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.
InputThe single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).
OutputOn 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.
Sample test(s) Input 1 1 Output 51 2 3 5 Input 2 2 Output 222 4 6 2214 18 10 16 NoteFor 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： 给你任意n,k,要你求出n组gcd(si,sj)=k的四个元素的组合.........
其实对于gcd(a,b)=k,我们只需要求出gcd(a,b)=1;然后进行gcd(a,b)*k=k;
不难发现这些数是固定不变的而且还有规律可循，即1,2,3,5 7 8 9 11 13 14 15 17 每一个段直接隔着2,段内前3个连续，后一个隔着2.....
代码：
1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 const int maxn=10005; 5 __int64 ans[maxn][4]; 6 void work() 7 { 8 __int64 k=1; 9 for(int i=0;i<10000;i++) 10 { 11 ans[i][0]=k++; 12 ans[i][1]=k++; 13 ans[i][2]=k++; 14 ans[i][3]=++k; 15 k+=2; 16 } 17 } 18 int main() 19 { 20 int n,k; 21 work(); 22 while(scanf("%d%d",&n,&k)!=EOF) 23 { 24 printf("%I64d\n",ans[n-1][3]*k); 25 for(int i=0;i<n;i++) 26 printf("%I64d %I64d %I64d %I64d\n",ans[i][0]*k,ans[i][1]*k,ans[i][2]*k,ans[i][3]*k); 27 } 28 return 0; 29 } View Code
转载于:https://www.cnblogs.com/gongxijun/p/4025638.html
