Java面向对象程序设计课后习题参考答案

习题3_4public class San_4 / 编写一个应用程序求1!+2!+.+20!/* * param args/本题考查对循环的运用和对算法的编写 */public static void main(String args) / TODO Auto-generated method stubint a;double b = 1, sum = 0;/ 定义变量for (a = 1; a = 20; a+) / for循环实现该功能b = b * a;/ 当a=1时，b=b*a=1，依此类推sum = sum + b;/ 当a=1时，sum=sum+b=1，依次类推for (int w = 1; w = 19; w+) / for循环打印出阶乘之和的形式来System.out.print(w + ! + +);System.out.println(20! + = + sum);/ 最终输出阶乘之和double sum1 = 0, a1 = 1;int i = 1;while (i = 20)/ 用while循环实现该功能sum1 = sum1 + a1;i+;a1 = a1 * i;System.out.println(sum1= + sum1);输出结果：1!+2!+3!+4!+5!+6!+7!+8!+9!+10!+11!+12!+13!+14!+15!+16!+17!+18!+19!+20!=2.5613274941118203E18sum1=2.5613274941118203E18习题3_5public class San_5 / 求100以内的素数/* * param args/本题考查对for循环、break语句、if语句的运用 */public static void main(String args) / TODO Auto-generated method stubint i, j;for (i = 2; i = 100; i+) / for循环从2开始，逐渐增1for (j = 2; j i / 2) / 如果ji/2，输出iSystem.out.print( + i);输出结果：2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97习题3_6public class San_6 / 分别用while和for循环计算1+1/2!+1/3!+1/4!.的前20项和/* * param args/本题考查同学们对while循环、for循环以及对输出字符串格式的运用 */public static void main(String args) / TODO Auto-generated method stubint i = 1, k;double j = 1, m = 1, sum1 = 0, sum2 = 0;while (i = 20) / 用while循环实现sum1 = sum1 + j;i+;j = j * (1.0 / i);System.out.print(1+);for (int r = 2; r = 19; r+) System.out.print(1/ + r + ! + +);System.out.print(1/20! + = + sum1);System.out.println();System.out.print(1+);for (int r = 2; r = 19; r+) System.out.print(1/ + r + ! + +);for (k = 1; k = 20; k+) / 用for循环实现m *= (1.0 / k);sum2 += m;System.out.println(1/20! + = + sum2);计算结果：1+1/2!+1/3!+1/4!+1/5!+1/6!+1/7!+1/8!+1/9!+1/10!+1/11!+1/12!+1/13!+1/14!+1/15!+1/16!+1/17!+1/18!+1/19!+1/20!=1.71828182845904551+1/2!+1/3!+1/4!+1/5!+1/6!+1/7!+1/8!+1/9!+1/10!+1/11!+1/12!+1/13!+1/14!+1/15!+1/16!+1/17!+1/18!+1/19!+1/20!=1.7182818284590455习题3_7public class San_7 / 一个数如果恰好等于它的因子之和，这个数就成为”完数“。/ 编写一个应用程序求1000以内的所有完数。/* * param args */public static void main(String args) / TODO Auto-generated method stubint i, j,sum;for (i = 1; i = 1000; i+) for (j = 1, sum= 0; j i; j+) if (i % j = 0) / 当i除以j的余数为0时，执行sum=sum+jsum = sum + j;if (sum = i) / 如果sum=i,执行下列输出语句System.out.println(完数： + i);计算结果：完数：6完数：28完数：496习题3_8import java.util.Scanner;public class San_8 / 编写应用程序，计算两个非零正整数的/ 最大公约数和最小公倍数，要求两个正整数从键盘输入/* * param args */public static void main(String args) / TODO Auto-generated method stubint a, b, c, d = 0, e = 0;System.out.println(请从键盘输入两个正整数：);Scanner nihao = new Scanner(System.in);System.out.print(你输入的第一个数为：);a = nihao.nextInt();System.out.print(你输入的第二个数为：);b = nihao.nextInt();if(a=0|b=0)/当输入的数有零存在时，输出错误提示信息System.out.println(输入有错，请确认后重新输入！);elsefor (c = 1; c 9999) / 当sum9999时，跳出while循环break;n+;a = n * a;System.out.println(满足条件的最大整数是: + (n - 1);System.out.println(1至 + (n - 1) + 的阶乘和为： + sum);计算结果：满足条件的最大整数是:71至7的阶乘和为：46233习题9_6public class A public static void main(String args) String s=aBcDeFgH;String t=你好!;String v=我是中国人;String a=s.toUpperCase();/将小写字母转换为大写字母String b=s.toLowerCase();/将大写字母转换为小写字母System.out.println(a);System.out.println(b);System.out.println(t.concat(v);/将字符串v连接到字符串t之后输出结果：ABCDEFGHabcdefgh你好!我是中国人习题9_7public class B public static void main(String args) String a = 书山有路勤为径;char b = a.charAt(0);/ 输出当前字符串的第一个字符char c = a.charAt(a.length() - 1);/ 输出当前字符串的最后一个字符System.out.println(b);System.out.println(c);输出结果： 书径习题9_8import java.util.Calendar;import java.util.Scanner;public class C /* * param args * 输出某年某月的日历页，通过键盘输入年份和月份 */public static void main(String args) / TODO Auto-generated method stubint m, n;System.out.println(请从键盘输入年份：);Scanner ym = new Scanner(System.in);m = ym.nextInt();/ 读入年份System.out.println(请从键盘输入月份：);n = ym.nextInt();/ 读入月份CalendarBean cb = new CalendarBean();cb.setYear(m);cb.setMonth(n);String a = cb.getCalendar(); / 返回号码的一维数组char str = 日一二三四五六.toCharArray();for (char c : str) System.out.printf(%4c, c);for (int i = 0; i a.length; i+) / 输出一维数组if (i % 7 = 0)System.out.println(); / 换行System.out.printf(%4s, ai);package xt9_8;import java.util.Calendar;public class CalendarBean String day;int year = 2008, month = 0;public void setYear(int year) this.year = year;public void setMonth(int month) this.month = month;public String getCalendar() String a = new String42;Calendar 日历 = Calendar.getInstance();日历.set(year, month - 1, 1);int 星期几 = 日历.get(Calendar.DAY_OF_WEEK) - 1;int day = 0;if (month = 1 | month = 3 | month = 5 | month = 7 | month = 8| month = 10 | month = 12)day = 31;if (month = 4 | month = 6 | month = 9 | month = 11)day = 30;if (month = 2) if (year % 4 = 0) & (year % 100 != 0) | (year % 400 = 0)day = 29;elseday = 28;for (int i = 0; i 星期几; i+)ai = ;for (int i = 星期几, n = 1; i 星期几 + day; i+) ai = String.valueOf(n);n+;for (int i = 星期几 + day; i a.length; i+)ai = ;return a;输出结果：请从键盘输入年份：2011请从键盘输入月份：12 日 一 二 三 四 五 六 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31习题9_9import java.util.Calendar;import java.util.Scanner;public class C /* * param args 计算某年、某月、某日与某年、某月、某日之间的天数间隔。 * 要求年、月、日从键盘上输入到程序中 */public static void main(String args) / TODO Auto-generated method stubint y1, y2, m1, m2, d1, d2;System.out.println(请从键盘输入起始年份:);Scanner ymd = new Scanner(System.in);y1 = ymd.nextInt();/读入数据System.out.println(请从键盘输入起始月份：);m1 = ymd.nextInt();System.out.println(请从键盘输入起始日：);d1 = ymd.nextInt();System.out.println(请从键盘输入终止年份：);y2 = ymd.nextInt();System.out.println(请从键盘输入终止月份：);m2 = ymd.nextInt();System.out.println(请从键盘输入终止日：);d2 = ymd.nextInt();Calendar calendar = Calendar.getInstance();/初始化日历对象calendar.set(y2, m2 - 1, d2);System.out.print(y2 + 年 + m2 + 月 + d2 + 日与);long time2 = calendar.getTimeInMillis();calendar.set(y1, m1 - 1, d1);System.out.print(y1 + 年 + m1 + 月 + d1 + 日);long time1 = calendar.getTimeInMillis();long 相隔天数 = (time2 - time1) / (1000 * 60 * 60 * 24);/毫秒*秒*分*时System.out.println(相隔 + 相隔天数 + 天);输出结果：请从键盘输入起始年份:1991请从键盘输入起始月份：8请从键盘输入起始日：8请从键盘输入终止年份：2011请从键盘输入终止月份：12请从键盘输入终止日：62011年12月6日与1991年8月8日相隔7425天习题9_10public class F /* * param args * Math类的常用方法 */public static void main(String args) / TODO Auto-generated method stubdouble a = -5, b = -4;double c1 = 1, c2 = 3;double g = Math.abs(a);/ 返回a的绝对值System.out.println(g);double d = Math.max(a, b);/ 返回a、b的最大值System.out.println(d);double m = Math.min(a, b);/ 返回a、b的最小值System.out.println(m);double x = Math.random();/ 返回01之间的随机数（不包括0和1）System.out.println(x);double r = Math.pow(c1, c2);/ 返回c1的c2次幂System.out.println(r);double s = Math.sqrt(c1)