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u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46276    Accepted Submission(s): 21237

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

#include<stdio.h>int jc(int n)
{
    int i,j=1;
    for(i=1;i<=n;i++)
        j=j*i;
        return j;
}
int main()
{
    int n,i;
    double e=2.5;    
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n");
    printf("1 2\n");
    printf("2 2.5\n");
    for(i=3;i<=9;i++)//为什么从3开始，是因为从3之后，小数点后有9位小数     {
     e=e+1.0/jc(i);
     printf("%d %.9lf\n",i,e);    
    }
    return 0;
}

Hdu 1012 u Calculate e

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原文地址：http://www.cnblogs.com/zhangliu/p/7057785.html