Emag eht htiw Em Pleh

 Emag eht htiw Em Pleh

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
大家可以去poj 看看原题
这提题是逆序输出

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
using namespace std;
char s[10][10];
void print(int t)
{
    if(t) printf(".");
    else printf(":");
}
int main()
{
    int xi,yi,i,j,t;
    char str[10],ch,x,y;
    for(i=1; i<9; i++)
        for(j=1; j<9; j++)
        {
            if((i+j)%2)
                s[i][j]='.';
            else s[i][j]=':';
        }
    for(i=0; i<2; i++)
    {
        scanf("%s",str);
        if(strcmp(str,"White:")==0)
        {
            while(scanf("%c",&ch)&&ch!='
')
            {
                if('A'<=ch&&ch<='Z')
                {
                    scanf("%c%c",&x,&y);
                    xi=x-'a'+1;
                    yi=y-'0';
                    s[yi][xi]=ch;
                }
                else if('a'<=ch&&ch<'i')
                {
                    xi=ch-'a'+1;
                    scanf("%c",&y);
                    yi=y-'0';
                    s[yi][xi]='P';
                }
            }
        }
        else
        {
            while(scanf("%c",&ch)&&ch!='
')
            {
                if('A'<=ch&&ch<='Z')
                {
                    scanf("%c%c",&x,&y);
                    xi=x-'a'+1;
                    yi=y-'0';
                    s[yi][xi]=ch+32;
                }
                else if('a'<=ch&&ch<'i')
                {
                    xi=ch-'a'+1;
                    scanf("%c",&y);
                    yi=y-'0';
                    s[yi][xi]='p';
                }
            }
        }
    }
    for(i=8; i>0; i--)
    {
        printf("+---+---+---+---+---+---+---+---+
");
        for(j=1; j<9; j++)
        {
            printf("|");
            t=(i+j)%2;
            print(t);
            printf("%c",s[i][j]);
            print(t);
        }
        printf("|
");
    }
    printf("+---+---+---+---+---+---+---+---+
");
    return 0;
}