HDU 3711 D - Binary Number

最新推荐文章于 2020-07-11 22:57:57 发布

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最新推荐文章于 2020-07-11 22:57:57 发布

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分类专栏：杂题文章标签： HDU-3711

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本博客介绍了一种算法，用于在两个非负整数集合中找到一对整数，使得它们二进制表示的位数差异最小。通过遍历一个集合中的每个元素并与另一个集合中的元素进行比较，该算法能够高效地解决此问题。

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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

在A里面找到一个数使得它和B里面的数二进制化时不一样的位数最大，如果有多个答案，取最小值，模拟。

代码：


#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int a[105];  
int count(int a)  
{  
    int ans = 0;  
    for(;a;a>>=1) if(a&1) ans++;  
    printf("ans=%d\n",ans);
    return ans;  
}  
int main()  
{  
    int b, i, j, n, m, k, minn, t,cas;  
    scanf("%d",&cas);  
    while(cas--)  
    {  
        scanf("%d%d",&n,&m);  
        for(i=0; i<n; i++) 
        scanf("%d",&a[i]);  
        for(i=0; i<m; i++)  
        {  
            scanf("%d",&b);  
            minn = count(b^a[0]);  
            k = 0;  
            for(j=1; j<n; j++)  
            {  
                t = count(b^a[j]);  
                if(t<minn||t==minn&&a[j]<a[k])  
                    { minn = t;k = j;}  
            }  
            printf("%d\n",a[k]);  
        }  
    }  
    return 0;  
}