java练习 计算n位可被浮点数精确表示的小数

原理，穷举生成N-1位定点二进制小数（第N位为1），然后转换成十进制的就行了。

import java.util.Scanner;public class GetDecimal {static Vector<int[]> record = null;public static void main(String[] args) {Scanner sc = new Scanner(System.in);System.out.println("输入n,输出n位可被浮点数精确表示的小数");int len = sc.nextInt();if (len == 1)System.out.println(0.5);elsef(len - 1);}public static void f(int len) {double dec = 0;int max = (int) Math.pow(2, len) - 1;for (int i = 0; i <= max; i++) {dec = getSum(getBinString(i, len) + "1");System.out.println(dec);}}// 转换成二进制字符串public static String getBinString(int i, int len) {StringBuffer str = new StringBuffer(Integer.toString(i, 2));for (int j = 0; str.length() != len; j++) {str.insert(0, "0");}return str.toString();}// 计算值public static double getSum(String s) {double sum = 0;for (int i = 0; i < s.length(); i++) {if (s.charAt(i) == '1') {sum += Math.pow(2, -(i + 1));}}return sum;}}