ZOJ 1028 Flip and Shift

Flip and Shift

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Time Limit: 2 Seconds                                     Memory Limit: 65536 KB                            

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This puzzle consists of a random sequence of m black disks and   n white disks on an oval-shaped track, with a turnstile capable of flipping   (i.e., reversing) three consecutive disks. In Figure 1, there are 8 black disks and 10  white disks on the track. You may spin the turnstile to flip the three disks   in it or shift one position clockwise for each of the disks on the track (Figure 1).

  Figure 1. A flip and a shift

  The goal of this puzzle is to gather the disks of the same color in adjacent   positions using flips and shifts. (Figure 2)

  Figure 2. A goal sequence

  You are to write a program which decides whether a given sequence can reach   a goal or not. If a goal is reachable, then write a message ��YES��; otherwise, write a message ��NO��.

Input

  The input consists of T test cases. The number of test cases ) (T is given in   the first line of the input. Each of the next T lines gives a test case. A  test case consists of an integer, representing the sum of m and n, and a sequence  of m+n 0s and 1s, representing an initial sequence. A 0 denotes a white disk and  a 1 denotes a black disk. The sum of m and n is at least 10 and does not exceed 30.  There is a space between numbers.


Output

  The output should print either ��YES�� or ��NO�� for each test case, one per line.   


Sample Input

  2
  18 0 0 1 0 1 1 1 1 0 1 0 0 1 0 0 0 0 1
  14 1 1 0 0 1 1 1 0 0 1 1 0 1 0

Output for the Sample Input

YES
  NO

 

找规律 分析的太棒了http://www.cnblogs.com/scau20110726/archive/2013/06/12/3133078.html，类似的题目 一定要注意分析问题的 实质，抓住问题的核心，就能找到突破口。

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int T,n,val;
    cin>>T;
    while(T--)
    {
        int tot0=0;
        int tot1=0;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>val;
            if(val&&i&1)
                tot1++;
            else if(val&&i%2==0)
                tot0++;
        }

        if(n&1||(n%2==0&&abs(tot1-tot0)<=1))
        {
            cout<<"YES"<<endl;continue;
        }
        else
            cout<<"NO"<<endl;
    }
    return 0;
}