HDU1012

最新推荐文章于 2024-07-19 14:00:00 发布

lx0208

最新推荐文章于 2024-07-19 14:00:00 发布

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分类专栏： AC历程文章标签： HDU1012

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本文介绍了一种通过数学公式计算自然常数e的近似值的方法，并提供了C++代码实现，该程序能够输出n从0到9时e的近似值。

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u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46913 Accepted Submission(s): 21522

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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练习格式输出的题

#include<cstdio>
int main()
{
    double n=1;
    int temp = 1;
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n");
    for (int i = 1; i < 10; i++)
    {
        temp *= i;
        n += 1.0 / temp;
        if (i == 1)printf("%d %.0f\n",i,n);
        else if (i == 2)printf("%d %.1f\n",i,n);
        else printf("%d %.9f\n", i, n);
    }
    return 0;
}