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

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<cstdlib>
 5 using namespace std;
 6 char s[10][10];
 7 void print(int t)
 8 {
 9     if(t) printf(".");
10     else printf(":");
11 }
12 int main()
13 {
14     int xi,yi,i,j,t;
15     char str[10],ch,x,y;
16     for(i=1; i<9; i++)
17         for(j=1; j<9; j++)
18         {
19             if((i+j)%2)
20                 s[i][j]='.';
21             else s[i][j]=':';
22         }
23     for(i=0; i<2; i++)
24     {
25         scanf("%s",str);
26         if(strcmp(str,"White:")==0)
27         {
28             while(scanf("%c",&ch)&&ch!='
')
29             {
30                 if('A'<=ch&&ch<='Z')
31                 {
32                     scanf("%c%c",&x,&y);
33                     xi=x-'a'+1;
34                     yi=y-'0';
35                     s[yi][xi]=ch;
36                 }
37                 else if('a'<=ch&&ch<'i')
38                 {
39                     xi=ch-'a'+1;
40                     scanf("%c",&y);
41                     yi=y-'0';
42                     s[yi][xi]='P';
43                 }
44             }
45         }
46         else
47         {
48             while(scanf("%c",&ch)&&ch!='
')
49             {
50                 if('A'<=ch&&ch<='Z')
51                 {
52                     scanf("%c%c",&x,&y);
53                     xi=x-'a'+1;
54                     yi=y-'0';
55                     s[yi][xi]=ch+32;
56                 }
57                 else if('a'<=ch&&ch<'i')
58                 {
59                     xi=ch-'a'+1;
60                     scanf("%c",&y);
61                     yi=y-'0';
62                     s[yi][xi]='p';
63                 }
64             }
65         }
66     }
67     for(i=8; i>0; i--)
68     {
69         printf("+---+---+---+---+---+---+---+---+
");
70         for(j=1; j<9; j++)
71         {
72             printf("|");
73             t=(i+j)%2;
74             print(t);
75             printf("%c",s[i][j]);
76             print(t);
77         }
78         printf("|
");
79     }
80     printf("+---+---+---+---+---+---+---+---+
");
81     return 0;
82 }

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