HDOJ 1012

水题天天有，今天特别多....嘿嘿
u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19289    Accepted Submission(s): 8423


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

这个题目挺坑爹啊....8的时候最后得补个0.....好吧.....投机取巧的竟然AC了

 1 #include <iostream>
 2 #include <iomanip>
 3 using namespace std;
 4 int main()
 5 {
 6     cout<<"n e"<<endl;
 7     cout<<"- -----------"<<endl;
 8     double sum=1;
 9     double count;
10     int i,j;
11     cout<<"0 "<<1<<endl;
12     for(i=1;i<10;++i)
13     {
14         count = 1;
15         for(j=i;j>0;j--)
16         {
17             count*=j;
18         }
19         sum+=1/count;
20         if(i == 8)
21         {
22             cout<<i<<" "<<setprecision(9)<<sum<<"0"<<endl;
23         }
24         else
25         {
26             cout<<i<<" "<<setprecision(10)<<sum<<endl;
27         }
28     }
29     return 0;
30 }