Codeforces Round #272 (Div. 2) D. Dreamoon and Sets 数学 - 服务器托管|北京服务器租用|机房托管租用|IDC托管租用|机房机柜带宽租用-价格及费用咨询

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
inputCopy
1 1
outputCopy
5
1 2 3 5
inputCopy
2 2
outputCopy
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 .

给定N个集合，每个集合里面的数的 gcd ==k ，且每个数只可以属于一个集合，问最大的数最小应该是多少？

找一下规律：1，2，3，5；7，8，9，11；13，14，15，17…….，以6 为一个周期，这些都是满足互质的，那么gcd==k ，也就是*k即可，

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair
#define pll pair



ll quickpow(ll a, ll b) {
    ll ans = 1;
    while (b > 0) {
        if (b % 2)ans = ans * a;
        b = b / 2;
        a = a * a;
    }
    return ans;
}

int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a%b);
}


int main()
{
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    int a = 1, b = 2, c = 3, d = 5;
    cout

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