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前言:
码蹄杯作为编程学习中经典的逻辑训练题型,是提升算法思维与代码实践能力的“磨刀石”。对于初入编程领域的学习者而言,从基础题入手拆解问题逻辑是快速入门的关键。本次分享将围绕码蹄杯基础题型展开,涵盖循环逻辑、条件判断、数组操作等核心知识点,通过典型例题解析与思路拆解,帮助大家掌握从问题建模到代码实现的完整流程。无论你是零基础的编程小白,还是希望巩固基础的学习者,都能在本次分享中收获解题技巧,为挑战更复杂的编程任务夯实基础。
一:实型数运算
题目:请编写一个简单程序,用户输入2个实型数据存储在变量中,并输出他们的乘积与商。(本题不考虑除数为0的情况)
题目详解
package demo5_2;import java.util.Scanner;/*** Created with IntelliJ IDEA.* Description:* User:Lenovo* Date:2025-05-11* Time:11:16*/
public class Main {public static void main(String[] args) {Scanner sc=new Scanner(System.in);float a=sc.nextFloat();float b=sc.nextFloat();
// System.out.println(String.format("%.6f",a)+"*"+String.format("%.6f",b)+"="+String.format("%.6f",a*b));
// System.out.println(String.format("%.6f",a)+"/"+String.format("%.6f",b)+"="+String.format("%.6f",a/b));System.out.printf("%.6f*%.6f=%.6f\n",a,b,a*b);System.out.printf("%.6f/%.6f=%.6f",a,b,a/b);}
}
二:平均分
题目详解
package demo5_2;import java.util.Scanner;/*** Created with IntelliJ IDEA.* Description:* User:Lenovo* Date:2025-05-13* Time:22:27*/
public class Main1 {public static void main(String[] args) {Scanner sc=new Scanner(System.in);double a=sc.nextDouble();double b=sc.nextDouble();double c=sc.nextDouble();double total=a+b+c;double avg=(a+b+c)/3;System.out.println(String.format("%.6f",total));System.out.println(String.format("%.6f",avg));}
}
三:圆球等的相关计算
请编写一个简单程序,输入半径和高,输出圆周长,圆面积,球面积,球体积,圆柱体积。(PI = 3.1415926)
代码详解:
package demo5_2;
import java.util.Scanner;/*** Created with IntelliJ IDEA.* Description:* User:Lenovo* Date:2025-05-15* Time:23:51*/
public class Main2 {public static final double PI = 3.14159;public static void main(String[] args) {// 输出圆周长,圆面积,球面积,球体积,圆柱体积。Scanner sc=new Scanner(System.in);double r=sc.nextDouble();double high=sc.nextDouble();double circleZ=PI*2.0*r;double circleM=PI*r*r;double qiuM=PI*r*r*4.0;double qiuT=PI*r*r*r*4*1.0/3.0;double circleZT=PI*r*r*high;System.out.println(String.format("%.2f",circleZ));System.out.println(String.format("%.2f",circleM));System.out.println(String.format("%.2f",qiuM));System.out.println(String.format("%.2f",qiuT));System.out.println(String.format("%.2f",circleZT));}
}
四:公式计算
代码详解:
package demo5_2;import java.util.Scanner;/*** Created with IntelliJ IDEA.* Description:* User:Lenovo* Date:2025-05-17* Time:8:58*/
public class Main3 {public static void main(String[] args) {//计算公式(1/2)∗(a∗x+(a+x)/(4∗a))Scanner sc=new Scanner(System.in);int x=sc.nextInt();int a=sc.nextInt();double s=(1.0/2)*(a*x+(a+x)*1.0/(4.0*a));System.out.println(String.format("%.2f",s));System.out.printf("%.2f\n",s);}
}
五:输入和输出字符型数据
请编写一个简单程序,用户输入2个的字符型数据存储在变量中,并分别以字符形式和整数形式输出在屏幕上。
代码详解:
package demo5_2;import java.util.Scanner;/*** Created with IntelliJ IDEA.* Description:* User:Lenovo* Date:2025-05-17* Time:9:25*/
public class Main4 {public static void main(String[] args) {Scanner sc=new Scanner(System.in);String s=sc.next();String []b=s.split(",");int A=s.charAt(0);int B=s.charAt(2);System.out.println("The ASCII code of "+ b[0] + " is " +A);System.out.println("The ASCII code of "+ b[1] + " is " +B);}
}
六:字符和整数
输出X、65的字符、十进制数据形式。
import java.util.Scanner;
import java.util.*;class Main {public static void main(String[] args) {Scanner input = new Scanner(System.in);// code hereSystem.out.println("X"+" "+(int)'X');System.out.println("A"+" "+(int)'A');input.close();}
}
七:各种类型长
请编写一个简单程序,输出int、float、double和char的大小。
详解代码:
/*** Created with IntelliJ IDEA.* Description:* User:Lenovo* Date:2025-05-17* Time:9:51*/
public class Main {public static void main(String[] args) {int a=Integer.SIZE;int b=Float.SIZE;int c=Double.SIZE;int d=Character.SIZE;System.out.println("Size of int: "+a/8+" "+"bytes");System.out.println("Size of float: "+b/8+" "+"bytes");System.out.println("Size of double: "+c/8+" "+"bytes");System.out.println("Size of char: "+d/16+" "+"byte");}
}
八:关键字long
请编写一个简单程序,输出int、long int、long long int、double和long double的大小
题目详解代码:
/*** Created with IntelliJ IDEA.* Description:* User:Lenovo* Date:2025-05-19* Time:22:17*/
public class Main {//请编写一个简单程序,输出int、long int、long long int、double和long double的大小public static void main(String[] args) {int a=Integer.SIZE;int b=Long.SIZE;int c=Long.SIZE;int d=Double.SIZE;int e=Double.SIZE;System.out.println("Size of int ="+" " +a/8+" bytes");System.out.println("Size of long int ="+" "+b/8+" bytes");System.out.println("Size of long long int ="+" "+c/8+" bytes ");System.out.println("Size of double ="+" "+d/8 +" bytes");System.out.println("Size of long double ="+" "+e*2/8+" bytes");}
}
九:输入分隔符
输入“a=22,b=b,c=14,d=d”给变量a、b、c、d,然后再输出他们。
代码详解:
package demo5_2;import java.util.Scanner;/*** Created with IntelliJ IDEA.* Description:* User:Lenovo* Date:2025-05-19* Time:22:34*/
public class Main7 {public static void main(String[] args) {
// 输入“a=22,b=b,c=14,d=d”给变量a、b、c、d,然后再输出他们。Scanner sc=new Scanner(System.in);String[] arr=sc.nextLine().split(",");String avalue="",bvalue="",cvalue="",dvalue="";for(String x:arr){String[] parts=x.split("=");if(parts.length==2){String key=parts[0].trim();String value=parts[1].trim();switch(key){case"a":avalue=value;break;case"b":bvalue=value;break;case"c":cvalue=value;break;case"d":dvalue=value;break;}}}System.out.printf("%s %s %s %s%n",avalue,bvalue,cvalue,dvalue);sc.close();}}
十:宽度与对齐
代码详解:
package demo5_2;/*** Created with IntelliJ IDEA.* Description:* User:Lenovo* Date:2025-05-20* Time:22:45*/
public class Main8 {//输出455、-123、987654,宽度为5,分别左对齐和右对齐public static void main(String[] args) {int a=455,b=-123,c=987654;System.out.printf("%-5d %5d\n",a,a);System.out.printf("%-5d %5d\n",b,b);System.out.printf("%-5d %5d\n",c,c);}
}
这篇的Java码蹄杯算法题就分享到这里了,我们下篇再见!🫰🫰🫰