hdu1012 Calculate e

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/* Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30131 Accepted Submission(s): 13439

Problem DescriptionA simple mathematical formula for e is

e=∑（0~n）1/(i！)

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
题目大意；输出i不同的时候，e的值；
题目难点：输出格式；
关键点：小数点后面的位数，以及3项后边的0
解题时间：2014,07,29
解题思路：前3项分别输出，其他再输出
体会：输出格式多练习****
*********/

#include<stdio.h>int main(){printf("n e\n");printf("- -----------\n");double e=0;int i,j,a;for(i=0;i<10;i++){a=1;if(i==0) a==1;else {for(j=1;j<=i;j++)a=a*j; }e+=1.0/a;if(i==0) printf("%d 1\n",i);else if(i==1) printf("%d 2\n",i);else if(i==2) printf("%d 2.5\n",i);else printf("%d %.9lf\n",i,e);}return 0;}

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