找到完美的正方形

Question

我想寫一個循環，調用一個方法來確定輸入的數字是否是一個完美的正方形。它編譯得很好，所以我必須有一個邏輯錯誤，對於我來說，儘管我找不到它。不管我輸入的數字是什麼，它總是返回false，導致我相信問題出在isPerfect（）方法中。然而，我對java還不夠了解，但還不知道該從哪裏出發。這裏是我到目前爲止的代碼：

public class Square 
{ 
    public static void main(String[] args) 
    { 
     int input = 0; //The default value for input 
     Scanner keyboard = new Scanner(System.in); 
     while (input != -1) 
     { 
      System.out.println("Please enter a number, or enter -1 to quit"); 
      input = keyboard.nextInt(); 
      if (isPerfect(input) == true) //Call isPerfect() to determine if is a perfect square 
      { 
       System.out.println(input + " is a perfect square."); 
      } 
      else if(input == -1) //If equals exit code, quit 
      { 
       break; 
      } 
      else //If it is not a perfect square... it's not a perfect square 
      { 
       System.out.println(input + " is not a perfect square."); 
      } 
     } 
    } 




    public static boolean isPerfect(int input) 
    {  
     double num = Math.sqrt(input); //Take the square root of the number passed 

     if (((num * num) == input) && (num%1 == 1)) //If the number passed = it's roots AND has no remainder, it must be a perfect sqaure 
     { 
      return true; //Return true to the call 
     } 
     else 
     { 
      return false; //Return false to the call 
     } 
    } 
} 

Answer 1

兩個潛在的問題。

1. 用雙打算術是不準確的。你可能想

   int num = Math.round(Math.sqrt(input)); 
   

2. 您的餘數不測試沒有做什麼你覺得... (num%1 == 1)只是測試n是否是奇數。而你並不真正需要它..你需要的是if(num*num == input) { ... }

Answer 2

因此具有固定的程序，這裏是代碼的全部：

import java.util.Scanner; 

public class Square 
{ 
    public static void main(String[] args) 
    { 
     int input = 0; //The default value for input 
     Scanner keyboard = new Scanner(System.in); 
     while (input != -1) 
     { 
      System.out.println("Please enter a number, or enter -1 to quit"); 
      input = keyboard.nextInt(); 
      if (isPerfect(input) == true) //Call isPerfect() to determine if is a perfect square 
      { 
       System.out.println(input + " is a perfect square."); 
      } 
      else if(input == -1) //If equals exit code, quit 
      { 
       System.out.println("Breaking!"); 
       break; 
      } 
      else //If it is not a perfect square... it's not a perfect square 
      { 
       System.out.println(input + " is not a perfect square."); 
      } 

     } 
    System.out.println("Main complete!"); 
    } 


    /** 
     The isPerfect() method returns whether or not a number is a perfect square. 
     @param input The input from the keyboard scanner, passed as an argument 
    */ 

    public static boolean isPerfect(int input) 
    {  
     int num = ((int)Math.sqrt(input)); //Take the square root of the number passed, as an integer 

     if (num*num == input) //If the number passed = it's roots AND has no remainder, it must be a perfect sqaure 
     { 
      return true; //Return true to the call 
     } 
     else 
     { 
      return false; //Return false to the call 
     } 
    } 
}