D. 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, sjfrom 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.
Sample test(s)
input
1 1
output
51 2 3 5
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：求出n个集合均为4个元素，且每个集合内任意两两，元素的最大公约数为k，集合不允许有交集，当n*4个元素最大值最小时，输出所有n个集合，
题解：我是暴力水过去的，正解是：
两两元素之间是互质的，然后为了满足元素最大值最小，
在除掉k后，任意集合内肯定是由3个奇数，1个偶数组成，
因为若少一个奇数，就会有至少一对数不互质，
若多一个奇数，元素最大值就会变大。
所以，从1开始构建所有元素集合即可。
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
int n,k;
cin>>n>>k;
cout<<(6*n-1)*k<<endl;
while(n--)
{
cout<<(6*n+1)*k<<" "<<(6*n+2)*k<<" "<<(6*n+3)*k<<" "<<(6*n+5)*k<<endl;
}
}
正解代码
///1085422276
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define inf 100000000
inline ll read()
{
ll 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;
}
//****************************************
///****************************************************************
/// Miller_Rabin 算法进行素数测试
///速度快，而且可以判断 <2^63的数
//****************************************************************
const int S=20;///随机算法判定次数，S越大，判错概率越小
///计算 (a*b)%c. a,b都是long long的数，直接相乘可能溢出的
/// a,b,c <2^63
long long mult_mod(long long a,long long b,long long c)
{
a%=c;
b%=c;
long long ret=0;
while(b)
{
if(b&1){ret+=a;ret%=c;}
a<<=1;
if(a>=c)a%=c;
b>>=1;
}
return ret;
}
///计算 x^n %c
long long pow_mod(long long x,long long n,long long mod)//x^n%c
{
if(n==1)return x%mod;
x%=mod;
long long tmp=x;
long long ret=1;
while(n)
{
if(n&1) ret=mult_mod(ret,tmp,mod);
tmp=mult_mod(tmp,tmp,mod);
n>>=1;
}
return ret;
}
///以a为基,n-1=x*2^t a^(n-1)=1(mod n) 验证n是不是合数
///一定是合数返回true,不一定返回false
bool check(long long a,long long n,long long x,long long t)
{
long long ret=pow_mod(a,x,n);
long long last=ret;
for(int i=1;i<=t;i++)
{
ret=mult_mod(ret,ret,n);
if(ret==1&&last!=1&&last!=n-1) return true;//合数
last=ret;
}
if(ret!=1) return true;
return false;
}
/// Miller_Rabin()算法素数判定
///是素数返回true.(可能是伪素数，但概率极小)
///合数返回false;
bool Miller_Rabin(long long n)
{
if(n<2)return false;
if(n==2)return true;
if((n&1)==0) return false;//偶数
long long x=n-1;
long long t=0;
while((x&1)==0){x>>=1;t++;}
for(int i=0;i<S;i++)
{
long long a=rand()%(n-1)+1;///rand()需要stdlib.h头文件
if(check(a,n,x,t))
return false;//合数
}
return true;
}
#define maxn 100000+5
int a[maxn],n,m;
vector<int >G;
int gcd(int M,int N )
{
int Rem;
while( N > 0 )
{
Rem = M % N;
M = N;
N = Rem;
}
return M;
}
int main(){
n=read(),m=read();
int ans[maxn];
int k=0;
int tmp=1;
G.clear();
for(int i=1;i<=n;i++){
while(G.size()!=4){
bool flag=1;
for(int j=0;j<G.size();j++){
if(gcd(G[j],m*tmp)!=m)flag=0;
}
if(flag) G.push_back(tmp*m);
tmp++;
}
for(int j=0;j<G.size();j++)
a[++k]=G[j];
G.clear();
if(!Miller_Rabin(tmp))tmp++;
}
cout<<a[k]<<endl;
for(int i=1;i<=k;i+=4){
cout<<a[i]<<" "<<a[i+1]<<" "<<a[i+2]<<" "<<a[i+3]<<endl;
}
return 0;
}
暴力代码