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

22 512123455 21000000999914233421013245353

Sample Output

1211199990

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

關鍵點：兩數抑或運算得到兩數二進位表示時不同的位，通過位移和與運算統計兩數不同在二進位表示時不同尾數的個數。

  1: #include<iostream>

  2: #include<vector>

  3: 

  4: using namespace std;

  5: /* 

  6:  * 尋找兩個數以二進位表示時的不同位元的個數

  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:  * 尋找A中最合格數字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;//用來存放容器A中的數字

 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: