我如何解决这个问题？我认为这与return语句有关。

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英文:

How do I resolve this issue ?? I think it has something to do with the return statement

问题

  1. import java.util.Scanner;
  3. public class theatre_square {
  4.     public static void main(String args[]) {
  5.         Scanner in = new Scanner(System.in);
  6.         long m = in.nextLong(), n = in.nextLong(), s = in.nextLong();
  7.         theatre_square ob = new theatre_square();
  8.         long x = ob.cal(m, s), y = ob.cal(n, s);
  9.         System.out.println(x * y);
  10.     }
  12.     long cal(long a, long b) {
  13.         if (a <= 0)
  14.             return 0;
  15.         else
  16.             return (1 + cal(a - b, b));
  17.     }
  18. }
  1. > 输入：1000000000 1000000000 1
  2. > 
  3. > 错误信息：
  4. > 
  5. > Exception in thread "main" java.lang.StackOverflowError
  6. > 	at theatre_square.cal(theatre_square.java:17)
  7. > 	at theatre_square.cal(theatre_square.java:17)
  8. > 	at theatre_square.cal(theatre_square.java:17)
  9. > 	... 以此类推 ...
  10. > 
  11. > 我认为问题与递归函数内的返回语句有关。
英文: 
  1. import java.util.Scanner;
  2. public class theatre_square
  3. {
  4.     public static void main(String args[])
  5.     {
  6.         Scanner in = new Scanner(System.in);
  7.         long m = in.nextLong(), n = in.nextLong(), s = in.nextLong();
  8.         theatre_square ob = new theatre_square();
  9.         long x = ob.cal(m,s), y = ob.cal(n,s);
  10.         System.out.println(x*y);
  11.     }
  12.     long cal(long a, long b)
  13.     {
  14.         if(a&lt;=0)
  15.             return(0);
  16.         else
  17.             return (1+cal(a-b,b));
  18.     }
  19. }

> Input: 1000000000 1000000000 1
>
> Error:
>
> Exception in thread "main" java.lang.StackOverflowError at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17) at
> theatre_square.cal(theatre_square.java:17)....... and so on.
>
>
>
> I think it has to do something with the return statement inside in the
> recursion function.

答案1

得分: 0

  1. 我认为你可以在不使用递归和for循环的情况下解决这个问题。
  3. import java.util.Scanner;
  4. public class theatre_square
  5. {
  6.     public static void main(String args[])
  7.     {
  8.         Scanner in = new Scanner(System.in);
  9.         long m = in.nextLong(), n = in.nextLong(), s = in.nextLong();
  10.         theatre_square ob = new theatre_square();
  11.         long x = ob.cal(m,s), y = ob.cal(n,s);
  12.         System.out.println(x*y);
  13.     }
  14.     
  15.     
  16.     long cal(long m, long a)    {
  17.         long count = m / a;
  18.         long rem = m % a;
  19.         if(rem  &gt; 0){
  20.             count++;
  21.         }
  22.         return count;
  23.     }
  24. }
英文:

i think you can solve the problem without recursion and for loop.

  1. import java.util.Scanner;
  2. public class theatre_square
  3. {
  4.     public static void main(String args[])
  5.     {
  6.         Scanner in = new Scanner(System.in);
  7.         long m = in.nextLong(), n = in.nextLong(), s = in.nextLong();
  8.         theatre_square ob = new theatre_square();
  9.         long x = ob.cal(m,s), y = ob.cal(n,s);
  10.         System.out.println(x*y);
  11.     }
  12.     
  13.     
  14.     long cal(long m, long a)    {
  15.         long count = m / a;
  16.         long rem = m % a;
  17.         if(rem  &gt; 0){
  18.             count++;
  19.         }
  20.         return count;
  21.     }
  22. }

huangapple
  • 本文由 发表于 2020年9月20日 19:53:10
  • 转载请务必保留本文链接：https://go.coder-hub.com/63978693.html
  • java
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