Java基本功练习二（显示日历，石头剪子布游戏，找完全数等）
本作者是基于Eclipse编写的程序，当然也可以用TextPad或NetBeans进行编程。不管用哪一种，学习的重点都是将具体问题转化成程序化语言的能力，学习的是这种思想。希望童鞋们和我一样，每天进步一点点！废话少说，进入正题
示例一：用户输入某一年，程序输出显示这一年每个月的日历。(你能自己设计出来吗？)请看运行效果图（只展示了前六月的六张图）。
这道题目有几个坑：
1)如何显示对齐以符合要求；
2)如何确定每月第一天是星期几；
3)如何判断某一月有多少天；
4)闰年的情况要考虑周到。
基于Eclipse的实现代码如下：
package shiyanPractice;
import java.util.Scanner;
public class ShiYanPractice {
public static void main(String[] args) {
//输入年份，显示相应年份的日历
Scanner input = new Scanner(System.in);
System.out.print("Enter the number of year: ");
int year = input.nextInt();//输入想要显示的年份
int month = 1;
int day = 0;
//12个月，所以循环12次，每次显示一个月的日历
for(int i = 0;i < 12;i++){
int h,q,m,j,k;
month = i+1;
day = 1; //只需算出每个月的第一天是周几
m = month;
q = day;
//当月份为1或2时，要将其视为上一年的13或14月，并且相应的年份改为上一年
if(m == 1 || m == 2){
m+=12;
year -= 1;
}
j = year/100; //求出h表示的星期几，其中0表示周日，1表示周六，2表示周一，直到6表示周五
k = year%100;
//泽勒一致性公式
h = (q+(int)(26*(m+1)/10)+k+(int)(k/4)+(int)(j/4)+5*j)%7;
//计算完h之后要将year的值改回来
if(month == 1 ||month == 2)
year += 1;
//用于将h转换成对应的星期几
String dayTranslate = "";
switch(h){
case 0:dayTranslate = "Saturday";break;
case 1:dayTranslate = "Sunday";break;
case 2:dayTranslate = "Monday";break;
case 3:dayTranslate = "Thusday";break;
case 4:dayTranslate = "Wednesday";break;
case 5:dayTranslate = "Thursday";break;
case 6:dayTranslate = "Friday";break;
}
//用以显示日历的每个月的抬头
String monthFor = "";
switch(month){
case 1:monthFor = "January";break;
case 2:monthFor = "February";break;
case 3:monthFor = "March";break;
case 4:monthFor = "April";break;
case 5:monthFor = "May ";break;
case 6:monthFor = "June ";break;
case 7:monthFor = "July ";break;
case 8:monthFor = "August";break;
case 9:monthFor = "September";break;
case 10:monthFor = "October";break;
case 11:monthFor = "November";break;
case 12:monthFor = "December";break;
}
//显示每个月的抬头
String title = "\t"+monthFor+" "+year+"\n"
+"Sun "+"Mon "+"Tue "+"Wed "+"Thu "+"Fri "+"Sat\n";
System.out.println(title);
//将h转换成相应的星期几的数字
int dayCount = 0;
switch(h){
case 0:dayCount = 6;break;
case 1:dayCount = 0;break;
case 2:dayCount = 1;break;
case 3:dayCount = 2;break;
case 4:dayCount = 3;break;
case 5:dayCount = 4;break;
case 6:dayCount = 5;break;
}
//某个月第一天在日历中位置的确认
for(int jj = 1;jj <= dayCount;jj++){
System.out.print(" ");
}
int countToAnotherLine = dayCount+1;//将第一天的位置数加1，方便用于后续转行的控制
//计算每个月的天数
int countDayPerMonth = 31;
switch(month){
case 1:countDayPerMonth = 31;break;
case 2:
if((year % 4 == 0 && year % 100 != 0)||(year % 400 ==0))
countDayPerMonth = 29;
else
countDayPerMonth = 28;
break;
case 3:countDayPerMonth = 31;break;
case 4:countDayPerMonth = 30;break;
case 5:countDayPerMonth = 31;break;
case 6:countDayPerMonth = 30;break;
case 7:countDayPerMonth = 31;break;
case 8:countDayPerMonth = 31;break;
case 9:countDayPerMonth = 30;break;
case 10:countDayPerMonth = 31;break;
case 11:countDayPerMonth = 30;break;
case 12:countDayPerMonth = 31;break;
}
//依次输出某个月的每一天
for(int ii = 1;ii <= countDayPerMonth;ii++){
if(countToAnotherLine % 7 == 0){
System.out.printf("%3d ",ii);
System.out.println();
countToAnotherLine++;
}
else{
System.out.printf("%3d ",ii);
countToAnotherLine++;
}
}
System.out.println();
}
}
}
此题的坑在“连续”两字上，要处理好交替赢和中间有平局的情况。下面是实现代码：
package shiyanPractice;
import java.util.Scanner;
public class ShiYanPractice {
public static void main(String[] args) {
//石头剪子布的游戏，电脑或用户赢两次以上就获胜，游戏结束
Scanner input = new Scanner(System.in);
int pc_user_Win = 0;//1 userwin,2 pcwin
int pc_user_WinLastTime = 0;
int pcData = 0;
int userData = 0;
System.out.println("石头、剪子、布游戏，由用户和电脑玩，谁连续赢两次就胜出！");
while(true){
pcData = (int)(Math.random()*3+1);
System.out.print("玩家请输入，1是剪刀，2是石头，3是布:");
userData = input.nextInt();
//统计谁赢，并将pc_user_Win标志位赋值，输出显示赢家
switch(userData){
case 1:{
if(userData - pcData == -2){
pc_user_Win = 1;
System.out.println("User Win!");
break;
}
else if(userData - pcData == -1){
pc_user_Win = 2;
System.out.println("Pc Win!");
break;
}
else{
System.out.println("User and Pc is same!");
break;
}
}
case 2:{
if(userData - pcData == 1){
pc_user_Win = 1;
System.out.println("User Win!");
break;
}
else if(userData - pcData == -1){
pc_user_Win = 2;
System.out.println("Pc Win!");
break;
}
else{
System.out.println("User and Pc is same!");
break;
}
}
case 3:{
if(userData - pcData == 1){
pc_user_Win = 1;
System.out.println("User Win!");
break;
}
else if(userData - pcData == 2){
pc_user_Win = 2;
System.out.println("Pc Win!");
break;
}
else{
System.out.println("User and Pc is same!");
break;
}
}
}
//判断是否连续赢两次，是就输出谁赢，并结束循环
if(pc_user_Win != 0){//排除因为等于零的情况的误判，因为等于零代表没有胜负
if(pc_user_WinLastTime == pc_user_Win && pc_user_Win == 1){
System.out.println("User Win the game!");
break;
}
else if(pc_user_WinLastTime == pc_user_Win && pc_user_Win == 2){
System.out.println("Pc Win the game!");
break;
}
}
//只有将非零的胜负标志值赋值给记录上一次胜负情况的标志位
//否则可能出现pc赢一次，平一次，再赢一次而不显示pc获胜的情况
if(pc_user_Win == 0);
else
pc_user_WinLastTime = pc_user_Win;
pc_user_Win = 0;//将标志位置为无效，否则将影响下一次判断
}
}
}
此题相对来说比较容易，但还是得自己动手实践，实践可以检查自己的问题并提高自我。运行效果图如下：
实现代码如下所示：
package shiyanPractice;
import java.util.Scanner;
public class ShiYanPractice {
public static void main(String[] args) {
//找出完全数的程序
Scanner input = new Scanner(System.in);
System.out.print("Enter an integer: ");
int data = input.nextInt();
System.out.println(0+"~"+data+"之间的完全数显示如下：");
for(int iData = 0;iData < data;iData++){
int number = 2;
int sumOfNumber = 1;
while(number <= iData){
if(iData % number == 0){
if(number != iData)
sumOfNumber += number;
number++;
}
else
number++;
}
if(iData == sumOfNumber &&iData != 1){
System.out.println(iData);
}
}
}
}
运行效果如图所示：
提示：此题的关键是对数据位操作是否熟悉。可以先自己尝试！
实现代码如下所示：
package shiyanPractice;
import java.util.Scanner;
public class ShiYanPractice {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("输入需要输入整型数的次数： ");
int n = input.nextInt();
for(int i = 0;i < n; i++){
System.out.print("Enter an integer: ");
int value = input.nextInt();
System.out.print("The 16 bits are: ");
int mask = 1;
for(int j = 15;j >=0; j--){
int temp = value>>j;
int bit = temp & mask;
System.out.print(bit);
}
System.out.println();
}
}
}