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  • 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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  • 原文地址:https://www.cnblogs.com/webmen/p/5739369.html
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