noj 1173 (宁波)Birdlike Angry Pig （暴力枚举）

• http://ac.nbutoj.com/Problem/view.xhtml?id=1173

• [1173] Birdlike Angry Pig

• 时间限制: 1000 ms　内存限制: 65535 K
• 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

• 思路：
• 题目的意思为 鸟猪 伪装在 鸟群里面，鸟猪的值为鸟群值的与集（&），设鸟猪为d，鸟群为a,b,c，则a&b&c=d；
• 现在题目给出一个序列，问是否鸟猪在这个序列之中，所以序列只要满足一个数的值等于其他数的相与集（&），则这个数就是鸟猪；
• 现在我先定义三个数组A，B，C；A里面存输入的数，B里面存着的是A数组里面前n项的与集（&），C里面存着的是A数组里面后n项的与集（&）；
• 假设A里面存着a,b,c,d;
• A ==>        a               b                 c                  d
• B ==>        a             a&b         a&b&c      a&b&c&d
• C ==> a&b&c&d   b&c&d         c&d               d

　　不难发现：如果A[i]==B[i-1]&C[i+1]，则A[i]即为鸟猪

代码：C++

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<string.h>
#include<string>
#include<ctime>
#include<queue>
#include<list>
#include<map>
#include<set>
#define INF 999999999
#define MAXN 10000000
using namespace std;

int a[100010],b[100010],c[100010];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int i;
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        b[0]=-1;
        for(i=1;i<=n;i++)
            b[i]=(b[i-1]&a[i]);
        c[n+1]=-1;
        for(i=n;i>0;i--)
            c[i]=(c[i+1]&a[i]);
        for(i=1;i<=n;i++)
            if((b[i-1]&c[i+1])==a[i])
            {
                printf("%d\n",a[i]);
                break;
            }
        if(i>n)
            printf("CAUTION: NO BIRDLIKE\n");
    }
    return 0;
}