ZOJ 1028 Flip and Shift

最新推荐文章于 2023-12-23 12:48:01 发布

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最新推荐文章于 2023-12-23 12:48:01 发布

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本文介绍了一种由黑白磁盘组成的环形谜题，玩家可通过翻转三个连续磁盘或整体顺时针移动来操作。目标是将相同颜色的磁盘聚集在一起。文章提供了一个判断初始状态是否能达成目标状态的程序实现。

Flip and Shift

Time Limit: 2 Seconds Memory Limit: 65536 KB

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;
}