Binary Number

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 2714    Accepted Submission(s): 1573

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

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

 

Author

CAO, Peng

 

Source

给n个数和m个数，判断对于m哪个n与它相差2进制数最少

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            #define INF 0x3f3f3f3fusing namespace std;const int maxn = 110;const int n = 100;bitset<20> a[maxn], b[maxn];int num[maxn];int f(int p, int q){ int ans = 0; for (int i = 0; i < 20; i++) { if (a[p][i] != b[q][i]) ans++; } return ans;}int main(){ int t; scanf("%d", &t); while (t--) { int n, m; scanf("%d%d", &n, &m); for (int i = 0; i < n; i++) { scanf("%d", num + i); a[i] = num[i]; } for (int j = 0; j < m; j++) { int p; scanf("%d", &p); b[j] = p; } for (int i = 0; i < m; i++) { int ans = INF, val = -1; for (int j = 0; j < n; j++) { int temp = f(j, i); if (temp < ans) { ans = temp; val = num[j]; } else if (temp == ans && (val == -1 || num[j] < val)) { val = num[j]; } } printf("%d\n", val); } }}