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.
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
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：
#include <bits/stdc++.h>
using namespace std;
inline double gcd(int n)
{
int sum=;
for(int i=;i<=n;i++)
sum*=i;
return sum;
}
int main()
{
cout<<'n'<<" "<<'e'<<endl;
cout<<"- -----------"<<endl;
cout<<<<" "<<<<endl;
cout<<<<" "<<<<endl;
cout<<<<" "<<2.5<<endl;
double sum=2.5;
for(int i=;i<=;i++)
{
sum+=(1.0/(double)gcd(i));
printf("%d %.9lf\n",i,sum);
}
return ;
}