NBUT 2012 Summer Training -1 / [B] Birdlike Angry Pig

• [B] Birdlike Angry Pig

  问题描述
  There's no end to revenge.Pigs like to eat birds' eggs, so birds decide to revenge.For this, pigs decide to revenge as well.Pigs give a task to a birdlike pig. They let he sneak into the birds group so that they can destroy the birds group.
  
  Every bird has a ID number (It may not be unique). To let pigs know, the birdlike pig makes the "AND operation" amount all the birds' ID number and sign himself to the answer.
  
  For example:
  Birds are signed as 5, 3, 4. So the birdlike pig signs himself as (5 & 3 & 4 = 0).
  
  One day, a group birds pass by pigs' home. So the angry pig want to know whether the birdlike pig is in.
  

输入

This problem has several cases. The first line of each case is an integer N (2 <= N <= 100 000).

Then follows a line with N integers, indicates the ID of each bird/birdlike pig.

输出

For each case, if you can find the birdlike pig then output the ID of birdlike pig. Or output 'CAUTION: NO BIRDLIKE'.

样例输入
5
4 0 3 4 5

样例输出

0

分析：

首先应该注意and运算的性质： 他有个必要条件：两个相同的数and后结果还等于那个数。因为最后就是birdlike和其他bird and 而birdlike又等于其他bird and后的结果，所以显然，如果存在birdlike，则birdlike的大小等于所有数从左往右and的结果。

但是这只是一个必要条件，不是充分条件，所以还需额外判断一下。（前几次提交忘了。。。T^T）比如63&7=7，可能你光看前面会觉得7是birdlike，但实际上他不是。这就需要再扫描一遍，找到“可能的”birdlike值，然后判断他到底是不是birdlike。（他本身等于除去他后剩下所有数and的结果）。

#include <cstdio>
#include <cmath>
using namespace std;


int main()
{
    int n;
    int ans=0;
    int a[100005];
    while (scanf("%d",&n)!=EOF)
    {
        ans=2147483647;
        for (int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            ans=ans&a[i];
        }
        int ok=0;
        for (int i=0;i<n;i++)
            if (a[i]==ans)
            {
                int nn=2147483647;
                for (int j=n-1;j>=0;j--)
                {
                    if (j!=i)   nn=nn&a[j];
                }
                if (nn==ans)
                {
                    printf("%d\n",ans);
                    ok=1;
                    break;
                }
            }
        if (!ok) printf("CAUTION: NO BIRDLIKE\n");
    }
    return 0;
}