自己的第一次容斥原理的实践（还有等差数列求和）终于会做数学题了

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#define ll long long
#define mod 376544743
using namespace std;

ll sn(ll a1,ll an,ll n)//等差数列求和
{
	return (a1+an)%mod*n%mod*188272372%mod; //当时忘了搞2对于mod的乘法逆元了，然后就wrong了
}
int count(int n)
{
	int ans;
	for (ans=0;n;n=n&(n-1)) ans++;
	return ans;
}
ll n,m,p[50],ans;
int main()
{
	scanf("%lld%lld",&n,&m);
	for (int i=1;i<=n;i++) scanf("%lld",&p[i]);
	for (int i=1;i<(1<<n);i++)//枚举子集
	{
		int j=i;
		ll k=count(j),g=0,t=1;
		for (int l=0;l<n;l++) if (j&(1<<l)) t*=p[l+1]; 
		if (t>m) continue;
		g=sn(t,(m/t)*t,m/t);
		if (k&1) ans=(ans+g)%mod; //容斥原理
		else ans=(ans+mod-g)%mod;	 
	}
	printf("%lld",ans);
	return 0;
}