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 }

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