#include<bits/stdc++.h>
using namespace std;
int n;
int getBits1(int n)//求取一个数的二进制形式中1的个数. 
{
	int res=0;
	while(n)
	{
		if(n&1) res++;
		n>>=1;
	}
	return res;
}

int getBits2(int n)
{
	int res=0;
	while(n)
	{
		res++;
		n=n&(n-1);//一次消去1个1哦 
	}
	return res;
}

int main()
{
	while(cin>>n)
	{
		cout<<getBits1(n)<<" "<<getBits2(n)<<endl;
	}
	return 0;
}   
// 1= 1    1
// 2= 10   1
// 3= 11    2
// 4=100    1

题目：

For 2 non-negative integers x and y, f(x, y) is defined as the number of different bits in the binary format of x and y. For example, f(2, 3)=1,f(0, 3)=2, f(5, 10)=4. Now given 2 sets of non-negative integers A and B, for each integer b in B, you should find an integer a in A such that f(a, b) is minimized. If there are more than one such integer in set A, choose the smallest one.

输入：

The first line of the input is an integer T (0 < T ≤ 100), indicating the number of test cases. The first line of each test case contains 2 positive integers m and n (0 < m, n ≤ 100), indicating the numbers of integers of the 2 sets A and B, respectively. Then follow (m + n) lines, each of which contains a non-negative integers no larger than 1000000. The first m lines are the integers in set A and the other n lines are the integers in set B.

输出：

For each test case you should output n lines, each of which contains the result for each query in a single line.

样例输入：

2
2 5
1
2
1
2
3
4
5
5 2
1000000
9999
1423
3421
0
13245
353
样例输出：

1
2
1
1
1
9999
0

#include<bits/stdc++.h>
using namespace std;
int f(int n)
{
	int res=0;
	while(n)
	{
		res++;
		n=n&(n-1);
	}
	return res;
}

int main()
{
	int a[105];
	int T,n,m,i,j,k,minn,b,t;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		for(i=0;i<n;i++) scanf("%d",&a[i]);
		for(i=0;i<m;i++)
		{
			scanf("%d",&b);
			minn=f(a[0]^b);
			k=0;
			for(j=1;j<n;j++)
			{
				t=f(a[j]^b);
				if(minn>t || minn==t&&a[k]>a[j])
				{
					minn=t;
					k=j;
				}
			}
			printf("%d\n",a[k]);
		}
	}
	return 0;
}