ZOJ1113

原创于 2018-10-13 18:48:46 发布 · 217 阅读

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本文介绍了一种使用简单数学公式来计算自然常数e的近似值的方法，并通过C语言实现，展示了从n=0到n=9时e的逐步逼近过程。代码中包含了阶乘的计算以及公式应用的具体实现。

Description

Background

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.

Example

Output

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

代码

#include<stdio.h>
#include<math.h>
int cal(int n)
{
    int ans=1;
    for(int i=n;i>=1;i--)
        ans*=i;
    return ans;
}
int main()
{
    printf("n e\n");
    printf("- -----------\n");
    for(int i=0;i<=9;i++)
    {
        if(i==0)
            printf("0 1\n");
        else if(i==1)
            printf("1 2\n");
        else if(i==2)
            printf("2 2.5\n");
        else
        {
            double ans=1;
            for(int k=1;k<=i;k++)
                ans+=1.0*1/cal(k);
            printf("%d %.9lf\n",i,ans);
        }
    }
    return 0;
}