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;}