请朋友们帮我指点一下（谢谢）：编C++程序求e的x次方的值执行相当怪异，

[诺希亚]

e(x)=1+x(1)/1!+x(2)/2!+.....+x(n)/n!
编程求e(x)的值
#include<iostream>
#include<iomanip>
#include<cmath>
#include<fstream>
using namespace std;


int main()
{
double product=1,quotient=0,x=0,value=0;
ofstream outfile("C:\\out.OUT");
while(x>-1&&x<1)
{
outfile<<"Enter x: "<<endl;
cin>>x;
outfile<<"x="<<setprecision(2)<<x<<endl;

for(int n=1;n<=500;n++)
{
product*=n;
quotient=pow(x,n)/product;
value+=quotient;
}

value+=1;
outfile<<"the sum of e of the x-the power is "<<setprecision(5)<<fixed<<value<<endl;
}
outfile.close();
return 0;
}

结果输出为：
Enter x:
x=0.01
the sum of e of the x-the power is 1.01005
Enter x:
x=0.02
the sum of e of the x-the power is 2.01005
Enter x:
x=0.03
the sum of e of the x-the power is 3.01005
Enter x:
x=0.04
the sum of e of the x-the power is 4.01005
Enter x:
x=0.05
the sum of e of the x-the power is 5.01005
Enter x:
x=0.06
the sum of e of the x-the power is 6.01005
Enter x:
x=0.07
the sum of e of the x-the power is 7.01005
Enter x:
x=0.08
the sum of e of the x-the power is 8.01005
Enter x:
x=0.09
the sum of e of the x-the power is 9.01005
Enter x:
x=0.10
the sum of e of the x-the power is 10.01005
Enter x:
x=0.20
the sum of e of the x-the power is 11.01005
Enter x:
x=0.30
the sum of e of the x-the power is 12.01005
Enter x:
x=0.40
the sum of e of the x-the power is 13.01005
Enter x:
x=0.50
the sum of e of the x-the power is 14.01005
Enter x:
x=0.60
the sum of e of the x-the power is 15.01005
Enter x:
x=0.70
the sum of e of the x-the power is 16.01005
Enter x:
x=0.80
the sum of e of the x-the power is 17.01005
Enter x:
x=0.90
the sum of e of the x-the power is 18.01005
Enter x:

另一次输出为：
Enter x:
x=0.1
the sum of e of the x-the power is 1.10517
Enter x:
x=0.20
the sum of e of the x-the power is 2.10517
Enter x:
x=0.30
the sum of e of the x-the power is 3.10517
Enter x:
x=0.40
the sum of e of the x-the power is 4.10517
Enter x:
x=0.50
the sum of e of the x-the power is 5.10517
Enter x:
x=0.60
the sum of e of the x-the power is 6.10517
Enter x:
x=0.70
the sum of e of the x-the power is 7.10517
Enter x:
x=0.80
the sum of e of the x-the power is 8.10517
Enter x:
x=0.90
the sum of e of the x-the power is 9.10517
Enter x:
哪位朋友帮我指点一下，哪里出错了
非常感谢

[ltc_mouse]

product能存得下500!吗？

[youdang45]

处理完第一个输入的x后product要恢复到1

[sagegz]

LZ哪里有问题呢?
您的第一次输入都是2位小数,第二次输入都是1位小数,结果当然2次不一样啦!

[youdang45]

似乎不是我说的问题

[sagegz]

当然还得考虑溢出问题了.

[youdang45]

试了下

product会溢出，用long double能解决

还有就是处理下一个输入的x前要将product置1， value置0

[waizqfor]

[Quote=引用楼主 huih303 的帖子:]
e(x)=1+x(1)/1!+x(2)/2!+.....+x(n)/n!
编程求e(x)的值
#include <iostream>
#include <iomanip>
#include <cmath>
#include <fstream>
using namespace std;


int main()
{
double product=1,quotient=0,x=0,value=0;
ofstream outfile("C:\\out.OUT");
while(x>-1&&x <1)
{
outfile < <"Enter x: " < <endl;
cin>>x;
outfile < <"x=" < <setprecision(2) < <x < <endl;

f…
[/Quote]
LZ变量定义的时候
double product=1,quotient=0,x=0,value=0;
product的范围太小了吧 精度会不够的吧 换个精度高的

[T_mon]

int main() 
{ 
  double product=1,quotient=0,x=0,value=0; 
  ofstream outfile("C:\\out.OUT"); 
  while(x>-1&&x <1) 
{ 
  outfile < <"Enter x: " < <endl; 
  cin>>x; 
  outfile < <"x=" < <setprecision(2) < <x < <endl; 
product=1,quotient=0,x=0,value=0; //这里需要重新赋值，并且没有必要算500项
  for(int n=1;n <=500;n++) 
  { 
    product*=n; 
quotient=pow(x,n)/product; 
    value+=quotient; 
  } 

  value+=1; 
  outfile < <"the sum of e of the x-the power is " < <setprecision(5) < <fixed < <value < <endl; 
} 
  outfile.close(); 
  return 0; 
} 

[T_mon]

int main() 
{ 
  double product=1,quotient=0,x=0,value=0; 
  ofstream outfile("C:\\out.OUT"); 
  while(x>-1&&x <1) 
{ 
  outfile < <"Enter x: " < <endl; 
  cin>>x; 
  outfile < <"x=" < <setprecision(2) < <x < <endl; 
product=1,quotient=0,x=0,value=0;//注意
  for(int n=1;n <=500;n++) 
  { 
    product*=n; 
quotient=pow(x,n)/product; 
    value+=quotient; 
  } 

  value+=1; 
  outfile < <"the sum of e of the x-the power is " < <setprecision(5) < <fixed < <value < <endl; 
} 
  outfile.close(); 
  return 0; 
}