hdoj1012--u Calculate e
where n is allowed to go to infinity.
This can actually yield very accurate approximations of e using relatively small
values of n.
formula for the values of n from 0 to 9. The beginning of your output should
appear similar to that shown below.
#include<iostream>
#include <iomanip> using namespace std; int main()
{
int i,n;
double s,t;
cout<<"n e"<<endl;
cout<<"- -----------"<<endl;
cout<<"0 1"<<endl;
cout<<"1 2"<<endl;
cout<<"2 2.5"<<endl;
s=2.5;
n=;
for(i=; i<=; i++)
{
n=i*n;
s+=1.0/n;
cout<<i<<" "<<fixed<<setprecision()<<s<<endl;
}
return ;
}
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