codeforces195c

link:http://codeforces.com/problemset/problem/336/C

从大到小枚举，如果对应的二进制位不等于0，就加进来，最后的sum如果%2^k==0那么就是合法的。

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <cctype>
 #include <algorithm>
 #include <queue>
 #include <deque>
 #include <queue>
 #include <list>
 #include <map>
 #include <set>
 #include <vector>
 #include <utility>
 #include <functional>
 #include <fstream>
 #include <iomanip>
 #include <sstream>
 #include <numeric>
 #include <cassert>
 #include <ctime>
 #include <iterator>
 const int INF = 0x3f3f3f3f;
 const int dir[][] = {{-,},{,},{,-},{,},{-,-},{-,},{,-},{,}};
 using namespace std;
 int a[];
 int main(void)
 {
     #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     #endif // ONLINE_JUDGE
     int n;
     while (~scanf("%d", &n))
     {
         for (int i=;i<n;++i) scanf("%d",a+i);
         for (int i=;i>=;--i)
         {
             int sad=<<i, sum=-, cnt=;
             for (int j=;j<n;++j)
             {
                 if((sad&a[j])!=)
                 {
                     if(sum==-) sum=a[j];
                     else sum = sum & a[j];
                     cnt++;
                 }
             }
             int tmp=;
             if (sum%sad==)
             {
                 printf("%d\n",cnt);
                 for (int j=;j<n;++j)
                 {
                     if((sad&a[j])!=)
                     {
                         tmp++;
                         if(tmp!=cnt)
                             printf("%d ",a[j]);
                         else printf("%d\n",a[j]);
                     }
                 }
                 break;
             }
         }
     }

     return ;
 }

很神奇的做法。

开始被这道题目吓到了。o(╯□╰)o