HDU 1012 u Calculate e

最新推荐文章于 2018-10-05 12:17:57 发布

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最新推荐文章于 2018-10-05 12:17:57 发布

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分类专栏： hdu ACM 水题128题文章标签： output go

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u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19373 Accepted Submission(s): 8458

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：根据公式计算e，注意前3个数据与后面精度保留不同，所以可以先独立输出。

代码：

#include<stdio.h>

int main()
{
    double a[10];
    int i;
    a[0]=1;
    for(i=1;i<10;i++)
        a[i]=a[i-1]*i;
    printf("n e\n- -----------\n");
    printf("0 1\n");
    printf("1 2\n");
    printf("2 2.5\n");
    double sum=2.5;
    for(i=3;i<10;i++)
    {
        sum+=1.0/a[i];
        printf("%d %.9f\n",i,sum);
    }
    return 0;
}