2010成都站c題||uvalive5006 二進位制

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

題目大意：對於B中的任意一個數在A中找出一個數，使兩個數的二進位制位不同的位數最少，輸出A中的數。

#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;
}*/