2010成都站c题||uvalive5006 二进制

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&category=381&page=show_problem&problem=3007

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 Asuch that f(a, b) is minimized. If there are more than one such integers in set A, choose the smallest one.

Input

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.

Output

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

Sample Input

2
2 5
1
2
1
2
3
4
5
5 2
1000000
9999
1423
3421
0
13245
353

Sample Output

1
2
1
1
1
9999
0

题目大意：对于B中的任意一个数在A中找出一个数，使两个数的二进制位不同的位数最少，输出A中的数。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
int a[105],b[105],m,n;
int judge(int a,int b)
{
    if(a<b)
        swap(a,b);
    int sum=0;
    while(b)
    {
        int x=a&1;
        int y=b&1;
        if(x!=y)
            sum++;
        a>>=1;
        b>>=1;
    }
    //printf("(%d %d %d)\n",a,b,sum);
    while(a)
    {
        if(a&1)
            sum++;
        a>>=1;
    }
   // printf("%d\n",sum);
    return sum;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&b[i]);
        sort(a,a+n);
        for(int i=0;i<m;i++)
        {
            int minn=0x3f3f3f3f;
            int cnt=0;
            for(int j=0;j<n;j++)
            {
                int ans=judge(b[i],a[j]);
                if(minn>ans)
                {
                    minn=ans;
                    cnt=a[j];
                }
            }
            printf("%d\n",cnt);
        }
    }
    return 0;
}
/*
int main()
{
    int x,y;
    while(~scanf("%d%d",&x,&y))
    {
        int ans=judge(x,y);
    }
    return 0;
}*/