joj1020

1020: u Calculate e

---

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	6098	1521	Standard

Background

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

#include<stdio.h>
double factoral(double);
double showsum(int );

int main(void)
{
    int n, i;
    double sum = 0;
    
    printf("n e\n");
    printf("- -----------\n");
    
    printf("0 1\n");
    printf("1 2\n");
    printf("2 2.5\n");
    
    
    for (i=3; i<=9; i++)
    {
        printf("%d ", i);
        printf("%.9lf\n",1+showsum(i));
    }
    
    return 0;
}

double factoral(double i)
{
    if (i == 0)
        return 1;
    else 
        return i*factoral(i-1);
}

double showsum(int n)
{    
    double sum=0;
    for(int i=1;i<=n;i++)
    {
        sum+=(double)1/factoral(i);
    }
    return sum;
}

 

---

This problem is used for contest: 76 182

---

Submit / Problem List / Status / Discuss