zju1113
u Calculate e
http://acm.zju.edu.cn/show_problem.php?pid=1113
Time limit: 1 Seconds
Memory limit: 32768K
Background
A simple mathematical formula for e is
\[ e=\sum_{i=0}^n\frac{1}{n!} \]
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
Output
Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.
Sample Output
n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333Problem Source: Greater New York 2000
/*Written by 洪峰*/
#include<iostream>
using namespace std;
int main()
{
long double t1=6, e=2.5;
int n;
cout<<"n"<<" "<<"e"<<endl;
cout<<"- -----------"<<endl;
cout<<0<<" "<<1<<endl;
cout<<1<<" "<<2<<endl;
cout<<2<<" "<<2.5<<endl;
for (n=3;n<10;n++)
{
e+=1/t1;
t1*=(n+1);
cout.precision(9);
cout.setf(ios::fixed);
cout<<n<<" "<<e<<endl;
}
return 0;
}