2022年09月16日Java上机

课上练习

public class Square {
    public static int getArea(int a) {
        return a;
    }
    public static int getGirth(int a) {
        return a*4;
    }
    public static void main(String[] args) {
        int a = 4;
        System.out.println(getArea(a));
        System.out.println(getGirth(a));
    }
}

示例1-3

JBT13

public class JBT13 {
	public static void main(String args[]) {
		JBT14 obj1=new JBT14();
		JBT15 obj2=new JBT15();
		int x=1,y=2;
		obj1.add(x,y);
		obj2.minus(x,y);
	}
}

JBT14

class JBT14 {
    void add(int x, int y) {
        System.out.println(x + "+" + y + "=" + (x + y));
    }
}

JBT15

public class JBT15 {
    void minus(int x, int y) {
        System.out.println(x + "-" + y + "=" + (x - y));
    }
}

示例2-2

public class JBT2202 {
    public static void main(String args[]) {
        //输出byte类型的最大值和最小值
        System.out.println("byte型最大值= " + Byte.MAX_VALUE);
        System.out.println("byte型最小值= " + Byte.MIN_VALUE);
        //输出short 类型的最大值和最小值
        System.out.println("short型最大值= " + Short.MAX_VALUE);
        System.out.println("short型最小值= " + Short.MIN_VALUE);
        //输出int 类型的最大值和最小值
        System.out.println("int型最大值= " + Integer.MAX_VALUE);
        System.out.println("int型最小值= " + Integer.MIN_VALUE);
        //输出long类型的最大值和最小值
        System.out.println("long型最大值= " + Long.MAX_VALUE);
        System.out.println("long型最小值= " + Long.MIN_VALUE);
        //输出float类型的最大值和最小值
        System.out.println("float型最大值= " + Float.MAX_VALUE);
        System.out.println("float型最小值= " + Float.MIN_VALUE);
        //输出double类型的最大值和最小值
        System.out.println("double型最大值= " + Double.MAX_VALUE);
        System.out.println("double型最小值= " + Double.MIN_VALUE);
    }
}

示例2-3

public class JBT2203 {
    public static void main(String args[]) {
        byte a = 120;
        short b = 255;
        int c = 2200;
        long d = 2147483649L;
        float f1, f2;
        double g = 0.1234567812345678;
        b = a;  //低精度赋值给高精度
        c = (int) d;    //高精度强制转换赋值给低精度
        f1 = a; //低精度赋值给高精度
        f2 = (float)g;   //高精度强制转换赋值给低精度
        System.out.println("a=" + a);
        System.out.println("b=" + b);
        System.out.println("c=" + c);
        System.out.println("d=" + d);
        System.out.println("f1=" + f1);
        System.out.println("f2=" + f2);
        System.out.println("g=" + g);
    }
}

示例2-6

import java.util.Scanner;
public class JBT2402 {
    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        System.out.print("输入岗位等级: ");
        int grade = stdin.nextInt();
        System.out.println("");
        //String grade stdin.nextLine();
        int wage = 0;
        if (grade == 1) {
            wage = 280;
        } else if (grade == 2) {
            wage = 1900;
        } else if (grade == 3) {
            wage = 1630;
        } else if (grade == 4) {
            wage = 1420;
        } else if (grade == 5) {
            wage = 1180;
        } else if (grade == 6) {
            wage = 1040;
        } else if (grade == 7) {
            wage = 930;
        } else if (grade == 8) {
            wage = 780;
        } else if (grade == 9) {
            wage = 730;
        } else if (grade == 10) {
            wage = 680;
        } else if (grade == 11) {
            wage = 620;
        } else if (grade == 12) {
            wage = 590;
        }
        System.out.printf("岗位等级%2d的岗位工资:%4d\n", grade, wage);
    }
}

示例2-9

public class JBT2404 {
    public static void main(String[] args) {
        int n = 10, m = 5;
        System.out.println("for循环求解下的结果是=" + sum1(n, m));
        System.out.println("while循环求解下的结果是=" + sum2(n, m));
        System.out.println("do while循环求解下的结果是=" + sum3(n, m));
    }

    //for
    public static long sum1(int n, int m) {
        int i = 0;
        long sum = 0;
        for (i = 1; i <= n; i++) {
            if (i % m == 0) {
                sum = sum + i;
            }
        }
        return sum;
    }

    //while
    public static long sum2(int n, int m) {
        int i = 0;
        long sum = 0;
        while (i <= n) {
            if (i % m == 0) {
                sum = sum + i;
            }
            i = i + 1;
        }
        return sum;
    }

    //do while
    public static long sum3(int n, int m) {
        int i = 0;
        long sum = 0;
        do {
            if (i % m == 0) {
                sum = sum + i;
            }
            i = i + 1;
        } while (i <= n);
        return sum;
    }
}