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 }