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