Codeforces Round #272 (Div. 2) D.Dreamoon and Sets 找规律

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

5
1 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<<(*n-)*k<<endl;

    while(n--)

    {

        cout<<(*n+)*k<<" "<<(*n+)*k<<" "<<(*n+)*k<<" "<<(*n+)*k<<endl;

    }

}

正解代码

///

#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=,f=;

    char ch=getchar();

    while(ch<''||ch>'')

    {

        if(ch=='-')f=-;

        ch=getchar();

    }

    while(ch>=''&&ch<='')

    {

        x=x*+ch-'';

        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=;///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=;

    while(b)

    {

        if(b&){ret+=a;ret%=c;}

        a<<=;

        if(a>=c)a%=c;

        b>>=;

    }

    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==)return x%mod;

    x%=mod;

    long long tmp=x;

    long long ret=;

    while(n)

    {

        if(n&) ret=mult_mod(ret,tmp,mod);

        tmp=mult_mod(tmp,tmp,mod);

        n>>=;

    }

    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=;i<=t;i++)

    {

        ret=mult_mod(ret,ret,n);

        if(ret==&&last!=&&last!=n-) return true;//合数

        last=ret;

    }

    if(ret!=) 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<)return false;

    if(n==)return true;

    if((n&)==) return false;//偶数

    long long x=n-;

    long long t=;

    while((x&)==){x>>=;t++;}

    for(int i=;i<S;i++)

    {

        long long a=rand()%(n-)+;///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 >  )

    {

        Rem = M % N;

        M = N;

        N = Rem;

}

    return M;

}

int main(){

   n=read(),m=read();

    int ans[maxn];

     int k=;

     int tmp=;

     G.clear();

     for(int i=;i<=n;i++){

            while(G.size()!=){

                    bool flag=;

                 for(int j=;j<G.size();j++){

                if(gcd(G[j],m*tmp)!=m)flag=;

                 }

                if(flag) G.push_back(tmp*m);

                tmp++;

            }

              for(int j=;j<G.size();j++)

                   a[++k]=G[j];

                   G.clear();

          if(!Miller_Rabin(tmp))tmp++;

     }

     cout<<a[k]<<endl;

     for(int i=;i<=k;i+=){

         cout<<a[i]<<" "<<a[i+]<<" "<<a[i+]<<" "<<a[i+]<<endl;

     }

  return ;

}

暴力代码

巴特西

Codeforces Round #272 (Div. 2) D.Dreamoon and Sets 找规律

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