杭电 u Calculate e

最新推荐文章于 2020-07-23 09:38:13 发布

runnerzyc

最新推荐文章于 2020-07-23 09:38:13 发布

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分类专栏：数学问题

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本文介绍了一种通过数学公式计算自然对数的底数e的近似值的方法，并提供了从n=0到n=9的近似值计算示例。代码使用C语言实现，涉及阶乘计算和输出格式化。

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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

Source

Greater New York 2000

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分析：此题本来感觉比较难，认真读题后，就是输出问题，此题需注意俩个点：

阶乘问题和输出的格式；

代码;

  #include<stdio.h>
  int f(int j)
  {
  	int i,sum=1;

  	for(i=1;i<=j;i++)
  	sum*=i;
  	return sum;
  }  
  int main()
  {
  	int n,i,j;
  	double num=2.5;
    double  e;
  	printf("n e\n");
  	printf("- -----------\n");
  	printf("0 1\n1 2\n2 2.5\n");
  	for(i=3;i<=9;i++)
  	{
	  e= 1.0*1/f(i);
  	  num+=e;
     printf("%d %.9lf\n",i,num);
    }
  	return 0;
  }