Prime numbers in sets (inclusion and exclusion principle and dfs)

• Note: Of course, this question can also be enumerated in binary, but 0 \small 0 is not $0$ this binary.
• Topic
• Question : See the question.
• Solution : The
  principle of tolerance and exclusion :
  
  Summary : The principle of tolerance and exclusion is odd plus even minus.
  Once you have the formula, just $dfs$ can be solved.
  The meaning of the variables:
  $\{\begin{array}{l}pos:Passescalendarnumber\; ofsetsofthe\; firstPOSbits\\ opt:Receivingrepellentoriginalrationaleof\; thesymbolnumber(+1or-1)\\ mul:The\; firstPOSbitofthe\; frontof\; theelectionof\; thenumberof\; themultiplicationproduct\; of\end{array}$
  Traverse each number from left to right, for this number, you can take it or not.
  Not taken : Except for the traversed position plus one, the rest remains unchanged.
  Take : add one to the traversed position, which can be obtained by the nature of the odd-plus-even-subtraction of the tolerance and exclusion principle. If one more number is taken, it must change sign, so $opt$ plus minus sign, the current number is multiplied by$u\left[pos\right]$ .
  You will find that the principle of tolerance and exclusion takes at least one number, and there is no case of not taking the number, so when traversing to$When\; pos$ , the contribution of the answer must be counted as$In\; the\; case\; of\; u\left[pos\right]$ , because$(mul:Onthe\; p-OSbitof\; thefrontof\; theelectionofthe\; numberofmultiplyaccumulate)$ , the contribution of the answer is:$m$对$mul\ast u\left[pos\right]$ is rounded down.
  Sorting pruning operation: find the problem, when the product of the number is greater than$When\; m$ , it has no contribution to the answer, so when traversing to$When\; pos$ has crossed the boundary, the following numbers can be traversed.
• AC code

//优化
#pragma GCC optimize(2)
//C
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
//C++
#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<istream>
#include<iomanip>
#include<climits>
#include<float.h>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
//宏定义
#define N 1010
#define DoIdo main
//#define scanf scanf_s
#define it set<ll>::iterator
#define TT template<class T>
#define cint const int 
//定义+命名空间
typedef long long ll;
typedef unsigned long long ull;
const int mod = 10007;
const ll INF = 1e18;
const int maxn = 1e6 + 10;
using namespace std;
//全局变量
ll n, m;
ll u[25];
ll ans = 0;
//函数区
ll max(ll a, ll b) {
    
     return a > b ? a : b; }
ll min(ll a, ll b) {
    
     return a < b ? a : b; }
ll loop(ll n, ll p) {
    
     return (n % p + p) % p; }
void dfs(ll pos, ll opt, ll mul) {
    
    
	//如果遍历的数超过输入的数，则返回
	if (pos == n + 1) return;
	//这就是u[pos] * mul <= m的变形
	//这样可以预防long long溢出
	if (u[pos] <= m / mul) {
    
    
		//加一下对答案的贡献
		ans += opt * m / (mul * u[pos]);
		//不取的情况
		dfs(pos + 1, opt, mul);
		//取的情况
		dfs(pos + 1, -opt, mul * u[pos]);
	}
	else return;
}
//主函数
int DoIdo() {
    
    

	ios::sync_with_stdio(false);
	cin.tie(NULL), cout.tie(NULL);

	cin >> n >> m;

	for (int i = 1; i <= n; i++) {
    
    
		cin >> u[i];
	}

	sort(u + 1, u + n + 1);
	dfs(1, 1, 1);

	cout << ans << endl;
	return 0;
}
//分割线---------------------------------QWQ
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