Help Me with the Game
Help Me with the GameCrawling in process... Crawling failed
Description
Input
Output
The description of the position of the pieces is a comma-separated list of terms describing the pieces of the appropriate player. The description of a piece consists of a single upper-case letter that denotes the type of the piece (except for pawns, for that this identifier is omitted). This letter is immediatelly followed by the position of the piece in the standard chess notation -- a lower-case letter between "a" and "h" that determines the column ("a" is the leftmost column in the input) and a single digit between 1 and 8 that determines the row (8 is the first row in the input).
The pieces in the description must appear in the following order: King("K"), Queens ("Q"), Rooks ("R"), Bishops ("B"), Knights ("N"), and pawns. Note that the numbers of pieces may differ from the initial position because of capturing the pieces and the promotions of pawns. In case two pieces of the same type appear in the input, the piece with the smaller row number must be described before the other one if the pieces are white, and the one with the larger row number must be described first if the pieces are black. If two pieces of the same type appear in the same row, the one with the smaller column letter must appear first.
Sample Input
+---+---+---+---+---+---+---+---+
|.r.|:::|.b.|:q:|.k.|:::|.n.|:r:|
+---+---+---+---+---+---+---+---+
|:p:|.p.|:p:|.p.|:p:|.p.|:::|.p.|
+---+---+---+---+---+---+---+---+
|...|:::|.n.|:::|...|:::|...|:p:|
+---+---+---+---+---+---+---+---+
|:::|...|:::|...|:::|...|:::|...|
+---+---+---+---+---+---+---+---+
|...|:::|...|:::|.P.|:::|...|:::|
+---+---+---+---+---+---+---+---+
|:P:|...|:::|...|:::|...|:::|...|
+---+---+---+---+---+---+---+---+
|.P.|:::|.P.|:P:|...|:P:|.P.|:P:|
+---+---+---+---+---+---+---+---+
|:R:|.N.|:B:|.Q.|:K:|.B.|:::|.R.|
+---+---+---+---+---+---+---+---+
Sample Output
White: Ke1,Qd1,Ra1,Rh1,Bc1,Bf1,Nb1,a2,c2,d2,f2,g2,h2,a3,e4
Black: Ke8,Qd8,Ra8,Rh8,Bc8,Ng8,Nc6,a7,b7,c7,d7,e7,f7,h7,h6
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
using namespace std;
struct node
{
int hang;
int lie;
int data;
int jb;
} white[],black[];
int sw(char p)//只是为了让级别好记,好排序
{
if(p=='K' || p=='k') return ;
if(p=='Q' || p=='q') return ;
if(p=='R' || p=='r') return ;
if(p=='B' || p=='b') return ;
if(p=='N' || p=='n') return ;
if(p=='P' || p=='p') return ;
return ;
}
int cmpw(const void*a,const void *b)//这个排序很是坑爹,最初一直没想这样写,结果还是这样,又快又方便
{
if(((node*)a)->jb!=((node*)b)->jb)
return ((node*)a)->jb-((node*)b)->jb;
if(((node*)a)->hang!=((node*)b)->hang)
return ((node*)a)->hang-((node*)b)->hang;
return ((node*)a)->lie-((node*)b)->lie;
}
int cmpb(const void*a,const void *b)
{
if(((node*)a)->jb!=((node*)b)->jb)
return ((node*)a)->jb-((node*)b)->jb;
if(((node*)a)->hang!=((node*)b)->hang)
return ((node*)b)->hang-((node*)a)->hang;
return ((node*)a)->lie-((node*)b)->lie;
}
int main()
{
int wi=,bi=,i;
char ch;
for(i=; i>; i--)
{
scanf("+---+---+---+---+---+---+---+---+\n");
int sum=;
while(scanf("%c",&ch)&&ch!='\n')
{
if('A'<=ch&&ch<='Z')
{
white[wi].hang=i;
white[wi].lie=sum;//明显第几列和“|”的数目挂钩
white[wi].data=ch;
white[wi].jb=sw(ch);
wi++;
}
else if('a'<=ch&&ch<='z')
{
black[bi].hang=i;
black[bi].lie=sum;
black[bi].data=ch;
black[bi].jb=sw(ch);
bi++;
}
else if(ch=='|')//再剩余的字符就不用看了
sum++;
}
}
scanf("+---+---+---+---+---+---+---+---+");
qsort(white,wi,sizeof(node),cmpw);//排序,这个让我纠结了好久,郁闷
qsort(black,bi,sizeof(node),cmpb);
wi--;
bi--;
printf("White: ");//这种坑爹的输出,好吧……只是很长,不复杂
for(i=; i<wi; i++)
{
if(white[i].data!='P')
printf("%c",white[i].data);
printf("%c",white[i].lie+'a'-);
printf("%d,",white[i].hang); }
if(white[i].data!='P')
printf("%c",white[i].data);
printf("%c",white[i].lie+'a'-);
printf("%d\n",white[i].hang);
printf("Black: ");
for(i=; i<bi; i++)
{
if(black[i].data!='p')
printf("%c",black[i].data-);
printf("%c",black[i].lie+'a'-);
printf("%d,",black[i].hang);
}
if(black[i].data!='p')
printf("%c",black[i].data-);
printf("%c",black[i].lie+'a'-);
printf("%d\n",black[i].hang);
return ;
}
随机推荐
- hadoop2.2 伪分布式环境
在安装JDK之前,请确认系统是32还是64,根据系统版本,选择JDK版本.Hadoop版本 下面是以在CentOS-6.5-x86_64系统上安装为例 安装前准备 在"/usr"下 ...
- 关于struts2中action请求会执行两次的问题
关于struts2中action请求会执行两次的问题 在struts2中发现,调用action中的方法,方法会被执行两次,后来发现调用的方法是get开头的,把它改为其他名称开头的后,就不会执行 ...
- linux 简单命令
很久没有接触linux了,很多命令也忘记了,现在自己独立安装一个linux,独立安装LAMP,让自己记录下来这段. 怎么进入命令行 init 3, 回到桌面 init 5在不是root用户情况下,切换 ...
- Tomcat-java.lang.ClassNotFoundException: org.apache.juli.logging.LogFactory
在我的MyEclipse中新建一个网站,并新建一个.jsp文件,配置server为Tomcat后,运行.jsp文件的时候,报错:java.lang.ClassNotFoundException: or ...
- js正则实现用户输入银行卡号的控制及格式化
//js正则实现用户输入银行卡号的控制及格式化 <script language="javascript" type="text/javascript"& ...
- javascript基础学习(二)
javascript的数据类型 学习要点: typeof操作符 五种简单数据类型:Undefined.String.Number.Null.Boolean 引用数据类型:数组和对象 一.typeof操 ...
- 找出整数中第k大的数
一 问题描述: 找出 m 个整数中第 k(0<k<m+1)大的整数. 二 举例: 假设有 12 个整数:data[1, 4, -1, -4, 9, 8, 0, 3, -8, 11, 2 ...
- 64位系统下System32文件系统重定向
前言 因为一次偶然的机会,需要访问系统目录“C:/Windows/System32“文件夹下的内容,使用的测试机器上预装了win7 64系统.在程序运行中竟然发生了该文件路径不存在的问题!!通过查看网 ...
- 【BZOJ2809】【splay启发式合并】dispatching
Description 在一个忍者的帮派里,一些忍者们被选中派遣给顾客,然后依据自己的工作获取报偿.在这个帮派里,有一名忍者被称之为 Master.除了 Master以外,每名忍者都有且仅有一个上级. ...
- 在CentOS 7中轻松安装Atomic应用(atomicapp)
sudo yum install docker atomic etcd kubernetes sudo systemctl enable docker.service sudo systemctl s ...