u Calculate e - 1517_1517 u calculate e-优快云博客

本文详细介绍了如何通过递归和递推两种方法计算数学常数e的近似值，适用于从n=0到n=9的不同情况。包括了基本输入输出格式说明和代码实现细节。

u Calculate e

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19242		Accepted: 11267		Special Judge

Description

A simple mathematical formula for e is
e=Σ_0<=i<=n1/i!
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Input

No input

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 Input

no input

Sample Output

n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333
...

Source

法1：由于题中就让输出10以内的n，采用递归比较简单：

/***********************************************************************
    Copyright (c) 2015,wangzp
    All rights no reserved.
  
    Name: 《u Calculate e》In PEKING UNIVERSITY ACM
    ID:   PROBLEM 1517
    问题简述: 根据n求e: e=Σ(1/i!),其中 0<=i<=n.
	      采用递归比较简单。
    Date: Sep 25, 2015 
 
***********************************************************************/
#include

double fac(int n)
{
       float f;
	   if(n == 0||n == 1)
	   {
		   f = 1;
	   }
	   else if(n > 1)
	   {
           f = fac(n - 1) * n;
	   }
	   return f;
}

int main(void)
{

	double a,sum=0;
	int i;
	printf("n e\n");
	printf("- -----------\n");
	for(i = 0;i < 10;i++)
	{ 
		a = 1 / fac(i);
		sum = a + sum;
		printf("%d %.9f\n",i,sum);    
	}
	return 0;
}

法2：采用递推也可以，如果n比较大的情况下，采用递推可以节省时间：

/***********************************************************************
    Copyright (c) 2015,wangzp
    All rights no reserved.
  
    Name: 《u Calculate e》In PEKING UNIVERSITY ACM
    ID:   PROBLEM 1517
    问题简述: 根据n求e: e=Σ(1/i!),其中 0<=i<=n.
	          采用递归比较简单，但是耗时，最好采用递推来做。
    Date: Sep 25, 2015 
 
***********************************************************************/
#include

int main(void)
{

	double sum = 1;
	double a,sum_f;
	int i,j;
	printf("n e\n");
	printf("- -----------\n");

	/*i = 0比较简单，单独输出*/
	printf("0 1\n");
	for(i = 1;i < 10;i++)
	{ 
		sum_f = 1;
		for (j = 1;j <= i;j++)
		{
			sum_f *= j;
		}

		a = 1 / sum_f;
		sum = a + sum;
		printf("%d %.9f\n",i,sum);
	}
	return 0;
}