hdu 1012 u Calculate e

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36110    Accepted Submission(s): 16298

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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这道题没有输入，开始看到的时候怎么看怎么觉得怪异~题目不难，读懂题就好，注意输出，前三个数的输出后后面的输出不同。

 

题意：按题目给的公式，输出n从0到9的结果，前3个lg输出，后面的保留9位小数。

 

附上代码：

#include <stdio.h>
double add(double t)
{
    double s=1;
    int i;
    if(t==0)   //0！=1
        return 1;
    for(i=1; i<=t; i++)  //求n！
        s*=i;
    return s;
}
int main()
{
    double e=0;
    int i,j;
    printf("n e
");
    printf("- -----------
");
    printf("0 1
");
    for(i=1; i<=9; i++)
    {
        e=0;
        for(j=0; j<=i; j++)
            e=e+1/add(j);
        if(i<=2)              //输出需要分开讨论
            printf("%d %lg
",i,e) ; 
        else
            printf("%d %.9lf
",i,e) ;
    }
    return 0;
}