zoj 1028 Flip and Shift(数学)

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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

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Source: Asia 2001, Taejon (South Korea)

      题意：在一个圆形的首尾相连的容器中放两种球，以任意一个球为中心，交换相邻两球的位置，问是否可以将颜色相同的球放在一起。

其实这题主要考察的是你运用数学知识分析问题的能力。设位置编号为1,2,3,...,n,n为小球的总数目。

     这样通过交换就可以使当前布局满足连续的必要性了

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cmath>
 7 #include <algorithm>
 8 #include <cctype>
 9 #include <string>
10 #include <map>
11 #define N 500015
12 #define INF 1000000
13 #define ll long long
14 using namespace std;
15 
16 int main(void)
17 {
18     int t,odd,even,n;
19     scanf("%d",&t);
20     while(t--)
21     {
22         even = odd = 0;
23         scanf("%d",&n);
24         for(int i = 0; i < n; i++)
25         {
26             int tp;
27             scanf("%d",&tp);
28             if(i & 1)
29                 odd += tp;
30             else
31                 even += tp;
32         }
33         if(n & 1)
34             printf("YES\n");
35         else
36         {
37             if(abs(odd - even) < 2)
38                 printf("YES\n");
39             else
40                 printf("NO\n");
41         }
42     }
43     return 0;
44 }

zoj 1028 Flip and Shift(数学)

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原文地址：http://www.cnblogs.com/henserlinda/p/4734322.html