poj 3711 Binary Number -- 据说是暴力（利用数位计算异或^水过）

Binary Number

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2189    Accepted Submission(s): 1288

Problem Description

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 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 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

 

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int x1[105],x2[105];int fun(int a,int b){    int num=0;    int temp = a^b;    while(temp){        int c = temp%2;        num+=c;        temp/=2;    }    return num;}int main(){    int t,m,n,i,j;    int flag;    scanf("%d",&t);    while(t--){        scanf("%d %d",&m,&n);        for(i=0;i<m;i++)            scanf("%d",&x1[i]);        for(i=0;i<n;i++)            scanf("%d",&x2[i]);        sort(x1,x1+m);             //不知道为啥排序，反正不排序就会wa        for(i=0;i<n;i++){            int min=0x3f3f3f3f;            for(j=0;j<m;j++){                int num = fun(x2[i],x1[j]);                if(num<min){                    min=num;                    flag=x1[j];                }            }            printf("%d\n",flag);        }    }    return 0;}