简单枚举类型——植物与颜色
Time Limit: 1000 ms
Memory Limit: 65536 KiB
Problem Description
请定义具有red, orange, yellow, green, blue, violet六种颜色的枚举类型color,根据输入的颜色名称,输出以下六种植物花朵的颜色:
Rose(red), Poppies(orange), Sunflower(yellow), Grass(green), Bluebells(blue), Violets(violet)。如果输入的颜色名称不在枚举类型color中,例如输入purple,请输出I don't know about the color purple.
Input
输入数据有多行,每行有一个字符串代表颜色名称,颜色名称最多30个字符,直到文件结束为止。
Output
输出对应颜色的植物名称,例如:Bluebells are blue. 如果输入的颜色名称不在枚举类型color中,例如purple, 请输出I don't know about the color purple.
Sample Input
blue yellow purple
Sample Output
Bluebells are blue. Sunflower are yellow. I don't know about the color purple.
package hello; import java.util.Scanner; enum Color{ red("Rose", "red"), //建立对应的枚举类型 orange("Poppies", "orange"), yellow("Sunflower","yellow"), green("Grass", "green"), blue("Bluebells", "blue"), violet("Violets", "violet"); String name; String color; Color(String name, String color){ //构造方法 this.name = name; this.color = color; } public String toString() { //重写toSting方法 return name+" are "+color+"."; } } public class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in); int flag = 1; while(cin.hasNext()) { String incolor = cin.next(); /*Color color = Color.valueOf(incolor); //本来想通过转化之后判断一个该颜色有没有在枚举序列中,当时如果没在序列中 for(Color a:color.values()) { //valueOf函数就会抛出异常,程序没法执行,所以用try来实现 if(a.equals(color)) { flag = 0; break; } } if(flag == 0) { System.out.println(color); } else { System.out.println("I don't know about the color "+incolor+"."); }*/ try{ //如果输入的颜色不在对应的颜色序列当中,就会抛出异常 Color color = Color.valueOf(incolor); System.out.println(color); } catch(Exception e){ System.out.println("I don't know about the color "+incolor+"."); } } cin.close(); } }