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
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 %.9lf\n",i,sum);
22 }
23 return 0;
24 }