A. Hanoi tower

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100114

Description

you the conditions of this task. There are 3 pivots: A, B, C. Initially, n disks of different diameter are placed on the pivot A: the smallest disk is placed on the top and every next one is placed in an increasing order of their diameters. The second and the third pivots are still empty. You have to move all the disks from pivot A to pivot B, using pivot C as an auxiliary. By one step you can take off 1 upper disk and put it either on an empty pivot or on another pivot over a disk with a bigger diameter. Almost all books on programming contain a recursive solution of this task. In the following example you can see the procedure, written in Pascal. Procedure Hanoi (A, B, C: integer; N:integer); Begin If N>0 then Begin Hanoi (A, C, B, N-1); Writeln(‘диск ’, N, ‘ from ‘, A, ‘ to ‘, B); Hanoi (C, B, A, N-1) End End; The number of step Disk From To Combination 0. AAA 1. 1 A B BAA 2. 2 A C BCA 3. 1 B C CCA 4. 3 A B CCB 5. 1 C A ACB 6. 2 C B ABB 7. 1 A B BBB It is well known that the solution given above requires (2n –1) steps. Taking into account the initial disposition we totally have 2n combinations of n disks disposition between three pivots. Thus, some combinations don’t occure during the algorithm execution. For example, the combination «CAB» will not be reached during the game with n = 3 (herein the smallest disk is on pivot C, the medium one is on pivot A, the biggest one is on pivot B). Write a program that establishes if the given combination is occurred during the game.

Input

The first line of an input file contains a single integer n – the number of disks, and the second line contains n capital letters (“A”, “B” or “C”) – the disposition of the n disks between the pivots. The first (leftmost) letter designates position (a pivot name) of the smallest disk, the second letter – position of the second lagest, and so on…
1 ≤ n ≤ 250

Output

The output file must contain “YES” or “NO” depending on the reachability of the disks disposition during the game.

Sample Input

3

ACB

Sample Output

YES

HINT

 

题意

汉诺塔,给你个状态,问你由题中所给的代码是否能跑到这个状态

题解:

找规律,找规律……

代码:

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm> int n,flag;
char s[]; int movef(int x,char a,char b,char c)
{
if(x==)
{
if(s[x-]==a||s[x-]==b) return ;
}
else
{
if(s[x-]==a) return movef(x-,a,c,b);
if(s[x-]==b) return movef(x-,c,b,a);
}
return ;
} main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
scanf("%d",&n);
scanf("%s",s);
flag=movef(n,'A','B','C');
if(flag) printf("YES\n");else printf("NO\n");
}

Codeforces Gym 100114 A. Hanoi tower 找规律的更多相关文章

  1. codeforces Gym 100418D BOPC 打表找规律,求逆元

    BOPCTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?c ...

  2. Codeforces Gym 100015A Another Rock-Paper-Scissors Problem 找规律

    Another Rock-Paper-Scissors Problem 题目连接: http://codeforces.com/gym/100015/attachments Description S ...

  3. Codeforces Gym 100425D D - Toll Road 找规律

    D - Toll RoadTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view ...

  4. Codeforces GYM 100114 C. Sequence 打表

    C. Sequence Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Description ...

  5. codeforces GYM 100114 J. Computer Network 无相图缩点+树的直径

    题目链接: http://codeforces.com/gym/100114 Description The computer network of “Plunder & Flee Inc.” ...

  6. codeforces GYM 100114 J. Computer Network tarjan 树的直径 缩点

    J. Computer Network Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Des ...

  7. Codeforces Gym 100114 H. Milestones 离线树状数组

    H. Milestones Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Descripti ...

  8. Codeforces GYM 100114 D. Selection 线段树维护DP

    D. Selection Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Descriptio ...

  9. Codeforces GYM 100114 B. Island 水题

    B. Island Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Description O ...

随机推荐

  1. 【JS】<a>标签调用js中函数的几种方法

    我们常用的在a标签中有点击事件: a href="javascript:js_method();" 这是我们平台上常用的方法,但是这种方法在传递this等参数的时候很容易出问题,而 ...

  2. 移动对meta的定义

    以下是meta每个属性详解 尤其要注意的是content里多个属性的设置一定要用分号+空格来隔开,如果不规范将不会起作用. 一.<meta http-equiv="Content-Ty ...

  3. php mysql事务

    这里记录一下php操作mysql事务的一些知识 要知道,MySQL默认的行为是在每条SQL语句执行后执行一个COMMIT语句,从而有效的将每条语句独立为一个事务.但是,在使用事务时,是需要执行多条sq ...

  4. mysql5.6子查询的优化

    https://dev.mysql.com/doc/refman/5.6/en/subquery-optimization.html Semi-join in MySQL 5.6   MySQL 5. ...

  5. Mem Cgroup目录无法清理问题分析

    http://blogs.360.cn/360xitong/2013/05/02/mem-cgroup%E7%9B%AE%E5%BD%95%E6%97%A0%E6%B3%95%E6%B8%85%E7% ...

  6. Android WebView常见问题的解决方案总结----例如Web page not available

    之前android虚拟机一直都可以直接联网,今天写了一个WebView之后,突然报出了Web page not available的错误,但是查看虚拟机自带的浏览器,是可以上网的,所以检查还是代码的问 ...

  7. Hadoop对文本文件的快速全局排序

    一.背景 Hadoop中实现了用于全局排序的InputSampler类和TotalOrderPartitioner类,调用示例是org.apache.hadoop.examples.Sort. 但是当 ...

  8. CSS基础知识——选择器

    选择器 元素选择器# 文档元素为最基本的选择器 例子:div{属性:值}; 选择器分组 例子:h2,p{属性:值}; 表示符合这两种规则的元素设置相同的属性值 通配选择器 表示所有元素 类选择器 应用 ...

  9. QT多线程笔记

    1.QT多线程涉及到主线程和子线程之间交互大量数据的时候,使用QThread并不方便,因为run()函数本身不能接受任何参数,因此只能通过信号和槽的交互来获取数据,如果只是单方面简单交互数据还过得去, ...

  10. C字符串和C++中string的区别 &amp;&amp;&amp;&amp;C++中int型与string型互相转换

    在C++中则把字符串封装成了一种数据类型string,可以直接声明变量并进行赋值等字符串操作.以下是C字符串和C++中string的区别:   C字符串 string对象(C++) 所需的头文件名称 ...