杭电1012

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40284    Accepted Submission(s): 18307

Problem Description

A simple mathematical formula for e is



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 11 22 2.53 2.6666666674 2.708333333

 

#include <iostream>#include <iomanip>using namespace std;double fun(double j){    double sum=1;    for(int i=1;i<=j;i++)    {        sum=sum*i;    }    return sum;}int main(){    double sum=0;    cout<<"n"<<" "<<"e"<<endl;    cout<<"-"<<" "<<"-----------"<<endl;    cout<<0<<" "<<1<<endl;    cout<<1<<" "<<2<<endl;    cout<<2<<" "<<2.5<<endl;    for(int i=3;i<=9;i++)    {        for(int j=0;j<=i;j++)        {            if(j==0)            {                sum=1;            }            else            {                sum=1/fun(j)+sum;            }        }        cout<<i<<" "<<setiosflags(ios::fixed)<<setprecision(9)<<sum<<endl;        sum=0;    }    return 0;}