hdu 3711 Binary Number

最新推荐文章于 2020-07-11 22:57:57 发布

原创最新推荐文章于 2020-07-11 22:57:57 发布 · 225 阅读

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本文介绍了一种解决两个非负整数集合A和B之间的最小化位差匹配问题的算法。对于B集合中的每个元素b，算法需找到A集合中使两者二进制位不同位数最少的元素a，并在多个选项中选择数值最小的那个。文章通过实例演示了如何通过异或运算和计数1的个数来高效解决问题。

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

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

开始以为要超时的，看了下数据觉得可以做下；
先异或，求出的是位数相同为0，不同为1 。对结果即求1的个数。

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
using namespace std;
int a[105],b[106];

int fun(int d)
{
        int count = 0;
        while (d)
        {
           count++;
           d=d&d-1;
        }
        return count;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&m,&n);
        for(int i=0;i<m;i++)
            scanf("%d",&a[i]);

        for(int i=0;i<n;i++)
            scanf("%d",&b[i]);

        for(int i=0;i<n;i++)
        {
            int ans=99999999,c=-1;
            for(int j=0;j<m;j++)
            {
                int sum,k;
                k=b[i]^a[j];
                sum=fun(k);
                 if(ans>sum)
                    ans=sum,c=a[j];
                 else if(ans==sum&&c>a[j])
                    c=a[j];
            }

            printf("%d\n",c);
        }
    }
    return 0;
}