如何使用逻辑循环打印出20个毕达哥拉斯数的非全等三角形

Question

How can I use logic loop to print out 20 Pythagorean numbers of non-congruent triangles. Without repeating numbers ie if I have 4,3,5 I can't have 3,4,5.

I was using "for" loops but I don't know how to remove the repeating answers.

for (k = 0; k < 50; k++)
    {
        for ( i = 0; i < 50; i++)
        {
            for ( j = 0; j < 50; j++)
            {
                if ( (k+1)*(k+1) + (i+1)*(i+1) == (j+1)*(j+1) )
                {
                    System.out.println( "\n\n\t\tThe numbers are : " + (k+1) + ", "
                                                                     + (i+1) + ", "
                                                                     + (j+1) );
                }
            }
        }
    }

Answer 1

Two (related) options come to mind: You could "keep" each of your triples in a sorted order, so when you find {4,3,5} you turn it into {3,4,5} before saving it for subsequent comparison with any others for uniqueness. Or, you could create a "Triple" class where you define a method like boolean equals(final Triple rval) that does the comparison of each element from lowest to highest. Of course, there are probably other ways this could be done as well.

UPDATE: Given the code you just added, if you only want to print them out, then you probably don't need to keep the triples you've found so far around as I was assuming above. The following modification to your code might work:

for (k = 0; k < 50; k++)
{
    for ( i = k; i < 50; i++)
    {
        for ( j = i; j < 50; j++)
        {
            if ( (k+1)*(k+1) + (i+1)*(i+1) == (j+1)*(j+1) )
            {
                System.out.println( "\n\n\t\tThe numbers are : " + (k+1) + ", "
                                                                 + (i+1) + ", "
                                                                 + (j+1) );
            }
        }
    }
}

Note that I changed the starting point of the inner loops to ensure the following holds of all of the triples you will find: k <= i <= j . I believe that this constraint will also ensure uniqueness.