hdu 3711 Binary Number(暴力 模拟)

Problem Description

For  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(, )=,f(, )=, f(, )=. Now given  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 ( < T ≤ ), indicating the number of test cases. The first line of each test case contains  positive integers m and n ( < m, n ≤ ), indicating the numbers of integers of the  sets A and B, respectively. Then follow (m + n) lines, each of which contains a non-negative integers no larger than . 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

 

Sample Output

 

Source

2010 Asia Chengdu Regional Contest

 

 

 

法一：求出b[i]、a[j]的二进制数后，再比较统计

 

 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<cmath>
 #include<stdlib.h>
 #include<map>
 using namespace std;
 #define N 106
 int n,m;
 int a[N];
 int b[N];
 int s[];
 int k1;

 void change(int x){
     memset(s,,sizeof(s));
     k1=;

     while(x){
         s[k1++]=x%;
         x/=;
     }

 }
 int s1[];
 int k2;
 void change1(int x){
     memset(s1,,sizeof(s1));
     k2=;

     while(x){
         s1[k2++]=x%;
         x/=;
     }

 }
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&m);
         for(int i=;i<n;i++){
             scanf("%d",&a[i]);
         }
         for(int i=;i<m;i++){
             scanf("%d",&b[i]);
         }
         sort(a,a+n);
         for(int i=;i<m;i++){
             change(b[i]);
             int minn=;
             int f=;
             for(int j=;j<n;j++){
                 change1(a[j]);
                 int ans=;
                 int q=max(k1,k2);
                 for(int w=;w<q;w++){
                     if(s[w]!=s1[w]){
                         ans++;
                     }
                 }
                 if(ans<minn){
                     minn=ans;
                     f=j;
                 }
             }
             printf("%d\n",a[f]);
         }
     }
     return ;
 }

法二：先求b[i]^a[j]，再一次性统计

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<set>
 #include<map>
 #include<vector>
 #include<algorithm>
 using namespace std;
 #define ll long long
 #define N 50006
 int a[];
 int b[];
 int n,m;
 int s1[];
 int solve(int x)
 {
     int k1=;
     while(x)
     {
         s1[k1++]=x%;
         x=x/;
     }
     int ans=;
     for(int i=;i<k1;i++)
     {
         if(s1[i]==)
          ans++;
     }
     return ans;
 }
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&m);
         for(int i=;i<n;i++)
         {
             scanf("%d",&a[i]);
         }
         for(int i=;i<m;i++)
         {
             scanf("%d",&b[i]);
             int minn=;
             int flag;
             for(int j=;j<n;j++)
             {
                 int tmp=solve(b[i]^a[j]);
                 //printf("---%d\n",b[i]^a[j]);
                 //printf("%d\n",tmp);
                 if(minn>tmp)
                 {
                     minn=tmp;
                     flag=a[j];
                 }
                 else if(minn==tmp)
                 {
                     if(flag>a[j])
                     {
                         flag=a[j];
                     }
                 }
             }
             printf("%d\n",flag);
         }
     }
     return ;
 }