Codeforces Round #681------Kids Seating

题目：

Today the kindergarten has a new group of n kids who need to be seated at the dinner table. The chairs at the table are numbered from 1 to 4n. Two kids can’t sit on the same chair. It is known that two kids who sit on chairs with numbers a and b (a≠b) will indulge if:

gcd(a,b)=1 or,
a divides b or b divides a.
gcd(a,b) — the maximum number x such that a is divisible by x and b is divisible by x.

For example, if n=3 and the kids sit on chairs with numbers 2, 3, 4, then they will indulge since 4 is divided by 2 and gcd(2,3)=1. If kids sit on chairs with numbers 4, 6, 10, then they will not indulge.

The teacher really doesn’t want the mess at the table, so she wants to seat the kids so there are no 2 of the kid that can indulge. More formally, she wants no pair of chairs a and b that the kids occupy to fulfill the condition above.

Since the teacher is very busy with the entertainment of the kids, she asked you to solve this problem.

Input
The first line contains one integer t (1≤t≤100) — the number of test cases. Then t test cases follow.

Each test case consists of one line containing an integer n (1≤n≤100) — the number of kids.

Output
Output t lines, which contain n distinct integers from 1 to 4n — the numbers of chairs that the kids should occupy in the corresponding test case. If there are multiple answers, print any of them. You can print n numbers in any order.

题意

有n个小朋友，有4*n个座位，坐座位不能出现以下情况
假定有两个座位号a， b
1、gcd(a, b) = 1
2、 a % b == 0 || b % a == 0
输入T组，输出T个答案，每一行输出n个数，表示每个小朋友的座位号

题解

从2 *n开始入座，每个小朋友之间间隔一个座位，刚刚满足条件

#include<iostream>
using namespace std;
int main()
{
	int t; cin >> t;
	while (t--) {
		int n; cin >> n;
		for (int i = 2 * n; i < 4 * n; i += 2)
			cout << i << " ";
		cout << endl;
	}
	return 0;
}