F - Binary Number

F - Binary Number
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status

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

水题，给了你两个容器A,B，对于B中的每个数找出A中的哪个数能够让F（a,b）最小，如果相等则取A的较小值

F（a,b）为两个数的二进制数在同一个位置值不同的数量。

思路，直接将两个数异或，然后求得到的数有几个1就是F（a,b）的数量，知道怎么求之后暴力扫一遍即可

#include"iostream"
#include"cstring"
#include"cstdio"
#include"algorithm"

#define INF 0x7f7f7f7f

using namespace std;

int N,M;
int A[105],B[105];

int compare(int a,int b)
{
int c = a^b;
int cnt = 0;
while(c > 0)
{
if(c&1 == 1) cnt ++;
c >>= 1;
}
return cnt;
}

int main(void)
{
int t;
scanf("%d",&t);
while(t--)
{
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 minn = INF;
int ans = -1;
for(int j = 0;j < M;j++)
{
int num = compare(B[i],A[j]);
if(minn > num || (minn == num && A[j] < ans))
{
minn = num;
ans = A[j];
}
}
printf("%d\n",ans);
}
}
return 0;
}