1335. White Thesis 會者不難 難者不會

# 1335 
n =  gets.chomp!.to_i

print "#{n*n+n} #{n*n+2*n} #{n*n}"

原文鏈接:  http://acm.timus.ru/problem.aspx?space=1&num=1335

背景  輸出一組triplet (A,B,C)  滿足A B的平方和是以的倍數. 

A B C 限制,  給定輸出的整數N,   A B C均在區間[N**2, (N+1)**2] 中. 

Input

contains one integer N (2 ≤ N ≤ 30000).

Output

Output three different integers A, B and C, that (A2 + B2) is a multiple of C and N2 ≤ A, B, C ≤ (N+1)2. If two or more such triplets exist, output any one. If there are no such triplets, then output "No solution".

Samples

input	output
2	8 6 4
1000	1000000 1000756 1000976

Problem Author: Den Raskovalov
Problem Source: The 10th Collegiate Programming Contest of the High School Pupils of the Sverdlovsk Region (October 16, 2004)

Tags: none

其實這個題沒有想象是那麼複雜.   -->  首先看這個公式

 (n+1)^2 =  n^2 + 2*n + 1  = (n+2)n + 1 

所以在給定區間中 n*n, n*(n+1) , n*(n+2) 是均存在的.   如果取後面兩個大數的平方和, 他們一定是n^2倍數.