poj 2993 Emag eht htiw Em Pleh

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.|
+---+---+---+---+---+---+---+---+

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <stack>
#include <algorithm>
#define LL long long
#define INF 0x3f3f3f3f

using namespace std;

char Map[100][100];


int main()
{
    char s[110];
    gets(s);
    int len = strlen(s);

    for(int i=0; i<len; i++)
    {
        if(s[i]>='0' && s[i]<='9')
        {
            if(s[i-2] != ',')
            {
                int xx = s[i] - '0';
                int yy = s[i-1] - 'a';
                Map[xx][yy] = s[i-2];
            }
            else
            {
                int xx = s[i] - '0';
                int yy = s[i-1] - 'a';
                Map[xx][yy] = 'P';
            }
        }
    }

    gets(s);
    len = strlen(s);

    for(int i=0; i<len; i++)
    {
        if(s[i]>='0' && s[i]<='9')
        {
            if(s[i-2] != ',')
            {
                int xx = s[i] - '0';
                int yy = s[i-1] - 'a';
                Map[xx][yy] = s[i-2] + 32;
            }
            else
            {
                int xx = s[i] - '0';
                int yy = s[i-1] - 'a';
                Map[xx][yy] = 'p';
            }
        }
    }

    for(int i=8; i>=1; i--)
    {
        bool flag;
        printf("+---+---+---+---+---+---+---+---+\n");
        printf("|");
        if(i%2 == 0)
            flag = false;
        else
            flag = true;
        for(int j=0; j<8; j++)
        {
            if(!flag)
            {
                if(Map[i][j])
                    printf(".%c.|",Map[i][j]);
                else
                    printf("...|");
                flag = !flag;
            }
            else
            {
                if(Map[i][j])
                    printf(":%c:|",Map[i][j]);
                else
                    printf(":::|");
                flag = !flag;
            }
        }
        printf("\n");
    }
    printf("+---+---+---+---+---+---+---+---+\n");

    return 0;
}