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不知還能不能再精簡了???

package?com;
public?abstract?class?Shape?{
static?double?Pi=3.14;
//抽象類方法定義
public?abstract?double?changzhouchang(double?a,double?b);
public?abstract?double?circlezhouchang(double?c);
public?abstract?double?changmianji(double?a,double?b);
public?abstract?double?circlemianji(double?c);
}
package?com;
public?class?Caculate?extends?Shape?{
//方法實現
public?double?changzhouchang(double?a,double?b)?{
return?(a+b)*2;
}
public?double?circlezhouchang(double?c)?{
return?2*Pi*c;
}
public?double?changmianji(double?a,double?b)?{
return?a*b;
}
public?double?circlemianji(double?c)?{
return?Pi*c*c;
}
}
package?com;
import?java.util.*;
public?class?Test?{
public?static?void?main(String[]?args)?{
//?TODO?Auto-generated?method?stub
double?a,b,c;
System.out.println("程序開始:");
Scanner?scanner=new?Scanner(System.in);
System.out.println("請輸入長方形的長,款,圓形的半徑:");
a=scanner.nextDouble();
b=scanner.nextDouble();
c=scanner.nextDouble();
Caculate?caculate?=new?Caculate();
System.out.println("長方形周長為:"+caculate.changzhouchang(a,?b)+"\n"+"長方形面積為:"+caculate.changmianji(a,?b)+"\n"+"圓形周長為:"+caculate.circlezhouchang(c)+"\n"+"圓形面積為:"+caculate.circlemianji(c));
scanner.close();
}
}


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2 回答

其實你的實現跟老師想讓實現的思路一點都不一樣。雖然你這樣也可以實現,但是代碼的健壯性很糟糕。

0 回復 有任何疑惑可以回復我~

差不多就這樣吧? 我覺得pi用final修飾更合適一些

0 回復 有任何疑惑可以回復我~

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不知還能不能再精簡了???

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