For those who getting WA. Just write the code for this problem and accept it :)

    Description

    A simple mathematical formula for e is 
                                                         e=?0<=i<=n1/i!

    where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small
    values of n.

    Input
    No input


    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 Input
    no input


    Sample Output

    n e
    - -----------
    0 1
    1 2
    2 2.5
    3 2.666666667
    4 2.708333333
    ...

Hi. Can anyone tell me why am I getting WA? This is my code:

#include<iostream>
#include<iomanip>
using namespace std;
unsigned long long int fact(int n)
{
     if (n==0)
        return 1;
        else
        return n*fact(n-1);
}
int main()
{
    int n;
    double ans;
    cout<<1<<endl<<2<<endl<<2.5<<endl;
    for (int k=3;k<=9;k++)
    {
        ans = 0;
        for (int i=0;i<=k;i++)
            ans+=(1/double(fact(i)));
        cout<<fixed<<setprecision(9)<<ans<<endl;
    }
    return 0;
}

･BECCA can make １＝＝３ anyway said:

For those who getting WA. Just write the code for this problem and accept it :)

    Description

    A simple mathematical formula for e is 
                                                         e=?0<=i<=n1/i!

    where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small
    values of n.

    Input
    No input


    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 Input
    no input


    Sample Output

    n e
    - -----------
    0 1
    1 2
    2 2.5
    3 2.666666667
    4 2.708333333
    ...

･BECCA can make １＝＝３ anyway said:

For those who getting WA. Just write the code for this problem and accept it :)

    Description

    A simple mathematical formula for e is 
                                                         e=?0<=i<=n1/i!

    where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small
    values of n.

    Input
    No input


    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 Input
    no input


    Sample Output

    n e
    - -----------
    0 1
    1 2
    2 2.5
    3 2.666666667
    4 2.708333333
    ...

Can you tell me why is this wrong?

cout<<"1"<<endl;
cout<<"2"<<endl;
cout<<"2.5"<<endl;
cout<<"2.666666667"<<endl;
cout<<"2.708333333"<<endl;
cout<<"2.716666667"<<endl;
cout<<"2.718055556"<<endl;
cout<<"2.718253968"<<endl;
cout<<"2.71827877"<<endl;
cout<<"2.718281526";

Hello!

No.

Your output should exactly look like this:

cout << "n e"           << endl;
cout << "- -----------" << endl;
cout << "0 1"           << endl;
cout << "1 2"           << endl;
cout << "2 2.5"         << endl;
cout << "3 2.666666667" << endl;
cout << "4 2.708333333" << endl;
cout << "5 2.716666667" << endl;
cout << "6 2.718055556" << endl;
cout << "7 2.718253968" << endl;
cout << "8 2.718278770" << endl;
cout << "9 2.718281526" << endl;

･BECCA can make １＝＝３ anyway said:

Hello!

No.

Your output should exactly look like this:

cout << "n e"           << endl;
cout << "- -----------" << endl;
cout << "0 1"           << endl;
cout << "1 2"           << endl;
cout << "2 2.5"         << endl;
cout << "3 2.666666667" << endl;
cout << "4 2.708333333" << endl;
cout << "5 2.716666667" << endl;
cout << "6 2.718055556" << endl;
cout << "7 2.718253968" << endl;
cout << "8 2.718278770" << endl;
cout << "9 2.718281526" << endl;

The problem did not say anything about this kind of output! Thanks. It worked.