# tctony

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# u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1645    Accepted Submission(s): 696

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```
#include<iostream>
using namespace std;

int main()
{

int f[10]={1,1,2},i;

double e[10]={1.0,2.0,2.5};
printf(
"n e\n");
printf(
"- -----------\n");
printf(
"0 1\n");
printf(
"1 2\n");
printf(
"2 2.5\n");

for(i=3; i<10++i){
f[i]
=f[i-1]*i;
e[i]
=e[i-1]+1.0/f[i];
printf(
"%d %.9lf\n",i,e[i]);
}

return 0;
}

posted on 2007-11-29 08:15 tctony 阅读(517) 评论(0)  编辑 收藏 引用