(y) - 1));
}
}

void main()
{
double x;
int y;
cout << "input x , y:" << endl;
cin >> x;
cin >> y;

double z1 = 0;
int z2 = 0;

y < 0 ? z1 =
Square<double>(x, y)
: z2 = Square<int>(x, y);

cout << "NAGATIVE:" << z1 << endl;
cout << "PULUS:" << z2 << endl;

}

[babyofbaby]

一楼很搞笑，四楼正解

[ouyh12345]

幂为负数很正常的

[jianjun7]

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
double x,y;
while(cin>>x>>y)
{
cout<<pow(x,y)<<endl;
}
return 0;
}

[tsocpp]

幂为负在数学上也是有定义的呀

x的-y次方应该返回x的y次方的倒数啊

[mjlxlx]

循环和判断就可以解决的问题……