Coprime AtCoder Beginner Contest 215

Coprime

AtCoder Beginner Contest 215

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

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.

Sample Input 1 Copy

Copy

3 12
6 1 5

Sample Output 1 Copy

Copy

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;

}