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;
}