HDU 1012 u Calculate e（简单阶乘计算）

Posted 2020-12-04 yinbiao

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http://acm.hdu.edu.cn/showproblem.php?pid=1012

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52607    Accepted Submission(s): 24106

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

 分析：

没有什么好说的

水题

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int f(int x)
{
    if(x==0)
        return 1;
    if(x==1)
        return 1;
    LL sum=1;
    for(int i=1;i<=x;i++)
    {
        sum*=i;
    }
    return sum;
}
int main()
{
    printf("n e
");
    printf("- -----------
");
    printf("0 1
1 2
2 2.5
");
    for(int i=3;i<=9;i++)
    {
        double result=0;
        for(int j=0;j<=i;j++)
        {
            result+=(1.0/(f(j)*1.0));
        }
        printf("%d %0.9lf
",i,result);
    }
    return 0;
}