AtCoder Beginner Contest 215 D - Coprime 2

D - Coprime 2 

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 400400 points

Problem Statement

Given a sequence of NN positive integers A=(A_1,A_2,\dots,A_N)A=(A1​,A2​,…,AN​), find every integer kk between 11 and MM (inclusive) that satisfies the following condition:

• \gcd(A_i,k)=1gcd(Ai​,k)=1 for every integer ii such that 1 \le i \le N1≤i≤N.

Constraints

• All values in input are integers.
• 1 \le N,M \le 10^51≤N,M≤105
• 1 \le A_i \le 10^51≤Ai​≤105

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Input

Input is given from Standard Input in the following format:

NN MM
A_1A1​ A_2A2​ \dots… A_NAN​

Output

In the first line, print xx: the number of integers satisfying the requirement.
In the following xx lines, print the integers satisfying the requirement, in ascending order, each in its own line.

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Sample Input 1 Copy

Copy

3 12
6 1 5

Sample Output 1 Copy

Copy

3
1
7
11

For example, 77 has the properties \gcd(6,7)=1,\gcd(1,7)=1,\gcd(5,7)=1gcd(6,7)=1,gcd(1,7)=1,gcd(5,7)=1, so it is included in the set of integers satisfying the requirement.
On the other hand, 99 has the property \gcd(6,9)=3gcd(6,9)=3, so it is not included in that set.
We have three integers between 11 and 1212 that satisfy the condition: 11, 77, and 1111. Be sure to print them in ascending order.

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;

bool vis[N];
bool vis1[N];
int a[N];
int prime[N];
int ans[N];
int cnt;
int len;
int n, m;
set<int> se;
void get_prime() //线性筛法
{

    for (int i = 2; i <= N; i++)
    {
        if (!vis[i])
            prime[cnt++] = i;
        for (int j = 0; prime[j] <= N / i; j++)
        {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0)
                break;
        }
    }
}
void find() // 因式分解，求所有的质因数
{
    for (int i = 0; i < n; i++)
    {
        int k = a[i];
        for (int j = 0; prime[j] <= k / prime[j]; j++)
        {
            if (k % prime[j] == 0)
            {
                se.insert(prime[j]);
                while (k % prime[j] == 0)
                    k = k / prime[j];
            }
        }
        if (k > 1)
            se.insert(k);
    }
}

void solve()
{
    scanf("%d%d", &n, &m);
    int i, j, k, l;
    for (i = 0; i < n; i++)
    {
        scanf("%d", &a[i]);
    }
    get_prime();
    find();
    for (auto &it : se) //利用埃式筛法的思想，去除所有质因数的倍数
    {
        if (it == 1)
            continue;
        for (j = it; j <= m; j += it)
            vis1[j] = 1;
    }
    for (i = 1; i <= m; i++)
    {
        if (!vis1[i])
            ans[len++] = i;
    }
    printf("%d\n", len);
    for (i = 0; i < len; i++)
    {
        printf("%d\n", ans[i]);
    }
}
signed main()
{
    solve();
    return 0;
}