一.循环结构与分支结构的嵌套练习
1.输入一个数，判断是不是素数(只能被1和自身整除的数)
import java.util.Scanner;
public class Text1 {
public static int sort(int count) {
count = (count + 1) * 2;
return count;
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in); System.out.println("请输入一个数");
*int a =input.nextInt(); boolean isPrime=true; for(int i =2;i<a;i++){
if(a%i==0){ isPrime=false; break; } } System.out.println(a
+(isPrime?"是":"不是")+"素数"); input.close();
2.打印100以内的质数；
for (int i = 1; i <= 100; i++) {
boolean isPrime = true;
for (int j = 2; j < Math.sqrt(i); j++) {
if (i % j == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
System.out.println(i);
}
}
3.九九乘法表
for(int i=1;i<=9;i++){
for(int j=1;j<=i;j++){
System.out.printf("%d*%d=%d\t",j,i,j*i);
}
System.out.println("");
}
4.猴子吃桃问题
int count = 1;
for (int i = 1; i < 10; i++) {
sort(count);
}
System.out.println(count);
}
}
.穷举法案列：100块钱买一百只鸡
public class HomeWork1 {
public static void main(String[] args) {
for (int gj = 0; gj <= 20; gj++) {
for (int mj = 0; mj <= 33; mj++) {
int xj = 100 - gj - mj;
if (5 * gj + 3 * mj + xj / 3 == 100 && (gj + mj + xj) == 100 && xj % 3 == 0) {
System.out.println("公鸡有 " + gj + " 母鸡有 " + mj + " 小鸡有 " + xj);
}
}
}
}
}
二.两个典型案列
1.Graps赌博游戏：
<1>我的代码
package comeDay12_01_2;
import java.util.Scanner;
public class Homework2 {
public static void main(String[] args) {
System.out.println("******************游戏规则**************************");
System.out.println("**玩家总共1000分，每次赢加100分，每次输扣100分，平局分数不变**");
System.out.println("**************************************************");
System.out.println("准备好开始了吗，按\"1\"退出，其他任意键继续:");
int fen = 200;
int num = 0;
int num1 = 0;
Scanner input = new Scanner(System.in);
String zb = input.nextLine();
if (zb.equals("1")) {
System.out.println("游戏结束，还剩" + fen + "分");
} else {
int a = (int) (Math.random() * 6 + 1);
int b = (int) (Math.random() * 6 + 1);
num = a + b;
if (num == 7 || num == 11) {
fen += 100;
System.out.println("玩家胜利，摇出了" + num + "点,当前分数" + fen);
} else if (num == 2 || num == 3 || num == 12) {
fen -= 100;
System.out.println("电脑胜利，摇出了" + num + "点,当前分数" + fen);
} else {
System.out.println("平局，摇出了" + num + "点,当前分数" + fen);
}
System.out.println("按\"1\"退出，其他任意键继续:");
String zb1 = input.nextLine();
if (zb1.equals("1")) {
System.out.println("游戏结束");
} else {
do {
int a1 = (int) (Math.random() * 6 + 1);
int b1 = (int) (Math.random() * 6 + 1);
num1 = a1 + b1;
if (num == num1) {
fen += 100;
System.out.println("玩家胜利，摇出了" + num1 + "点,当前分数" + fen);
} else if (num1 == 7) {
fen -= 100;
System.out.println("电脑胜利，摇出了" + num1 + "点,当前分数" + fen);
} else {
System.out.println("平局，摇出了" + num1 + "点,当前分数" + fen);
}
System.out.println("按\"1\"退出，其他任意键继续:");
String zb2 = input.nextLine();
if (zb2.equals("1")) {
System.out.println("游戏结束");
} else {
int a2 = (int) (Math.random() * 6 + 1);
int b2 = (int) (Math.random() * 6 + 1);
a1 = a2;
b1 = b2;
}
} while (fen > 0);
System.err.println("你已经输完了！！！无法继续了！！！");
}
}
input.close();
}
}
<2>教师代码
- switch case 结构中，case后面若不跟break，程序会继续执行下去，直到出现break为止
public class HomeWork__ {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int money = 1000;
do {
System.out.println("你还有"+money+"元！");
int bet;
do {
System.out.println("请下注");
bet=input.nextInt();
} while (bet<=0||bet>money);
int face1 = (int) (Math.random() * 6 + 1);
int face2 = (int) (Math.random() * +1);
int firsstPoint = face1 + face2;
System.out.println("玩家摇出" + firsstPoint + "点");
boolean needsGoOn = false;
switch (firsstPoint) {
case 7:
case 11:
System.out.println("玩家胜");
money+=bet;
break;
case 2:
case 3:
case 12:
System.out.println("庄家胜");
money-=bet;
break;
default:
needsGoOn = true;
}
while (needsGoOn) {
face1 = (int) (Math.random() * 6 + 1);
face2 = (int) (Math.random() * 6 + 1);
int currentPoint = face1 + face2;
System.out.println("玩家摇出了" + currentPoint + "点");
if (currentPoint == 7) {
System.out.println("庄家胜");
money-=bet;
needsGoOn = false;
} else if (currentPoint == firsstPoint) {
System.out.println("玩家胜");
money+=bet;
needsGoOn = false;
}
}
} while (money>0);
System.out.println("你已经破产");
input.close();
}
2.输入年月日，输出这是这一年的第几天
- 如果程序中出现了重复的或者相对独立的功能，那么应该将这些功能单独写到一个方法中
package comDay12_02_2;
import java.util.Scanner;
public class Text3 {
// year作为因变量时，此方法需要返回一个判断是否为闰年的结果（是or不是），属于布尔类型；
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("请输入年 月 日");
int year = input.nextInt();
int month = input.nextInt();
int day = input.nextInt();
int sum = 0;
for (int i = 1; i < month; i++) {
sum += daysOfMonth(year, i);
}
System.out.printf("有%d天", (sum + day));
input.close();
}
//isLeapYear方法判断输入的年份是否属于闰年
public static boolean isLeapYear(int year) {
return year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
}
//daysOfMonth用于判断当前月份的天数
public static int daysOfMonth(int year, int month) {
if (month == 4 || month == 6 || month == 9 || month == 11) {
return 30;
} else if (month == 2) {
if (isLeapYear(year)) {
return 29;
} else {
return 28;
}
} else {
return 31;
}
}
}
}
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