ACM1012：u Calculate e

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

----------------------------------------------------------------------------------------------------------

#include<stdio.h>

int main()
{
	double e[10] = {1};
	double sum = 1;
	double ans = 0;
	int i;
	for (i = 1; i < 10; i++)
	{
		sum *= (double)i;
		e[i] = 1 / sum;
	}

	printf("n e\n");
	printf("- -----------\n");
	for (i = 0; i < 10; i++)
	{
		ans += e[i];
		if (i <= 2)
			printf("%d %g\n", i, ans);
		else
			printf("%d %.9f\n", i, ans);
	}
	return 0;
}