poj2993模拟

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

Source

CTU Open 2005

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int main()
{
	char s1[260],s2[260],s[260][260];
	while(gets(s1)!=NULL)
	{   //getchar();
		gets(s2);
		for(int i=0;i<17;)
		{
			for(int j=0;j<33;j++)
				if(j%4==0)
				s[i][j]='+';
			else
				s[i][j]='-';
			i+=2;
		}
		int kkk=0,num=0;
		for(int i=1;i<17;)
		{
			   num++;
			   kkk=3*num-1;
			for(int j=0;j<33;j++)
				if(j%4==0)
				s[i][j]='|';
			   else
			   {kkk++;
				if((kkk/3)%2)
				s[i][j]='.';
				else
				s[i][j]=':';
			   }
			i+=2;
		}

		for(int i=7;i<strlen(s1);i++)
		{
			if(s1[i]>='a'&&s1[i]<='z')
			{
				if(s1[i-1]>='A'&&s1[i-1]<='Z')
				{
					int a=17-(int)((s1[i+1]-'0')*2);
					int b=(int)((s1[i]-'a')*4)+2;
					s[a][b]=s1[i-1];
				}
				else
				{
					int a=17-(int)((s1[i+1]-'0')*2);
					int b=(int)((s1[i]-'a')*4)+2;
					s[a][b]='P';
				}
			}
		}

		for(int i=7;i<strlen(s2);i++)
		{
			if(s2[i]>='a'&&s2[i]<='z')
			{
				if(s2[i-1]>='A'&&s2[i-1]<='Z')
				{
					int a=17-(int)((s2[i+1]-'0')*2);
					int b=(int)((s2[i]-'a')*4)+2;
					s[a][b]=s2[i-1]-'A'+'a';         //格式转换
				}
				else
				{
					int a=17-(int)((s2[i+1]-'0')*2);
					int b=(int)((s2[i]-'a')*4)+2;
					s[a][b]='p';
				}
			}
		}
		for(int i=0;i<17;i++)
		{
			for(int j=0;j<33;j++)
			{
				cout<<s[i][j];
			}
			cout<<endl;
		}
	}
}

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