考试中的艰巨任务：如何将 "IF" 语句“ "for nested" 循环”内的两个数组的总和复制到新数组

Question

我需要帮助解决以下问题：

创建程序计算并打印以下方程的所有可能对（您可以在解决方案中使用 Math.sqrt(double) 但不能使用 Math.square ）

n>=a>1

n>=b>1

其中 (a^2 + b^2 = c^2)

例如，如果 n = 4，那么我正在打印：

System.out.println(a = 3 , b =4 or a = 4 and b =3 ; 3^2 + 4^2=25; 我们找到了一对) 因为： 4>=4>1

4>=3>1

4^2 +3^2 =25

16+9=25 ，所以这是一对 a 和 b (a= 3 and b= 4 and a= 4 和 b= 3 是一样的，所以它只给了我一对 )

另一个例子是： n =144 ，然后 144>=144>1 和 144>=17>1

144^2 + 17^2 = 21025 ，所以我们找到了第一对 /.

n = 144 ，然后 144 >=143>1 和 144 >=24>1

143^2 + 24^2 = 21025 ，我们找到了第二对 /.

（n=144 给出了所有前面的数字对，而不仅仅是我提到的那些）

所以我从 Method main 开始并完成了一些事情，但最近无法复制两个数组的总和： (cel1 and cel2_copy , their sum is equals to ) 到另一个数组 (tempofsum1) 正确。

   public static void main(String[] args)

 {

  Scanner inp = new Scanner (System.in);
  System.out.println(" Input the value of n  ");
  int n = inp.nextInt();
  int counter  = 0 ;
  //counts the possible number of digits that gonna insert into my original array

 int counter2 = 0; 
//counts the number of times that our sum is calculated

    int i , j,newroot; // newroot of 25 = 5 , root of 25 = 5.0
    int sumofsquare  = 0; //(a^2 + b^2)
      int  defaultsum  =0; // (a+b)
    int [] tempofsum2,tempofsum1 ;


    for (i = 2; i<=n;i++) 
    {
//counts number of digits that i'm gonna insert into my original array


     counter ++;            
}
     System.out.println("We have to sort out " + counter + " numbers inside our original array \n" ); 


      int [] cel1 = new int [counter] ;// Our Original Array:
//all numbers in our array are equal to zero at the moment

    for ( j = 0 ,  i =2  ; j<cel1.length  ;j++ ,i++ )
      {

//loop that insersts the numbers that I calculated      
// i = 2 is first number in our array ,becasue numbers in the range of  n > 1 

    cel1[j] = i ;

//number at the index of 0 is now 2 , this row is sorting our array numbers in the ascending order

    System.out.println("Value of the original array at the index of # " +j+ " is "  +cel1[j]);
}
    System.out.println("   Our  N is  " + n );


// I'm creating my second array,

      int [ ] cel2_copy = new int [cel1.length];

      for (i = 0;i< cel1.length;i++) 
     { 

//copying values from Original Array to this new .  

        cel2_copy [i] = cel1 [i] ;      
}     

//Our For "Loop", that calculates all possible summary of cel2_copy  and cel1 arrays  , pythagoras triplets, their roots 

    for (i = 0 ; i<cel1.length ;i++) 
      { 
      for ( j = 0 ; j < cel1.length; j++  )

//cel1[i] = a 和 cel2_copy[j] = b（在我们的毕达哥拉斯方程中）

sumofsquare = cel1[i]*cel1[i] + cel2_copy[j]*cel2_copy[j];

//summing our a^2 +  b^2

    defaultsum = cel1[i] + cel2_copy[j];

//same as previous row but a + b , not a^2 + b^2

    double root = Math.sqrt (sumofsquare);

//calcutates the root of a^2 + b^2

    newroot = (int ) root;

/*
cel1[i] = a , cel2_copy[j]=b
Our If statemenet that gonna check if ( n >= a >1  && n>= b > 1  )and if (our 
integer root of a^2 +b^2  equals to its double value)if it is ,that means 
that we have sum that can be rooted properly.
*/

      if ((((n > cel1[i] || n == cel1[i]) && ( ( cel1[i] >1 ) 
            &&  (n > cel2_copy[j] || n == cel2_copy[j]) 
            && ( (cel2_copy[j] >1 )))
            &&( newroot * newroot == root * root ))))
               {

这是我的 Originofsum 数组，它获取 "cel1" 数组和 "cel2_copy" 数组

的摘要

        originofsum [i]  = sumofsquare ;

     counter2 ++;

    //counter2 calculates the number of times that we get summary` 

    System.out.println("The sum is calulated " +counter2  + " times "  );
    System.out.println(sumofsquare  + " is sumofsquare " + " at the index of i = "  +i  +" and the index of J = " +j ); 
    System.out.println("Our sum of a = " +cel1[i]+ " and b = " + cel2_copy[j] + " is a+b =" +defaultsum);
    System.out.println("The double root of sumofsquare =" +root );
    System.out.println("The int root of sumofsquare = " +newroot  );
    System.out.println("\n");
     }// end of If                 
   }//end for J
 }// end for I

所以我在这里尝试将我的总和值从数组 "OriginofSum" 复制到 "tempofsum1"，但这不起作用，

在控制台中（如果您在程序开始时按 4（4 是 N ）打印以下数字：0 25

我们得到两个数字，因为我只有两个在 2 到 4 范围内的和（n = 4 和 n >=a>1 和 n>=b>1），给我正确的平方根是 a= 3 ,b= 4 and a= 4 and b= 3 (16+9=25) and (9+16=25)

所以我的数组有一次漏掉了数字 25。

如果你按 14 ，你得到总和 8 次： 0 25 25 169 100 0 100 225

所以现在我的数组错过了一次 169 和一次 225。（在它们的位置是 0 值）。

System.out.println("Sum is calculted " +counter2 + " times " );
tempofsum2 = new int [counter2];
tempofsum1 = new int [ counter2];
   for ( i = 0 ; i <tempofsum1.length;i++ ) 
       {


       tempofsum1 [i] = originofsum[i];

       System.out.println(originofsum[i]);

    } // end of For i Loop
 }// end of Main    
}// end of Class

我很高兴得到你的帮助，但请记住这是一道测试题，所以我应该坚持一个基础。提前谢谢你。）

Answer 1

如果您的目标只是找到与另一个整数形成勾股三元组的数字，这对我来说似乎非常简单：

private static void printPairs(int n) {
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      long sum = (i * i) + (j * j);
      double root = Math.sqrt(sum);
      if (root == ((int) root)) {
        System.out.printf("Found pair (%d, %d): %d^2 + %d^2 = %d^2\n",
            i, j, i, j, (int) root);
      }
    }
  }
}

示例输出：

Found pair (3, 4): 3^2 + 4^2 = 5^2

Found pair (4, 3): 4^2 + 3^2 = 5^2

演示：https://ideone.com/dj1w0M

这不需要任何数组或任何非常复杂的东西。我错过了什么吗？

考试中的艰巨任务：如何将 "IF" 语句“ "for nested" 循环”内的两个数组的总和复制到新数组

Monstrous task from the Exam : How to copy the sum of two arrays that are inside "IF" statement" that is inside "for nested" loop " to a new array

java

arrays

for-loop

nested-loops