CodeForces 1395C-Boboniu and Bit Operations（位运算-暴力）

题目链接：https://codeforces.com/problemset/problem/1395/C
博客园食用链接： https://www.cnblogs.com/lonely-wind-/p/13494811.html

Boboniu likes bit operations. He wants to play a game with you.

Boboniu gives you two sequences of non-negative integers $a_1,a_2,\dots ,a_n$ and $b_1,b_2,\dots ,b_m$.

For each $i(1\le i\le n)$, you’re asked to choose $a_j(1\le j\le m)$ and let $c_i=a_i\&b_j$, where & denotes the bitwise AND operation. Note that you can pick the same j for different i’s.

Find the minimum possible $c_1|c_2|\dots |c_n$, where | denotes the bitwise OR operation.

Input
The first line contains two integers n and m (1≤n,m≤200).

The next line contains n integers $a_1,a_2,\dots ,a_n(0\le a_i<29)$.

The next line contains m integers $b_1,b_2,\dots ,b_m(0\le b_i<29)$.

Output
Print one integer: the minimum possible $c_1|c_2|\dots |c_n$.

input
4 2
2 6 4 0
2 4
output
2

input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output
0

input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
output
147

Note
For the first example, we have $c1=a1\&b2=0,c2=a2\&b1=2,c3=a3\&b1=0,c4=a4\&b1=0$.Thus $c1|c2|c3|c4=2$, and this is the minimal answer we can get.

题目大意：给你数列a,b,必须对每一个$a_i$从$b_j$中任选一个数得$c_i=a_i\&b_j$，最后使得$c_1|c_2|...|c_n$最小

emmmm，刚开始的思路就是暴力，也写对了，就是出了点小BUG，后面结果改成DP了。。。然后赛后就被fst了。。。写了个假DP。。。

它的数据范围很小，所以我们可以直接考虑枚举答案，不难看出答案的区间为$[0,1<<2^9)$，然后我们就来检查答案了，两层for过去，然后取一下位运算与得$x$，接下来就是对二进制下的$x$的每一位是否符合答案，如果$x$的某一位存在1，而答案不存在，那么就说明了这个$a_i\&b_j$是不合法的。那么只要有一个$a_i\&b_j$是合法的，那么$c_i$就是合法的。于是我们很容易写出代码了。

以下是AC代码：

#include 
using namespace std;

int a[300],b[300];

int main(int argc, char const *argv[])
{
	int n,m;
	scanf ("%d%d",&n,&m);
	for (int i=1; i<=n; i++)
		scanf ("%d",&a[i]);
	for (int i=1; i<=m; i++)
		scanf ("%d",&b[i]);
	int ans=0;
	for (int i=0; i<(1<<9); i++){
		int mk=0;
		for (int j=1; j<=n; j++){
			int flag=0;
			for (int k=1; k<=m; k++){
				int x=a[j]&b[k];
				for (int g=0; g<=9; g++){
					if ((x&(1<<g)) && !(i&(1<<g))) {flag++; break;}
				}
			}
			if (flag==m) {mk=1; break;}
		}
		if (!mk) {ans=i; break;}
	}
	printf("%d\n",ans);
	return 0;
}