CodeForce-476D-Dreamoon and Sets

原题：
Dreamoon likes to play with sets, integers and . is defined as the largest positive integer that divides both a and b.

Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, .

Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.

Input
The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).

Output
On the first line print a single integer — the minimal possible m.

On each of the next n lines print four space separated integers representing the i-th set.

Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.

Examples
Input
1 1
Output
5
1 2 3 5
Input
2 2
Output
22
2 4 6 22
14 18 10 16
Note
For the first example it’s easy to see that set {1, 2, 3, 4} isn’t a valid set of rank 1 since .
题意：
给出n和k，要求找出n组4个数字组成的序列，要求每一组的数字两两之间的最大公约数是k，并且每组之间的数字不能够重复。要求ｎ组数字中的最大值尽可能小。
题解：
这道题目一开始我开始类似枚举剪枝的去写，结果ＴＬＥ了。之后看了网上的代码才发现有规律性的。
要求是四个数字且最大公约数为ｋ，我可以先找出四个互质的数字：
１　２　３　５　最大公约数为１
此时如果我想最大公约数为ｋ，只需要将其乘以ｋ倍即可，最大公约数必定是ｋ，
如果要得到ｎ组的话，由于我选取得互质的最大值为５，所以我可以以６为一个循环，每一个６中间都可以找出这样的四个最大公约数是１的数字：
７　８　９　１１
１３　１４　１５　１７
总结规律就可以写出答案。
附上ＡＣ代码：

#include <iostream>
#include <cstring>
using namespace std;
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        cout<<(6*n-1)*k<<endl;
        for(int i=1;i<=n;i++)
        {
            cout<<(6*i-5)*k<<" "<<(6*i-4)*k<<" "<<(6*i-3)*k<<" "<<(6*i-1)*k<<endl;
        }
    }
    return 0;
}

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CodeForce-476D-Dreamoon and Sets

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