运用位运算会比较简单 Binary Numb…

Binary Number

Time Limit : 4000/2000ms(Java/Other)   MemoryLimit : 65536/65536K (Java/Other)

Total Submission(s) :20   AcceptedSubmission(s) : 6

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Problem Description

For 2 non-negative integers x and y, f(x, y) is defined as thenumber of different bits in the binary format of x and y. Forexample, f(2, 3)=1,f(0, 3)=2, f(5, 10)=4. Now given 2 sets ofnon-negative integers A and B, for each integer b in B, you shouldfind an integer a in A such that f(a, b) is minimized. If there aremore than one such integer 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 eachtest case contains 2 positive integers m and n (0 <m, n ≤ 100), indicating the numbers of integers of the 2 sets A andB, respectively. Then follow (m + n) lines, each of which containsa non-negative integers no larger than 1000000. The first m linesare the integers in set A and the other n lines are the integers inset B.

Output

For each test case you should output n lines, each of whichcontains 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

#include<stdio.h>

#include<string.h>

#define MAX 101

int a[MAX],b[MAX];

int data(int num)

{

     int sum=0;

  while(num)

  {

           sum+=num%2;

         num/=2;

     }

   return sum;

}

void solve(int *a,int *b,int m,int n)

{

   int i,j,num,min,temp;

       for(i=1;i<=n;i++)

    {

           min=9999999;

                for(j=1;j<=m;j++)

            {

                   temp=data(a[j]^b[i]);

                               if(min>temp)

                         {

                                   min=temp;

                                   num=a[j];

                           }

                           else if(min==temp)

                          {

                                   num=num>=a[j]?a[j]:num;

                              }

           }

           printf("%d\n",num);

 }

}

int main()

{

        int cas, m, n; 

    int i;

      scanf("%d", &cas);

  while (cas--)

       {

           scanf("%d%d", &m, &n);      

            for (i=1;i<=m;i++)

           {

                   scanf("%d", a+i);

           }

                for (i=1;i<=n;i++)

           {

                   scanf("%d",b+i);

            }

           solve(a,b,m,n);

     }

   return 0;

}