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

最新推荐文章于 2019-07-20 18:15:00 发布

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本文介绍了一道算法题目，任务是找到两个集合中元素的二进制表示下不同的位数最少的配对，并选择集合A中最小的元素进行匹配。输入包括多组测试数据，每组数据包含两个整数集合A和B，输出则是对于B集合中的每个元素，从A集合中找到与其二进制位差异最小的元素。

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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

Sample Output

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;
}