u Calculate e

u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35137 Accepted Submission(s): 15824

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 1
1 2
2 2.5
3 2.666666667
4 2.708333333
开始错了几次，后来在网上说第八个后面有个０，最后决定用ｐｒｉｎｔｆ控制格式

#include <iostream>#include <iomanip>#include <cstdio>using namespace std;double a;double sum[10];int main(){    a=1;    sum[0]=1;    sum[1]=2;    for(int i=2; i<=9; i++)    {        a=a*(1.0/i);        sum[i]=a+sum[i-1];    }    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++)    {        printf("%d %.9f\n",i,sum[i]);    }    return 0;}