[题解]gdfzoj 726 | Easy Number

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二分答案套路题，其中check用容斥瞎搞即可

代码：

#include<bits/stdc++.h>
namespace my_std
{
    using namespace std;
    #define int long long
    #define R register
    #define rep(i,a,b) for (R int i=(a);i<=(b);i++)
    #define drep(i,a,b) for (R int i=(a);i>=(b);i--)
    #define go(u) for (R int i=head[(u)];i;i=e[i].nxt)
    #define pf printf
    #define writeln(x) write(x),putchar('\n')
    #define writesp(x) write(x),putchar(' ')
    #define mem(x,v) memset(x,v,sizeof(x))
    typedef long long ll;
    const int INF=1e12;
    inline int read()
    {
        int sum=0,f=1;
        char c=0;
        while (!isdigit(c))
        {
            if (c=='-') f=-1;
            c=getchar();
        }
        while (isdigit(c))
        {
            sum=(sum<<1)+(sum<<3)+(c^48);
            c=getchar();
        }
        return sum*f;
    }
    void write(int k)
    {
        if (k<0) putchar('-'),k=-k;
        if (k>=10) write(k/10);
        putchar(k%10+'0');
    }
    bool bet(int bot,int x,int top)
    {
    	return x>=bot&&x<=top;
    }
}
using namespace my_std;
namespace ka_chang
{
	#pragma GCC optimize(2)
	#pragma GCC optimize(3)
	#pragma GCC optimize("Ofast")
	#pragma GCC optimize("inline")
	#pragma GCC optimize("-fgcse")
	#pragma GCC optimize("-fgcse-lm")
	#pragma GCC optimize("-fipa-sra")
	#pragma GCC optimize("-ftree-pre")
	#pragma GCC optimize("-ftree-vrp")
	#pragma GCC optimize("-fpeephole2")
	#pragma GCC optimize("-ffast-math")
	#pragma GCC optimize("-fsched-spec")
	#pragma GCC optimize("unroll-loops")
	#pragma GCC optimize("-falign-jumps")
	#pragma GCC optimize("-falign-loops")
	#pragma GCC optimize("-falign-labels")
	#pragma GCC optimize("-fdevirtualize")
	#pragma GCC optimize("-fcaller-saves")
	#pragma GCC optimize("-fcrossjumping")
	#pragma GCC optimize("-fthread-jumps")
	#pragma GCC optimize("-funroll-loops")
	#pragma GCC optimize("-fwhole-program")
	#pragma GCC optimize("-freorder-blocks")
	#pragma GCC optimize("-fschedule-insns")
	#pragma GCC optimize("inline-functions")
	#pragma GCC optimize("-ftree-tail-merge")
	#pragma GCC optimize("-fschedule-insns2")
	#pragma GCC optimize("-fstrict-aliasing")
	#pragma GCC optimize("-fstrict-overflow")
	#pragma GCC optimize("-falign-functions")
	#pragma GCC optimize("-fcse-skip-blocks")
	#pragma GCC optimize("-fcse-follow-jumps")
	#pragma GCC optimize("-fsched-interblock")
	#pragma GCC optimize("-fpartial-inlining")
	#pragma GCC optimize("no-stack-protector")
	#pragma GCC optimize("-freorder-functions")
	#pragma GCC optimize("-findirect-inlining")
	#pragma GCC optimize("-frerun-cse-after-loop")
	#pragma GCC optimize("inline-small-functions")
	#pragma GCC optimize("-finline-small-functions")
	#pragma GCC optimize("-ftree-switch-conversion")
	#pragma GCC optimize("-foptimize-sibling-calls")
	#pragma GCC optimize("-fexpensive-optimizations")
	#pragma GCC optimize("-funsafe-loop-optimizations")
	#pragma GCC optimize("inline-functions-called-once")
	#pragma GCC optimize("-fdelete-null-pointer-checks")
}
using namespace ka_chang;
const int N=100010;
int n,k,a[N];
int gcd(int x,int y)
{
	if (y==0) return x;
	return gcd(y,x%y);
}
#define lcm(a,b) (a*b/gcd(a,b))
inline int get_sum(int mid)
{
	int res=0,cnt,tmp;
	rep(s,0,(1<<n)-1)
	{
		cnt=0,tmp=1;
		rep(i,1,n) if (s&(1<<(i-1)))
		{
			tmp=lcm(tmp,a[i]);
			cnt++;
		}
		if (cnt)
		{
			if (cnt&1) res+=mid/tmp;
			else res-=mid/tmp;
		}
	}
	return res;
}
signed main()
{
	n=read(),k=read();
	rep(i,1,n) a[i]=read();
	int l=1,r=INF;
	while (l<r)
	{
		int mid=(l+r)>>1;
		if (get_sum(mid)>=k) r=mid;
		else l=mid+1;
	}
	writeln(r);
}

（我是全队跑得最慢的，我也不知道为什么）

update : 我还是把容斥的过程写一下吧：

容斥遵循最基本的原则：奇加偶减

然后在 get_sum 函数里， s 枚举的是一个集合，代表 a 数组有没有被选中

再把集合中的数的最小公倍数求出即可

cnt 是集合 s 中的元素个数，即 popcount ,可以用 lowbit 预处理

最后奇加偶减即可