Codeforces Round 931 (Div. 2) |
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Finished |
Given an integer $$$n$$$, you construct an array $$$a$$$ of $$$n$$$ integers, where $$$a_i = i$$$ for all integers $$$i$$$ in the range $$$[1, n]$$$. An operation on this array is defined as follows:
After all the operations $$$a_i \le 10^{18}$$$ should hold for all $$$1 \le i \le n$$$.
We can show that an answer always exists.
The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^2$$$) — the number of test cases. The description of the test cases follows.
The first and only line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 3 \cdot 10^{4}$$$) — the length of the array.
It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^{4}$$$.
The first line should contain an integer $$$k$$$ ($$$0 \leq k \leq \lfloor \frac{n}{6} \rfloor + 5$$$) — where $$$k$$$ is the number of operations.
The next $$$k$$$ lines should contain the description of each operation i.e. $$$3$$$ integers $$$i$$$, $$$j$$$ and $$$k$$$, where $$$1 \leq i, j, k \leq n$$$ and all must be distinct.
3347
1 1 2 3 1 1 3 4 3 3 5 7 5 6 7 2 3 4
In the third test case, $$$a = [1, 2, 3, 4, 5, 6, 7]$$$.
First operation:
$$$i = 3$$$, $$$j = 5$$$, $$$k = 7$$$
$$$x = 3$$$, $$$y = 5$$$, $$$z = 7$$$.
$$$a = [1, 2, \operatorname{lcm}(y,z), 4, \operatorname{lcm}(x,z), 6, \operatorname{lcm}(x,y)]$$$ = $$$[1, 2, \color{red}{35}, 4, \color{red}{21}, 6, \color{red}{15}]$$$.
Second operation:
$$$i = 5$$$, $$$j = 6$$$, $$$k = 7$$$
$$$x = 21$$$, $$$y = 6$$$, $$$z = 15$$$.
$$$a = [1, 2, 35, 4, \operatorname{lcm}(y,z), \operatorname{lcm}(x,z), \operatorname{lcm}(x,y)]$$$ = $$$[1, 2, 35, 4, \color{red}{30}, \color{red}{105}, \color{red}{42}]$$$.
Third operation:
$$$i = 2$$$, $$$j = 3$$$, $$$k = 4$$$
$$$x = 2$$$, $$$y = 35$$$, $$$z = 4$$$.
$$$a = [1, \operatorname{lcm}(y,z), \operatorname{lcm}(x,z), \operatorname{lcm}(x,y), 30, 105, 42]$$$ = $$$[1, \color{red}{140}, \color{red}{4}, \color{red}{70}, 30, 105, 42]$$$.
Subsequences whose GCD equal to $$$i$$$ is as follows:
$$$\gcd(a_1, a_2) = \gcd(1, 140) = 1$$$
$$$\gcd(a_3, a_4) = \gcd(4, 70) = 2$$$
$$$\gcd(a_5, a_6, a_7) = \gcd(30, 105, 42) = 3$$$
$$$\gcd(a_2, a_3) = \gcd(140, 4) = 4$$$
$$$\gcd(a_2, a_4, a_5, a_6) = \gcd(140, 70, 30, 105) = 5$$$
$$$\gcd(a_5, a_7) = \gcd(30, 42) = 6$$$
$$$\gcd(a_2, a_4, a_6, a_7) = \gcd(140, 70, 105, 42) = 7$$$
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