POJ 2996
发布日期:2021-06-30 15:30:53 浏览次数:2 分类:技术文章

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Help Me with the Game

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 5473   Accepted: 3393

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,e4Black: Ke8,Qd8,Ra8,Rh8,Bc8,Ng8,Nc6,a7,b7,c7,d7,e7,f7,h7,h6

Source

大致题意:

给你一个棋盘,白棋用大写字母表示,黑棋用小写字母表示。让你按照K,Q,R,B,N,P的方式输出棋子,K,Q,R,B,N输出的格式为字母+位置,P输出的格式仅为位置。白棋的位置输出方式是列由大到小,行由小到大。黑棋的位置输出方式是列和行都是由小到大。

代码:

#include
#include
#include
#include
using namespace std;int k,l;char ss[17][33]; char b[5] = {'K','Q','R','B','N'}; char g[5] = {'k','q','r','b','n'}; int flag=0;void Find(char P,int k,int l){ for(int i=15; i>=1; i-=2) { k++; l=0; for(int j=2; j<33; j+=4) { l++; if(ss[i][j]==P) { if(flag) { printf(",%c%c%d",P,96+l,k); } else { flag=1; printf("%c%c%d",P,96+l,k); } } } }}void key(char P,int k,int l){ for(int i=1; i<=15; i+=2) { k++; l=0; for(int j=2; j<33; j+=4) { l++; if(ss[i][j]==P) { if(flag) { printf(",%c%c%d",P-32,96+l,9-k); } else { flag=1; printf("%c%c%d",P-32,96+l,9-k); } } } }}int main(){ for(int i=0; i<17; i++) { for(int j=0; j<33; j++) { scanf("%c",&ss[i][j]); } getchar(); } printf("White: "); flag=0; for(int h=0; h<5; h++) { Find(b[h],0,0); } k=0; for(int i=15; i>=1; i-=2) { k++; l=0; for(int j=2; j<33; j+=4) { l++; if(ss[i][j]=='P') { printf(",%c%d",96+l,k); } } } printf("\n"); printf("Black: "); flag=0; for(int h=0; h<5; h++) { key(g[h],0,0); } l=0; for(int j=2; j<33; j+=4) { l++; k=0; for(int i=1; i<=15 ; i+=2) { k++; if(ss[i][j]=='p') { printf(",%c%d",96+l,9-k); } } } printf("\n"); return 0;}

 

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[***.8.128.20]2024年05月03日 09时35分02秒