Emag eht htiw Em Pleh

Emag eht htiw Em Pleh

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

SubmitStatus Practice POJ 2993

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

根据字母的位置打印表格，注意是从下往上数（从1到8）

#include <stdio.h>#include <string.h>int main(){    char qp[18][34] =    {        "+---+---+---+---+---+---+---+---+",        "|...|:::|...|:::|...|:::|...|:::|",        "+---+---+---+---+---+---+---+---+",        "|:::|...|:::|...|:::|...|:::|...|",        "+---+---+---+---+---+---+---+---+",        "|...|:::|...|:::|...|:::|...|:::|",        "+---+---+---+---+---+---+---+---+",        "|:::|...|:::|...|:::|...|:::|...|",        "+---+---+---+---+---+---+---+---+",        "|...|:::|...|:::|...|:::|...|:::|",        "+---+---+---+---+---+---+---+---+",        "|:::|...|:::|...|:::|...|:::|...|",        "+---+---+---+---+---+---+---+---+",        "|...|:::|...|:::|...|:::|...|:::|",        "+---+---+---+---+---+---+---+---+",        "|:::|...|:::|...|:::|...|:::|...|",        "+---+---+---+---+---+---+---+---+",    };    char w[100],b[100],ch;    int lw,lb,i,j,k,ll,hh;    gets(w);    lw = strlen(w);    gets(b);    lb = strlen(b);    int bj = 0;    for(i = 7; i <= lw; i++)    {        if(bj == 0)            ch = 'P';        if(w[i] >= 'a' && w[i] <= 'z')        {            ll = 4*(w[i] - 'a' + 1) - 2;        }        else if(w[i] >= '1' && w[i] <= '9')        {            hh = (9-(w[i]-'0')) * 2 - 1;        }        else if(w[i] >= 'A' && w[i] <= 'Z')        {            ch = w[i];            bj = 1;        }        else if(w[i] == ',' || w[i] == '\0')        {            qp[hh][ll] = ch;            bj = 0;        }    }    bj = 0;    for(i = 7; i <= lb; i++)    {        if(bj == 0)            ch = 'p';        if(b[i] >= 'a' && b[i] <= 'z')        {            ll = 4*(b[i] - 'a' + 1) - 2;        }        else if(b[i] >= '1' && b[i] <= '9')        {            hh = (9-(b[i]-'0')) * 2 - 1;        }        else if(b[i] >= 'A' && b[i] <= 'Z')        {            ch = b[i] + 32;            bj = 1;        }        else if(b[i] == ',' || b[i] == '\0')        {            qp[hh][ll] = ch;            bj = 0;        }    }    for(i = 0; i<17; i++)    {        printf("%s\n",qp[i]);    }    return 0;}