HDU 3711 Binary Number

Binary Number

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 472    Accepted Submission(s): 313

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

 

Source

2010 Asia Chengdu Regional Contest

 

比较简单的一个题，就是判断指定两数字转化成二进制后相应位上数字不同的个数。

代码如下：

#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<cmath>#include<ctime>using namespace std;int INT_cmp(const void *a, const void *b){    return *(int*)a - *(int*)b;}int compare(int n, int m){    int ct = 0;    int num_n, num_m;    while(n || m)    {        num_n = n % 2;        num_m = m % 2;        m /= 2;        n /= 2;        if(num_n != num_m)            ++ct;    }    return ct;}int main(){#ifdef test    freopen("in.txt", "r", stdin);#endif    int t, m, n;    int M[102], N[102];    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &m, &n);        for(int i = 0; i < m; i++)            scanf("%d", &M[i]);        for(int i = 0; i < n; i++)            scanf("%d", &N[i]);        qsort(M, m, sizeof(M[0]), INT_cmp);        for(int i = 0; i < n; i++)        {            int min = 0x7FFFFFFF,min_num;            for(int j = 0; j < m; j++)            {                int ans = compare(N[i], M[j]);                if(ans < min)                {                    min = ans;                    min_num = M[j];                }            }            printf("%d\n",min_num);        }    }    return 0;}