​   A children’s puzzle that was popular 30 years ago consisted of a 5×5 frame which contained 24 small squares of equal size. A unique letter of the alphabet was printed on each small square. Since there were only 24 squares within the frame, the frame also contained an empty position which was the same size as a small square. A square could be moved into that empty position if it were immediately to the right, to the left, above, or below the empty position. The object of the puzzle was to slide squares into the empty position so that the frame displayed the letters in alphabetical order.

​   The illustration below represents a puzzle in its original configuration and in its configuration after the following sequence of 6 moves:

  1. The square above the empty position moves.

  2. The square to the right of the empty position moves.

  3. The square to the right of the empty position moves.

  4. The square below the empty position moves.

  5. The square below the empty position moves.

  6. The square to the left of the empty position moves.

​   Write a program to display resulting frames given their initial configurations and sequences of moves

Input

​   Input for your program consists of several puzzles. Each is described by its initial configuration and the sequence of moves on the puzzle. The first 5 lines of each puzzle description are the starting configuration. Subsequent lines give the sequence of moves.

​   The first line of the frame display corresponds to the top line of squares in the puzzle. The other lines follow in order. The empty position in a frame is indicated by a blank. Each display line contains exactly 5 characters, beginning with the character on the leftmost square (or a blank if the leftmost square is actually the empty frame position). The display lines will correspond to a legitimate puzzle.

​   The sequence of moves is represented by a sequence of As, Bs, Rs, and Ls to denote which square moves into the empty position. A denotes that the square above the empty position moves; B denotes that the square below the empty position moves; L denotes that the square to the left of the empty position moves; R denotes that the square to the right of the empty position moves. It is possible that there is an illegal move, even when it is represented by one of the 4 move characters. If an illegal move occurs, the puzzle is considered to have no final configuration. This sequence of moves may be spread over several lines, but it always ends in the digit 0. The end of data is denoted by the character Z.

Output

​   Output for each puzzle begins with an appropriately labeled number (Puzzle #1, Puzzle #2, etc.). If the puzzle has no final configuration, then a message to that effect should follow. Otherwise that final configuration should be displayed.

​   Format each line for a final configuration so that there is a single blank character between two adjacent letters. Treat the empty square the same as a letter. For example, if the blank is an interior position, then it will appear as a sequence of 3 blanks — one to separate it from the square to the left, one for the empty position itself, and one to separate it from the square to the right.

​   Separate output from different puzzle records by one blank line.

​   Note: The first record of the sample input corresponds to the puzzle illustrated above.

Sample Input

TRGSJ
XDOKI
M VLN
WPABE
UQHCF
ARRBBL0
ABCDE
FGHIJ
KLMNO
PQRS
TUVWX
AAA
LLLL0
ABCDE
FGHIJ
KLMNO
PQRS
TUVWX
AAAAABBRRRLL0
Z

Sample Output

Puzzle #1:
T R G S J
X O K L I
M D V B N
W P A E
U Q H C F
Puzzle #2:
A B C D
F G H I E
K L M N J
P Q R S O
T U V W X
Puzzle #3:
This puzzle has no final configuration.

HINT

​   这道题真是把我这个菜鸡给整吐了。我不知道如果空格再最后到底会不会录入到数组中,就加上一个判断数组长度,一开始使用gets()结果不可以使用然后自己写了一个类似的对于这个题来说够用的函数来代替,还是不可以。最后看到VJ下面评论说最后一个没有空行,就加上一个判断(设置一个flag1为0,如果是第一次输入就将flag1设置为1,否则在输入之前输出换行),结果还是不行。。。。求大佬指点 我还是个孩子啊,呜呜~

Presentation error

#include<stdio.h>
#include<stdlib.h>
#include<string.h> void swaps(char* a, char* b)
{
char t = *a;
*a = *b;
*b = t;
}
void ingets(char* s)
{
int i = 0;
char temp;
while (1)
{
temp = getchar();
while(!i&&temp=='\n')temp = getchar();
if (temp != '\n')
s[i++] = temp;
else break;
}
}
int main()
{
char arr[5][5];
int h, k;
int sum = 0,flag=0;
int flag1 = 0;
while (1)
{
if (!flag1)flag1++;
else printf("\n");
flag = 0;
for (int i = 0;i < 5;i++)
{
memset(arr[i], '\0', 5);
ingets(arr[i]);
while (strlen(arr[i])==0)
ingets(arr[i]);
if (i == 0 && arr[i][0] == 'Z')
{
exit(0);
flag = 1;
break;
}
if (strlen(arr[i]) == 4)
arr[i][4] = ' ';
for (int j = 0;j < 5;j++)
{
if (arr[i][j] == ' ')
{
h = i;
k = j;
}
}
}
if (flag==1)break;
char move;
while (1)
{
move = getchar();
if (move == '0')break;
if (move == 'A')
{
if (h == 0)
{
flag = 1;
while(move!='0')
move = getchar();
break;
} swaps(&arr[h][k], &arr[h-1][k]);
h--;
}
else if (move == 'B')
{
if (h == 4)
{
flag = 1;
while (move != '0')
move = getchar();
break;
}
swaps(&arr[h][k], &arr[h+1][k]);
h++;
}
else if (move == 'L')
{
if (k == 0)
{
flag = 1;
while (move != '0')
move = getchar();
break;
}
swaps(&arr[h][k], &arr[h][k-1]);
k--;
}
else if (move == 'R')
{
if (k == 4)
{
flag = 1;
while (move != '0')
move = getchar();
break;
}
swaps(&arr[h][k], &arr[h][k+1]);
k++;
}
}
if (flag)
printf("Puzzle #%d:\nThis puzzle has no final configuration.", ++sum);
else
{
printf("Puzzle #%d:\n", ++sum);
for (int i = 0;i < 5;i++)
printf("%c %c %c %c %c\n", arr[i][0], arr[i][1], arr[i][2], arr[i][3], arr[i][4]);
}
} }

Puzzle UVA - 227 PE代码求大佬指点的更多相关文章

  1. Hibernate入门(八)级联保存或更新(含问题在末尾,求大佬指点..)

    级联保存或更新CASCADE 级联保存或更新: 作用就是:保存一方的数据的时候,会把关联的对象也同时保存. 级联保存或更新的配置: 属性名:cascade 属性值: 1.none:所有情况下均不进行关 ...

  2. uva 227 Puzzle (UVA - 227)

    感慨 这个题实在是一个大水题(虽然说是世界决赛真题),但是它给出的输入输出数据,标示着老子世界决赛真题虽然题目很水但是数据就能卡死你...一直pe pe直到今天上午AC...无比感慨...就是因为最后 ...

  3. 记一次dotnet拆分包,并希望得大佬指点

    记一次dotnet拆分包,并希望得大佬指点 之前做了一个用于excel导入导出的包, 定义了一些接口, 然后基于 NPOI EPPlus MiniExcel 做了三种实现 接口大概长下面这样(现在可以 ...

  4. JavaScript忍者秘籍——运行时代码求值

    1. 代码求值机制 JavaScript中,有很多不同的代码求值机制. ● eval()函数 ● 函数构造器 ● 定时器 ● <script>元素 - 用eval()方法进行求值 作为定义 ...

  5. 震惊!Vector两行代码求逆序对,六行代码过普通平衡树

    Vector两行代码求逆序对 背景:济南集训Day7上午T2,出了一道逆序对的裸题,SB的我没看出是逆序对来,于是现场推了一个很刁钻的求逆序对的方法 首先我们想一下冒泡排序的过程,我们不难发现,对于每 ...

  6. java代码求阶乘n!

    面试过程中总是遇到要求写一段Java代码求阶乘.下面就是就是两种求阶乘 n! 的方法: 1.使用递归求解n! public int doFactorial(int n){ if(n<0){ re ...

  7. UVA大模拟代码(白书训练计划1)UVA 401,10010,10361,537,409,10878,10815,644,10115,424,10106,465,10494

    白书一:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=64609#overview 注意UVA没有PE之类的,如果PE了显示WA. UVA ...

  8. centos7中文显示为小方块~~啊啊啊 求大佬们解答

    这个问题困扰我很久了,刚好前几天注册了博客园,就想问问大佬们是怎么解决中文显示小方块的? 我试了很多办法,包括但不限于修改i18n配置文件,locale.conf,添加中文字体库等等等... 但都没有 ...

  9. 某整形数组中除了两个单身整数外, 其余的整数都是成对出现的, 利用C/C++代码求出这两个单身整数。 要求: 时间复杂度o(n), 空间复杂度o(1)------某公司招聘试题

    先看看这个题目:某整形数组中除了两个单身整数外, 其余的整数都是成对出现的, 利用C代码求出这两个单身整数. 要求: 时间复杂度o(n), 空间复杂度o(1). 我们先用最傻瓜的方式来做吧: #inc ...

随机推荐

  1. 开源OA办公平台搭建教程:基于nginx的快速集群部署——端口分发

    主机信息 主机1:172.16.98.8(linux) 主机2:172.16.98.9(linux) 集群需求 172.16.98.8:WEB服务器,应用服务器,文件存储服务器,中心服务器 172.1 ...

  2. Vuex理解与使用

    1.Vuex是什么 Vuex 是一个专为 Vue.js 应用程序开发的状态管理模式,用于管理页面的数据状态.提供统一数据操作的生态系统.在组件中可以任意获取仓库中的数据.和Vuex类似的还有redux ...

  3. mac 下如何轻松安装神器 Anaconda

    本文推荐使用homebrew 安装 1.打开终端执行 brew cask install anaconda3 然后就可以喝一杯咖啡了,终端会自动执行安装好 如果终端卡在update homebrew ...

  4. Cloud Alibabab笔记问世,全网详解仅此一份手慢无

    转: Cloud Alibabab笔记问世,全网详解仅此一份手慢无 什么是Spring cloud alibaba Spring Cloud Alibaba 是阿里巴巴提供的微服务开发一站式解决方案, ...

  5. 《C++ Primer》笔记 第13章 拷贝控制

    拷贝和移动构造函数定义了当用同类型的另一个对象初始化本对象时做什么.拷贝和移动赋值运算符定义了将一个对象赋予同类型的另一个对象时做什么.析构函数定义了当此类型对象销毁时做什么.我们称这些操作为拷贝控制 ...

  6. iOS之CoreBluetooth

    思路 手机与设备间的通讯方式CoreBluetooth是比较常见且通用的.在iOS开发中需明晰以下几点 蓝牙4.0最多可联机7个设备,iPhone6以上都是蓝牙4.0 两台iPhone并不直接通过蓝牙 ...

  7. 一文读懂Java动态代理

    作者 :潘潘 日期 :2020-11-22 事实上,对于很多Java编程人员来说,可能只需要达到从入门到上手的编程水准,就能很好的完成大部分研发工作.除非自己强主动获取,或者工作倒逼你学习,否则我们好 ...

  8. celery 与 flask 实现异步任务调度

    Flask 定了2中上下文,来实现机遇线程\协程的,wsgi服务的请求(request.session)和存储(g,current_app )过程,通过栈来完成不同线程和协程的上下文切换,在与cele ...

  9. WNN48T6X 54端口国产化万兆交换机

    WNN48T6X是基于盛科CTC5160设计的国产化三层万兆交换机,提供48路千兆电口和6路万兆光口,采用龙芯 2K1000处理器,支持双冗余可插拔电源供电.支持常规的L2/L3协议,支持Telnet ...

  10. android分析之智能指针

    智能指针是一个包装类,该类有一个指针指向真正的类对象 引用计数型智能指针,该引用计数是在应该被真正类所持有,而非包装类(智能指针) 为了方便,会将引用计数单独实现在一个类中,这样所有继承它的类都有计数 ...