1012

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 11 22 2.53 2.6666666674 2.708333333
 

 

Source
Greater New York 2000
 

 

Recommend
JGShining
 1 //控制输出
 2  
 3  /*Description: to 
 4  * 
 5  * Author: Vincent
 6  * 
 7  * Date:2010/04/
 8  * Contact:agilely@126.com 
 9  * Zju CS Lab
10  */
11 #include <iostream>
12  using namespace std;
13 
14  #define MAX 1000
15 
16  int main()
17 {
18     int ca,i=1;
19     double e=1;
20     cout<<"n e"<<endl<<"- -----------"<<endl;
21     cout<<"0 1"<<endl<<"1 2"<<endl<<"2 2.5"<<endl;
22     cout.setf(ios::fixed);
23     cout.precision(9);
24     for(ca=1;ca<10;ca++)
25     {
26         i*=ca;
27         e+=1.0/i;
28         if(ca>2)
29             cout<<ca<<" "<<e<<endl;
30     }
31     return 0;
32 }
33 
34 
35  

 

posted @ 2010-04-15 22:01  にんじゃ  阅读(122)  评论(0)    收藏  举报