POJ 2993 Emag eht htiw Em Pleh(模拟)

Emag eht htiw Em Pleh

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1695		Accepted: 1161

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,e4
Black: 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.|
+---+---+---+---+---+---+---+---+

Source

CTU Open 2005

解题报告：这是一道模拟题，以国际象棋为背景，给我们分别描述白棋和黑棋的位置，让我们在固定的格式中填充，我们首先把图表打出来（在没有白棋和黑棋的情况下），在根据所给的信息，填充即可，剩下的就是找规律了，用到的是我们高中所学的等差数列的知识推出所给信息和row（行），col（列）之间的关系；

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 110;
char str[N];
char map[17][34]={//打表
                  "+---+---+---+---+---+---+---+---+",
                  "|...|:::|...|:::|...|:::|...|:::|",
                  "+---+---+---+---+---+---+---+---+",
                  "|:::|...|:::|...|:::|...|:::|...|",
                  "+---+---+---+---+---+---+---+---+",
                  "|...|:::|...|:::|...|:::|...|:::|",
                  "+---+---+---+---+---+---+---+---+",
                  "|:::|...|:::|...|:::|...|:::|...|",
                  "+---+---+---+---+---+---+---+---+",
                  "|...|:::|...|:::|...|:::|...|:::|",
                  "+---+---+---+---+---+---+---+---+",
                  "|:::|...|:::|...|:::|...|:::|...|",
                  "+---+---+---+---+---+---+---+---+",
                  "|...|:::|...|:::|...|:::|...|:.:|",
                  "+---+---+---+---+---+---+---+---+",
                  "|:::|...|:::|...|:::|...|:::|...|",
                  "+---+---+---+---+---+---+---+---+"
    };
int Find1(char c)//确定所在行的函数
{
    int row = c - '0';
    return 17 - row *2;
}
int Find2(char c)//确定所在列的函数
{
    int col = c - 'a';
    return 4 * col + 2;
}
int main()
{
    int i, j, row, col;// row表示行，col表示列
    gets(str);//因为输入时有空格，所以用gets比较方便
    for (i = 7; str[i] >= 'A' && str[i] <= 'Z'; i += 4)
    {
        row = Find1(str[i + 2]);
        col = Find2(str[i + 1]);
        map[row][col] = str[i];

    }
    for (j = i; str[j] >= 'a' && str[j] <= 'z'; j +=3)
    {
        row = Find1(str[j + 1]);
        col = Find2(str[j]);
        map[row][col] = 'P';//白棋是大写字母
    }
    memset(str, 0,sizeof(str));
    gets(str);
    for (i = 7; str[i] >= 'A' && str[i] <= 'Z'; i += 4)
    {
        row = Find1(str[i + 2]);
        col = Find2(str[i + 1]);
        map[row][col] = str[i] + 32; //因为黑棋是小写字母，而str[i]中存的是大写字母
    }
    for (j = i; str[j] >= 'a' && str[j] <='z'; j += 3)
    {
        row = Find1(str[j + 1]);
        col = Find2(str[j]);
        map[row][col] = 'p';
    }
    for (i = 0 ; i < 17; ++i)
    {
        for (j = 0; j < 33; ++j)
        {
            printf("%c", map[i][j]);
        }
        printf("\n");
    }
    return 0;
}