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
Source
法一：求出b[i]、a[j]的二进制数后，再比较统计
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<queue> 6 #include<cmath> 7 #include<stdlib.h> 8 #include<map> 9 using namespace std; 10 #define N 106 11 int n,m; 12 int a[N]; 13 int b[N]; 14 int s[1000]; 15 int k1; 16 17 void change(int x){ 18 memset(s,0,sizeof(s)); 19 k1=0; 20 21 while(x){ 22 s[k1++]=x%2; 23 x/=2; 24 } 25 26 } 27 int s1[1000]; 28 int k2; 29 void change1(int x){ 30 memset(s1,0,sizeof(s1)); 31 k2=0; 32 33 while(x){ 34 s1[k2++]=x%2; 35 x/=2; 36 } 37 38 } 39 int main() 40 { 41 int t; 42 scanf("%d",&t); 43 while(t--){ 44 scanf("%d%d",&n,&m); 45 for(int i=0;i<n;i++){ 46 scanf("%d",&a[i]); 47 } 48 for(int i=0;i<m;i++){ 49 scanf("%d",&b[i]); 50 } 51 sort(a,a+n); 52 for(int i=0;i<m;i++){ 53 change(b[i]); 54 int minn=1000000000; 55 int f=0; 56 for(int j=0;j<n;j++){ 57 change1(a[j]); 58 int ans=0; 59 int q=max(k1,k2); 60 for(int w=0;w<q;w++){ 61 if(s[w]!=s1[w]){ 62 ans++; 63 } 64 } 65 if(ans<minn){ 66 minn=ans; 67 f=j; 68 } 69 } 70 printf("%d ",a[f]); 71 } 72 } 73 return 0; 74 }
法二：先求b[i]^a[j]，再一次性统计
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<set> 5 #include<map> 6 #include<vector> 7 #include<algorithm> 8 using namespace std; 9 #define ll long long 10 #define N 50006 11 int a[106]; 12 int b[106]; 13 int n,m; 14 int s1[1000]; 15 int solve(int x) 16 { 17 int k1=0; 18 while(x) 19 { 20 s1[k1++]=x%2; 21 x=x/2; 22 } 23 int ans=0; 24 for(int i=0;i<k1;i++) 25 { 26 if(s1[i]==1) 27 ans++; 28 } 29 return ans; 30 } 31 int main() 32 { 33 int t; 34 scanf("%d",&t); 35 while(t--) 36 { 37 scanf("%d%d",&n,&m); 38 for(int i=0;i<n;i++) 39 { 40 scanf("%d",&a[i]); 41 } 42 for(int i=0;i<m;i++) 43 { 44 scanf("%d",&b[i]); 45 int minn=100000000; 46 int flag; 47 for(int j=0;j<n;j++) 48 { 49 int tmp=solve(b[i]^a[j]); 50 //printf("---%d ",b[i]^a[j]); 51 //printf("%d ",tmp); 52 if(minn>tmp) 53 { 54 minn=tmp; 55 flag=a[j]; 56 } 57 else if(minn==tmp) 58 { 59 if(flag>a[j]) 60 { 61 flag=a[j]; 62 } 63 } 64 } 65 printf("%d ",flag); 66 } 67 } 68 return 0; 69 }