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,e4Black: 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.|+---+---+---+---+---+---+---+---+

水题，模拟一下即可，代码如下

#include<stdio.h>#include<string.h>int main(){char map[][50]={"+---+---+---+---+---+---+---+---+", "|...|:::|...|:::|...|:::|...|:::|", "+---+---+---+---+---+---+---+---+", "|:::|...|:::|...|:::|...|:::|...|", "+---+---+---+---+---+---+---+---+", "|...|:::|...|:::|...|:::|...|:::|", "+---+---+---+---+---+---+---+---+", "|:::|...|:::|...|:::|...|:::|...|", "+---+---+---+---+---+---+---+---+", "|...|:::|...|:::|...|:::|...|:::|", "+---+---+---+---+---+---+---+---+", "|:::|...|:::|...|:::|...|:::|...|", "+---+---+---+---+---+---+---+---+", "|...|:::|...|:::|...|:::|...|:::|", "+---+---+---+---+---+---+---+---+", "|:::|...|:::|...|:::|...|:::|...|", "+---+---+---+---+---+---+---+---+"};char a[100];gets(a);for(int i=0;a[i]!='\0';i++){if(a[i]==' '||a[i]==','){if(a[i+1]=='a'||a[i+1]=='b'||a[i+1]=='c'||a[i+1]=='d'||a[i+1]=='e'||a[i+1]=='f'||a[i+1]=='g'||a[i+1]=='h')map[(8-(a[i+2]-'0'))*2+1][4*(a[i+1]-'a'+1)-1-1]='P';elsemap[(8-(a[i+3]-'0'))*2+1][4*(a[i+2]-'a'+1)-1-1]=a[i+1]; }}gets(a);for(int i=0;a[i]!='\0';i++){if(a[i]==' '||a[i]==','){if(a[i+1]=='a'||a[i+1]=='b'||a[i+1]=='c'||a[i+1]=='d'||a[i+1]=='e'||a[i+1]=='f'||a[i+1]=='g'||a[i+1]=='h')map[(8-(a[i+2]-'0'))*2+1][4*(a[i+1]-'a'+1)-1-1]='p';elsemap[(8-(a[i+3]-'0'))*2+1][4*(a[i+2]-'a'+1)-1-1]=a[i+1]+32; }}for(int i=0;i<17;i++)puts(map[i]);return 0;}