HDOJ 1012 u Calculate e

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

简单的阶乘运算。
对于小数大于9位的，保留9位小数，四舍五入。

public class Main { public static void main(String[] args) { //打表输出： System.out.println("n e"); System.out.println("- -----------"); System.out.println("0 1"); System.out.println("1 2"); System.out.println("2 2.5"); //3-9 的数： for(int i=3;i<10;i++){ double a=0; for(int k=0;k<=i;k++){ a = a+fact(a,k); } System.out.print(i+" "); //默认为四舍五入 System.out.printf("%.9f",a); System.out.println(); } } //返回数为i的阶乘分之一 private static double fact(double a, int i) { double e = 1; if(i==0){ return 1; } for(int j=1;j<=i;j++){ e = e*j; } return 1/e; }

}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/50937152