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.
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
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 ",i,sum); 22 } 23 return 0; 24 }