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[][];
 void print(int t)
 {
     if(t) printf(".");
     else printf(":");
 }
 int main()
 {
     int xi,yi,i,j,t;
     char str[],ch,x,y;
     for(i=; i<; i++)
         for(j=; j<; j++)
         {
             if((i+j)%)
                 s[i][j]='.';
             else s[i][j]=':';
         }
     for(i=; i<; i++)
     {
         scanf("%s",str);
         if(strcmp(str,"White:")==)
         {
             while(scanf("%c",&ch)&&ch!='\n')
             {
                 if('A'<=ch&&ch<='Z')
                 {
                     scanf("%c%c",&x,&y);
                     xi=x-'a'+;
                     yi=y-'';
                     s[yi][xi]=ch;
                 }
                 else if('a'<=ch&&ch<'i')
                 {
                     xi=ch-'a'+;
                     scanf("%c",&y);
                     yi=y-'';
                     s[yi][xi]='P';
                 }
             }
         }
         else
         {
             while(scanf("%c",&ch)&&ch!='\n')
             {
                 if('A'<=ch&&ch<='Z')
                 {
                     scanf("%c%c",&x,&y);
                     xi=x-'a'+;
                     yi=y-'';
                     s[yi][xi]=ch+;
                 }
                 else if('a'<=ch&&ch<'i')
                 {
                     xi=ch-'a'+;
                     scanf("%c",&y);
                     yi=y-'';
                     s[yi][xi]='p';
                 }
             }
         }
     }
     for(i=; i>; i--)
     {
         printf("+---+---+---+---+---+---+---+---+\n");
         for(j=; j<; j++)
         {
             printf("|");
             t=(i+j)%;
             print(t);
             printf("%c",s[i][j]);
             print(t);
         }
         printf("|\n");
     }
     printf("+---+---+---+---+---+---+---+---+\n");
     return ;
 }