POJ 2993, Emag eht htiw Em Pleh

模拟类

Description
This problem is a reverse case of the problem 2996. You are given the output of the problem H and your task is to find the corresponding input.

 

Input
according to output of problem 2996.

 

Output
according to input of problem 2996.

 

Sample Input
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

 

Sample Output
+---+---+---+---+---+---+---+---+
|.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.|
+---+---+---+---+---+---+---+---+

 

Source
CTU Open 2005

---

 

// POJ2993.cpp : Defines the entry point for the console application.
//

#include<iostream>
#include<string>
using namespace std;

int main(int argc, char* argv[])
{
    char chessboard[17][33];

    //init chessboard
    for (int j = 0; j < 33; ++j)
        for (int i = 0; i < 17; ++i)
        {
            if ((i & 1) == 0) chessboard[i][j] = (j & 3) == 0 ? '+': '-';
            else if ((j & 3) == 0) chessboard[i][j] = '|';
            else chessboard[i][j] = ((j >> 2) + (i >> 1)) & 1 ? ':':'.';
        };

    string line;
    for (int k = 0; k < 2; ++k)
    {
        cin >> line;
        cin >> line;
        int x, y;
        const string CAP = "KQRBN";
        bool white = k == 0 ? true: false;
        char ch;
        int i = 0;
        while (i < line.size())
        {
            if (CAP.find(line[i]) != -1)
            {
                ch = white ? line[i] : line[i] -'A' +'a';
                ++i;
            }
            else
            {
                int z = CAP.find(line[i]);
                ch = white ? 'P' : 'p';
            }

            x = line[i] - 'a';
            y = line[i + 1] - '1';

            x = 2 + x * 4;
            y = 16 - (1 + y * 2);

            chessboard[y][x] = ch;

            i += 3;
        }
    }

    for (int i = 0; i < 17; ++i)
    {
        for (int j = 0; j < 33; ++j)
            cout << chessboard[i][j];
        cout <<"\n";
    }
    return 0;
}