Help Me with the Game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3630   Accepted: 2319

Description

Your task is to read a picture of a chessboard position and print it in the chess notation.

Input

The input consists of an ASCII-art picture of a chessboard with chess pieces on positions described by the input. The pieces of the white player are shown in upper-case letters, while the black player's pieces are lower-case letters. The letters are one of "K" (King), "Q" (Queen), "R" (Rook), "B" (Bishop), "N" (Knight), or "P" (Pawn). The chessboard outline is made of plus ("+"), minus ("-"), and pipe ("|") characters. The black fields are filled with colons (":"), white fields with dots (".").

Output

The output consists of two lines. The first line consists of the string "White: ", followed by the description of positions of the pieces of the white player. The second line consists of the string "Black: ", followed by the description of positions of the pieces of the black player.

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

Source

 
 #include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<algorithm>
using namespace std;
char st[] , chess[][] ;
int a[][] ; struct white
{
char ch , row , col;
int pow ;
}wh[]; struct black
{
char ch , row , col ;
int pow ;
}bl[]; bool cmp2 (white a , white b)
{
if (a.pow < b.pow)
return ;
if (a.pow > b.pow)
return ;
if (a.pow == b.pow ) {
if (a.row == b.row)
return a.col < b.col ;
else
return a.row < b.row ;
}
} bool cmp1 (black a , black b)
{
if (a.pow < b.pow)
return ;
if (a.pow > b.pow)
return ;
if (a.pow == b.pow ) {
if (a.row == b.row)
return a.col < b.col ;
else
return a.row > b.row ;
}
} int main ()
{
// freopen ("a.txt" , "r" , stdin) ;
for (int i = ; i <= ; i++) {
if (i & )
gets (st) ;
else
gets (chess[i / ]) ;
}
int k , f; for (int i = ; i <= ; i++) {
k = ;
for (int j = ; j < ; j += , k++) {
switch ( chess[i][j] )
{
case 'k' : a[i][k] = ; break ;
case 'q' : a[i][k] = ; break ;
case 'r' : a[i][k] = ; break ;
case 'b' : a[i][k] = ; break ;
case 'n' : a[i][k] = ; break ;
case 'p' : a[i][k] = ; break ;
case 'K' : a[i][k] = - ; break ;
case 'Q' : a[i][k] = - ; break ;
case 'R' : a[i][k] = - ; break ;
case 'B' : a[i][k] = - ; break ;
case 'N' : a[i][k] = - ; break ;
case 'P' : a[i][k] = - ; break ;
case ':' : a[i][k] = ; break ;
case '.' : a[i][k] = ; break ;
default : break ;
}
}
}
/* for (int i = 1 ; i <= 8 ; i++) {
for (int j = 1 ; j <= 8 ; j++) {
printf ("%d\t" , a[i][j]) ;
}
puts ("") ;
}*/
k = , f = ;
for (int i = ; i <= ; i++) {
for (int j = ; j <= ; j++) {
if (a[i][j] > ) {
bl[k].pow = a[i][j] ;
bl[k].row = '' + - i ;
bl[k].col = 'a' + j - ;
switch (a[i][j])
{
case : bl[k].ch = 'K' ; break ;
case : bl[k].ch = 'Q' ; break ;
case : bl[k].ch = 'R' ; break ;
case : bl[k].ch = 'B' ; break ;
case : bl[k].ch = 'N' ; break ;
default : break ;
}
k++ ;
}
if (a[i][j] < ) {
wh[f].pow = -a[i][j] ;
wh[f].row = '' + - i ;
wh[f].col = 'a' + j - ;
switch (-a[i][j])
{
case : wh[f].ch = 'K' ; break ;
case : wh[f].ch = 'Q' ; break ;
case : wh[f].ch = 'R' ; break ;
case : wh[f].ch = 'B' ; break ;
case : wh[f].ch = 'N' ; break ;
default : break ;
}
f++ ;
}
}
} sort (bl , bl + k , cmp1) ;
sort (wh , wh + f , cmp2) ; printf ("White: ") ;
for (int i = ; i < f ; i++) {
// printf ("wh[%d].pow = %d\n" , i , wh[i].pow) ;
if (wh[i].pow != )
printf ("%c" , wh[i].ch) ;
printf ("%c%c" , wh[i].col , wh[i].row ) ;
if (i != f - )
printf (",") ;
}
puts ("") ; printf ("Black: ") ;
for (int i = ; i < k ; i++) {
if (bl[i].pow != )
printf ("%c" , bl[i].ch) ;
printf ("%c%c" , bl[i].col , bl[i].row ) ;
if (i != k - )
printf (",") ;
}
puts ("") ;
return ;
}

Help Me with the Game(imitate)的更多相关文章

  1. Emag eht htiw Em Pleh(imitate)

    Emag eht htiw Em Pleh Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2901   Accepted:  ...

  2. Robot Motion(imitate)

    Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11065   Accepted: 5378 Des ...

  3. Crashing Robots(imitate)

    Crashing Robots Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8124   Accepted: 3528 D ...

  4. Parencodings(imitate)

    Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20679   Accepted: 12436 De ...

  5. [JavaScript] Imitate String.Format() in c#

    Definition if (!String.prototype.format) { String.prototype.format = function () { var args = argume ...

  6. Basic Tutorials of Redis(8) -Transaction

    Data play an important part in our project,how can we ensure correctness of the data and prevent the ...

  7. 第一届山东省ACM——Balloons(java)

    Description Both Saya and Kudo like balloons. One day, they heard that in the central park, there wi ...

  8. Essential controls for web app

    AUTO-COMPLETE/AUTO-SUGGEST Auto-complete using Vaadin Offer auto-suggest or auto-complete to help yo ...

  9. sqlmap用户手册 | WooYun知识库

    sqlmap用户手册 说明:本文为转载,对原文中一些明显的拼写错误进行修正,并标注对自己有用的信息. 原文:http://drops.wooyun.org/tips/143  ============ ...

随机推荐

  1. Swift与Objective-c 混编CocoaPods 引入第三方库遇到的问题 (一)

    最近Swift 这么火也想尝试着用一下.考虑到Swift 出来的时间也不长.还有就是就是苹果更新的过于平凡 暂时还是不要将现有项目都用swift开发. 先来看看我遇到的问题: 问题一.

  2. [USACO 1.5.4]checker(水题重做——位运算(lowbit的应用))

    描述 检查一个如下的6 x 6的跳棋棋盘,有六个棋子被放置在棋盘上,使得每行.每列有且只有一个,每条对角线(包括两条主对角线的所有平行线)上至多有一个棋子. 0 1 2 3 4 5 6 ------- ...

  3. Spring 事务配置管理,简单易懂,详细 [声明式]

    Spring 事务配置说明 Spring 如果没有特殊说明,一般指是跟数据存储有关的数据操作事务操作:对于数据持久操作的事务配置,一般有三个对象,数据源,事务管理器,以及事务代理机制: Spring ...

  4. DOM(二)使用DOM

    在了解DOM(文本对象模型)的框架和节点后,最重要的是使用这些节点处理html网页 对于一个DOM节点node,都有一系列的属性和方法可以使用.常用的有下表. 完善:http://www.w3scho ...

  5. MongoDB 客户端 MongoVue

    直接上图片,图片是按顺序来的 软件下载地址(Windows下的MongoDB客户端MongoVUE 这是最后一个全功能的不收费的版本): http://pan.baidu.com/s/1skYIEq5

  6. MVC4 code first 增加属性,对应自动修改列的方法笔记

    VS工具>库程序包管理器>程序包管理控制台,然后输入以下命令 enable-migrations -contexttypename Mvc4Application1.Models.Movi ...

  7. java设计模式--原始模型模式

    简介 原始模型模式属于对象的创建模式.通过一个原型对象来指明要创建对象的类型,然后用复制原型对象的方法来创建出更多同类型的对象. Java所有的类都是从java.lang.Object类继承来的,Ob ...

  8. easyui_动态添加隐藏toolbar按钮

    目标:动态添加隐藏toolbar,比如根据权限动态显示新增.修改.删除按钮等 思路:先初始化toolbar的所有按钮,加载datagrid其它信息,再根据权限显示隐藏toolbar按钮 步骤: 1.加 ...

  9. 【CodeForces 624D】Array GCD

    题 You are given array ai of length n. You may consecutively apply two operations to this array: remo ...

  10. Java编程思想学习(十四) 枚举

    关键字enum可以将一组具名的值有限集合创建一种为新的类型,而这些具名的值可以作为常规的程序组件使用. 基本enum特性 调用enum的values()方法可以遍历enum实例,values()方法返 ...