HDU1012

u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46913 Accepted Submission(s): 21522

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

Source
Greater New York 2000

Recommend
JGShining | We have carefully selected several similar problems for you: 1013 1002 1021 1017 1020

---

练习格式输出的题

#include<cstdio>int main(){    double n=1;    int temp = 1;    printf("n e\n");    printf("- -----------\n");    printf("0 1\n");    for (int i = 1; i < 10; i++)    {        temp *= i;        n += 1.0 / temp;        if (i == 1)printf("%d %.0f\n",i,n);        else if (i == 2)printf("%d %.1f\n",i,n);        else printf("%d %.9f\n", i, n);    }    return 0;}