Hdu 3816 To Be number one number theory is the main idea

To Be NUMBER ONE
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 383 Accepted Submission (s): 190
Special Judge

Problem Description
One is an interesting integer. This is also an interesting problem. You are assigned with a simple task.
Find N (3 <= N <= 18) different positive integers Ai (1 <= I <= N), and

 

Any possible answer will be accepted.
 

Input
No input file.
 

Output
Your program's output shoshould contain 16 lines:
The first line contain 3 integers which are the answer for N = 3, separated by a single blank.
The following 15 lines are the answer for N = 4 to 18, as the same format.
The sample output is the first two lines of a possible output.
 

Sample Output
2 3 6
2 4 6 12
 
Question
The reciprocal sum of numbers 3 to 18 is equal to 1.
And each number must be less than or equal to the number + 1 square
For example, three numbers are used.
The three numbers must meet the condition that these three numbers are less than or equal to (3 + 1) * (3 + 1) and the reciprocal sum is 1.
 
Ideas:
A reciprocal can be divided into the reciprocal sum of two numbers.
1/n = 1/(a * B) = 1/a * (a + B) + 1/B * (a + B)
 
Therefore, when the upper layer is known, the number of the upper layer can be directly divided into two, and so on.
 
I did not write a program
[Cpp]
# Include <stdio. h>
Int main ()
{

Printf ("2 3 6 \ n ");
Printf ("2 4 6 12 \ n ");
Printf ("2 5 6 12 20 \ n ");
Printf ("2 5 7 12 20 42 \ n ");
Printf ("3 5 6 7 12 20 42 \ n ");
Printf ("3 5 6 8 12 20 42 56 \ n ");
Printf ("3 5 6 8 16 20 42 48 56 \ n ");
Printf ("3 5 6 12 16 20 24 42 48 56 \ n ");
Printf ("3 5 6 16 20 21 24 28 42 48 56 \ n ");
Printf ("3 5 6 16 21 24 28 36 42 45 48 56 \ n ");
Printf ("3 5 6 16 24 28 30 36 42 45 48 56 70 \ n ");
Printf ("3 5 6 16 28 30 33 36 42 45 48 56 70 88 \ n ");
Printf ("3 5 6 16 28 33 36 39 42 45 48 56 70 88 130 \ n ");
Printf ("3 5 6 16 33 36 39 42 45 48 56 70 77 88 130 \ n ");
Printf ("3 5 6 20 33 36 39 42 45 48 56 70 77 80 88 130 \ n ");
Printf ("3 5 6 24 33 36 39 42 45 48 56 70 77 80 88 120 130 \ n ");

Return 0;
}