HDU1012

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46106    Accepted Submission(s): 21178

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
  
  
  
  

   
   //利用公式计算0-9次循环中 e 的值
  
  
  
  

   
   
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iomanip>
using namespace std;

int main(int argc, char const *argv[])
{
    double result = 1;
    double a[10];
    cout << "n e" << endl;
    cout << "- -----------" << endl;
    for(int i = 0;i <10; ++i)
    {
        if(i == 0)
            a[i] = 1;
        else
        {
            double t = 1;
            for(int j = 1;j <= i; ++j)
                t *= j;
            a[i] = 1/t;
            result += a[i];
        }
        if(i >=  3)
        {
            cout << setiosflags(ios::fixed);
            cout << setprecision(9);
            cout << i << " " << result << endl;
        }
        else
            cout << i << " " << result << endl;
    }
    return 0;
}
   
   
//或者直接枚举
  
  
  
  

   
   
# include<stdio.h>
int main()
{
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n1 2\n2 2.5\n");
    printf("3 2.666666667\n");
    printf("4 2.708333333\n");
    printf("5 2.716666667\n");
    printf("6 2.718055556\n");
    printf("7 2.718253968\n");
    printf("8 2.718278770\n");
    printf("9 2.718281526\n");
    
    return 0;
}