大家好，又见面了，我是全栈君，今天给大家准备了Idea注册码。
Binary Number
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1287 Accepted Submission(s): 807
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
AC代码例如以下：
#include <stdio.h>
int a[105];
int count(int x)
{
int c = 0;
for(;x;x>>=1) if(x&1) c++;
return c;
}
int main()
{
int b, i, j, n, m, k, min, t,cases;
scanf("%d",&cases);
while(cases--)
{
scanf("%d%d",&n,&m);
for(i=0; i<n; i++) scanf("%d",&a[i]);
for(i=0; i<m; i++)
{
scanf("%d",&b);
min = count(b^a[0]);
k = 0;
for(j=1; j<n; j++)
{
t = count(b^a[j]);
if(t<min||t==min&&a[j]<a[k])
{ min = t;k = j;}
}
printf("%d\n",a[k]);
}
}
return 0;
}
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