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            The 2010 ACM-ICPC Asia Chengdu Regional Contest - C Binary Number 水題,枚舉~

            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
            
            不解釋了,N2
            
             1# include <iostream>
             2# include <algorithm>
             3using namespace std;
             4int a[101],n,m;
             5int main()
             6{
             7    int t;
             8    cin>>t;
             9    while(t--)
            10    {
            11        cin>>n>>m;
            12        for(int i=0;i<n;i++)
            13           cin>>a[i];
            14        sort(a,a+n);
            15        for(int i=0;i<m;i++)
            16        {
            17           int t,res=0xfffffff,num;
            18           cin>>t;
            19           for(int j=0;j<n;j++)
            20           {
            21              int tmp1=a[j],tmp2=t,diff=0;
            22              while(tmp1||tmp2)
            23              {
            24                 diff+=((tmp1&1)!=(tmp2&1));
            25                 tmp1>>=1;
            26                 tmp2>>=1;
            27              }

            28              if(diff<res) res=diff,num=a[j];
            29           }

            30           cout<<num<<endl;
            31        }

            32    }

            33  //  system("pause");
            34    return 0;
            35}

            36

            posted on 2010-11-16 00:35 yzhw 閱讀(343) 評論(0)  編輯 收藏 引用 所屬分類: simple problem~

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