ZOJ Problem Set - 3418 Binary Number

Time Limit: 2 Seconds Memory Limit: 65536 KB

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 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

Author: CAO, Peng
Source: The 2010 ACM-ICPC Asia Chengdu Regional Contest

关键点：两数抑或运算得到两数二进制表示时不同的位，通过位移和与运算统计两数不同在二进制表示时不同尾数的个数。

1: #include<iostream>

2: #include<vector>

3:

4: using namespace std;

5: /*

6:  * 寻找两个数以二进制表示时的不同位数的个数

7: */

8: unsigned  countDef(unsigned a,unsigned b){

9:     int result = 0;

10:

11:     for(int c = a^b; c > 0;c = c >> 1){

12:         if(c&1 > 0){

13:             result ++;

14:         }

15:     }

16:     return result;

17: }

18: /*

19:  * 寻找A中最符合条件的数字a

20: */

21: unsigned findMinCountInA(const vector<unsigned>& set, unsigned b){

22:     unsigned min = set[0];

23:     unsigned count = countDef(set[0], b);

24:     unsigned count_ = 0;

25:     for(unsigned i = 1;i < set.size(); i++){

26:         count_ = countDef(set[i], b);

27:         if(count_ < count){

28:             min = set[i];

29:             count = count_;

30:         }else if(count_ == count && set[i] < min){

31:             min = set[i];

32:         }

33:     }

34:     return min;

35: }

36: int main(){

37:     unsigned cases = 0;

38:     cin >> cases;

39:     unsigned aSize = 0, bSize = 0;

40:     while(cases--){

41:         cin>>aSize>>bSize;

42:         vector<unsigned> aSet;//用来存放容器A中的数字

43:         unsigned elemIna = 0;

44:         while(aSize--){

45:             cin>>elemIna;

46:             aSet.push_back(elemIna);

47:         }

48:         unsigned elemInb = 0;

49:         while(bSize--){

50:             cin>>elemInb;

51:             cout<<findMinCountInA(aSet, elemInb)<<endl;

52:         }

53:     }

54: }

55: