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

题目链接：https://codeforces.com/problemset/problem/1395/C
CSDN食用链接：https://blog.csdn.net/qq_43906000/article/details/107973093

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

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

For each \(i (1≤i≤n)\), you're asked to choose \(a_j (1≤j≤m)\) 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 \(c_1|c_2|…|c_n\), 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 \(a_1,a_2,…,a_n (0≤a_i<29)\).

The next line contains m integers \(b_1,b_2,…,b_m (0≤b_i<29)\).

Output
Print one integer: the minimum possible \(c_1|c_2|…|c_n\).

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 <bits/stdc++.h>
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;
}