u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 46844    Accepted Submission(s): 21489

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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1012

分析：暴力打表就好了，因为数据范围只有10个，按照格式打出来就好了，一个简单的求阶层的题目！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 inline double gcd(int n)
 4 {
 5     int sum=1;
 6     for(int i=1;i<=n;i++)
 7         sum*=i;
 8     return sum;
 9 }
10 int main()
11 {
12     cout<<'n'<<" "<<'e'<<endl;
13     cout<<"- -----------"<<endl;
14     cout<<0<<" "<<1<<endl;
15     cout<<1<<" "<<2<<endl;
16     cout<<2<<" "<<2.5<<endl;
17     double sum=2.5;
18     for(int i=3;i<=9;i++)
19     {
20         sum+=(1.0/(double)gcd(i));
21         printf("%d %.9lfn",i,sum);
22     }
23     return 0;
24 }