HDOJ 1012

水題每天有，今天特別多....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

這個題目挺坑爹啊....8的時候最後得補個0.....好吧.....投機取巧的居然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 }