ACM-简单题之u Calculate e——hdu1012

u Calculate e
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 2 Accepted Submission(s) : 2

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

这道题就是求e了，e的算法图中也给明了，假如i=4，e=1+1/1!+1/2!+1/3!+1/4!

就是注意一下格式，我是用printf，好控制一些

#include <stdio.h>
#include <math.h>

double arr[10];

void cal(void)
{
	int i;
	double ft,s;
	ft=arr[0]=s=1.0;
	for(i=1;i<10;++i)
	{
		s*=i;
		arr[i]=1.0/double(s)+ft;
		ft=arr[i];
	}
}

int main()
{
	int i;
	printf("n e\n");
	printf("- -----------\n");
	
	cal();

	printf("0 1\n");
	printf("1 2\n");
	printf("2 2.5\n");
	for(i=3;i<10;++i)
		printf("%d %.9f\n",i,arr[i]);

	return 0;
}