Emag eht htiw Em Pleh

Emag eht htiw Em Pleh

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3314		Accepted: 2176

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 <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
int flag[10][10];
char s[100];
void in(int t)
{
scanf("%s",s) ;
scanf("%s",s);
int l=strlen(s);
int i;
for(i=0;i<l;++i)
{
if(isupper(s[i]))
{
int y=s[i+1]-'a';
int x=s[i+2]-'0';
if(t==1)
flag[x][y]=s[i];
else flag[x][y]=tolower(s[i]);
i+=3;
continue;
}
else
if(islower(s[i]))
{
int y=s[i]-'a';
int x=s[i+1]-'0';
if(t==1)
flag[x][y]='P';
else flag[x][y]='p';
i+=2;
continue;
}
}
}
void printl()
{
printf("+---+---+---+---+---+---+---+---+\n");
}
void print()
{
int i,j;
for(i=8;i>=1;--i)
{
printl();
for(j=0;j<8;++j)
{
if((j+i)%2==0)
{
if(flag[i][j])
printf("|.%c.",flag[i][j]);
else printf("|...");
}
else
{
if(flag[i][j])
printf("|:%c:",flag[i][j]);
else printf("|:::");
}

}
printf("|\n");
}
printl();
}
int main()
{
in(1);
in(2);
print();
return 0;
}