Puzzle UVA - 227 PE代码求大佬指点
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:
The square above the empty position moves.
The square to the right of the empty position moves.
The square to the right of the empty position moves.
The square below the empty position moves.
The square below the empty position moves.
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 主键策略+mybayisPlus自动增长
主键策略: 1.自动增长 有一点小缺陷:例如当一张表里的数据过于庞大时我们会进行分表操作,若是用自动增长策略,那么除了第一张表外的每一张表都必须知道上一张的表的的最后ID值.这个操作便会造成效率的变低 ...
- JDBC概念理解
##JDBC: 概念:Java DataBase Connectivity Java 数据库连接 Java语言操作数据库 JDBC本质:其实是官方(sun公司)定义的一套操作所有关系型数据库的规则 ...
- 学习String源码的部分方法
先看构造器: private final char value[]; //char类型的数组 以下均会用到 private int hash; //缓存字符串的哈希值 //以下均会用到 public ...
- 如何用Eggjs从零开始开发一个项目(3)
上一篇中我们编写了用户注册登录.登录的代码,学习了如何进行用户的认证(JWT),如何安全地存储用的密码(hash).这一篇我们有以下2个任务: 获取token中的数据: 通过model来同步数据库. ...
- go语言-csp模型-并发通道
[前言]go语言的并发机制以及它所使用的CSP并发模型 一.CSP并发模型 CSP模型是上个世纪七十年代提出的,用于描述两个独立的并发实体通过共享的通讯 channel(管道)进行通信的并发模型. C ...
- springMvc+AJAX+JSON的增删改查
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"% ...
- 后端程序员之路 24、Redis hiredis
Redishttps://redis.io/ Redis快速入门 - Redis教程http://www.yiibai.com/redis/redis_quick_guide.html wget ht ...
- MySQL深入研究--学习总结(3)
前言 接上文,继续学习后续章节.细心的同学已经发现,我整理的并不一定是作者讲的内容,更多是结合自己的理解,加以阐述,所以建议结合原文一起理解. 第九章<普通索引和唯一索引,如何选择> 从查 ...
- golang调用shell命令(实时输出, 终止等)
背景 是这样的,最近在研究一个定时任务系统的改造,可能有点像jenkins做到的那种吧. 可以输入shell命令,也可以执行py脚本等等,相比之前来说,也要能够及时停止! 但是遇到了这么个问题,gol ...
- rest framework renderers
渲染器 前TemplateResponse实例可以被返回给客户端,它必须被渲染.渲染过程需要模板和上下文的中间表示,并把它变成能够提供给客户端的最后一个字节流. - Django文档 REST框架包含 ...