AtCoder Beginner Contest 215 D - Coprime 2 (质因子分解 gcd)

​​原题链接​​

题意：

给定一个长度为$n$的序列$a$和整数$m$，求有多少$k$满足$1<=k<=m$并且$gcd(k,a_i)=1,1<=i<=n$
$n,m,a_i<=1e5$

思路：

$gcd(k,a_i)=1$说明$k$和每一个$a_i$都没有相同的质因子。
对每个$a_i$进行质因子分解，对分解出来的因子，类似于质数筛的方法筛去$[1,m]$里因子的倍数。最后剩下的数就是合法的$k$.
时间复杂度$O(n\sqrt n)$

代码：

// Problem: D - Coprime 2
// Contest: AtCoder - AtCoder Beginner Contest 215
// URL: https://atcoder.jp/contests/abc215/tasks/abc215_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
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;}

inline void write(ll x){if (x < 0) x = ~x + 1, putchar('-');if (x > 9) write(x / 10);putchar(x % 10 + '0');}

#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)

ll ksm(ll a, ll b,ll mod){ll res = 1;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}


const int maxn=1e6+7;

int n,m,a[maxn];
vector<int>res,tmp;
map<int,bool>mp;

int main(){
  n=read,m=read;
  rep(i,1,n){
    ll x=read;
    tmp.clear();    
    for(ll j=2;j*j<=x;j++)
      if(x%j==0){
        tmp.push_back(j);
        while(x%j==0) x/=j;
      }
    if(x>1) tmp.push_back(x);
    for(auto t:tmp)
      if(!mp[t]){
        for(ll j=t;j<=m;j+=t) mp[j]=1;
      }
  }
  rep(i,1,m) 
    if(!mp[i]) res.push_back(i);
  printf("%d\n",res.size());
  for(auto t:res) printf("%d\n",t);
  
  return 0;
}