「CF477B」Dreamoon and Sets
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 msuch 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.
The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).
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.
1 |
1 1 |
1 2 |
5 1 2 3 5 |
1 |
2 2 |
1 2 3 |
22 2 4 6 22 14 18 10 16 |
For the first example it’s easy to see that set {1, 2, 3, 4} isn’t a valid set of rank 1 since .
题解
本质就是求n个集合,元素互不相同,每个集合4个互质的数,要求最大值最小
公约数k即所有元素都乘上k即可
每个集合显然只能有一个偶数,而且发现每三个奇数都是互质的。。。
然后偶数也要跟这些奇数互质。。
那么只要
2 1 3 5
8 7 9 11
14 13 15 17
这样构造就可以了
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 |
#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cmath> #include<cstring> #define inf 1000000000 #define ll long long #define mod 1000000007 using namespace std; int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,k,tot; int ans[10005][5]; int main() { n=read();k=read(); for(int i=1;i<=n;i++) { int k=i*6-4; ans[i][1]=k;ans[i][2]=k-1; ans[i][3]=k+1;ans[i][4]=k+3; } tot=(n*6-4)+3; printf("%d\n",tot*k); for(int i=1;i<=n;i++) { for(int j=1;j<=4;j++) printf("%d ",ans[i][j]*k); printf("\n"); } return 0; } |