k-LCM (easy version) (Codeforces 1497C1)
Tags: Math, constructive algorithms

It is the easy version of the problem. The only difference is that in this version $$$k = 3$$$.You are given a positive integer $$$n$$$. Find $$$k$$$ positive integers $$$a_1, a_2, \ldots, a_k$$$, such that:  $$$a_1 + a_2 + \ldots + a_k = n$$$  $$$LCM(a_1, a_2, \ldots, a_k) \le \frac{n}{2}$$$ Here $$$LCM$$$ is the least common multiple of numbers $$$a_1, a_2, \ldots, a_k$$$.We can show that for given constraints the answer always exists.

Input: The first line contains a single integer $$$t$$$ $$$(1 \le t \le 10^4)$$$  — the number of test cases.The only line of each test case contains two integers $$$n$$$, $$$k$$$ ($$$3 \le n \le 10^9$$$, $$$k = 3$$$).

Output: For each test case print $$$k$$$ positive integers $$$a_1, a_2, \ldots, a_k$$$, for which all conditions are satisfied.


Sample Inputs:
3
3 3
8 3
14 3

Sample Outputs:
1 1 1
4 2 2
2 6 6

Solution (Python):
for s in [*open(0)][1:]:
    (n, _) = map(int, s.split())
    if n & 1:
        print(n // 2, n // 2, 1)
    elif n // 2 & 1:
        print(n // 2 - 1, n // 2 - 1, 2)
    else:
        print(n // 2, n // 4, n // 4)

Solution (Python):
for s in [*open(0)][1:]:
    (n, _) = map(int, s.split())
    if n & 1:
        print(n // 2, n // 2, 1)
    elif n // 2 & 1:
        print(n // 2 - 1, n // 2 - 1, 2)
    else:
        print(n // 2, n // 4, n // 4)

Solution (Python):
for s in [*open(0)][1:]:
    (n, _) = map(int, s.split())
    if n & 1:
        print(n // 2, n // 2, 1)
    elif n // 2 & 1:
        print(n // 2 - 1, n // 2 - 1, 2)
    else:
        print(n // 2, n // 4, n // 4)