HDU 1012 u Calculate e (格式呀)

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40587    Accepted Submission(s): 18448

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
   
   
   
   

 

Source

Greater New York 2000

题意:

就是根据公式求e的值

PS:最后的结果保留的小数是关键!!!

AC代码:

#include <stdio.h>
int fac(int n);
int main ()
{
    int i;
    double e=0;
    printf("n e\n");
    printf("- -----------\n");
    for(i=0;i<=9;i++)
    {
        e+=1.0/fac(i);
        if (i<3)
        printf("%d %g\n",i,e);
        else
        printf("%d %.9lf\n",i,e);
    }
    return 0;
}
int fac(int n)
{
    int i,s=1;
    for (i=1;i<=n;i++)
    {
        s*=i;
    }
    return s;
}

AC代码2;

# 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;
}