码迷,mamicode.com
首页 > 其他好文 > 详细

模拟 + 打表 --- Emag eht htiw Em Pleh

时间:2014-05-26 06:21:53      阅读:297      评论:0      收藏:0      [点我收藏+]

标签:des   style   class   blog   c   code   

Emag eht htiw Em Pleh
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2578   Accepted: 1731

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

【题目来源】

CTU Open 2005

【题目大意】

这题和上一题是相关联的,上一题是给棋盘让你输出棋子的坐标,这题时给坐标让你输出棋盘。

【题目分析】

先将整个棋盘的初始状态打表存放起来,然后每次都初始化,输入坐标后更新数组的值,最后输出。

思路清晰就能1A。

 

bubuko.com,布布扣
#include<cstdio>
#include<cstring>
char Map[17][33];
char Graph[17][33]=
{
+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,
|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,
+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,
|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,
+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,
|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,
+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,
|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,
+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,
|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,
+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,
|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,
+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,
|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,
+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,
|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,:,:,:,|,.,.,.,|,
+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,-,-,-,+,
};

void make_table()
{
    for(int i=0;i<17;i++)
        for(int j=0;j<33;j++)
            Map[i][j]=Graph[i][j];
}

char w[500];
char b[500];

void update1(char c,char x1,char y1)
{
    int x=(8-(y1-0)+1)*2-1;
    int y=(x1-a+1)*4-2;
    Map[x][y]=c;
}

void update2(char c,char x1,char y1)
{
    int x=(8-(y1-0)+1)*2-1;
    int y=(x1-a+1)*4-2;
    Map[x][y]=c+32;
}

void update3(char x1,char y1)
{
    int x=(8-(y1-0)+1)*2-1;
    int y=(x1-a+1)*4-2;
    Map[x][y]=P;
}

void update4(char x1,char y1)
{
    int x=(8-(y1-0)+1)*2-1;
    int y=(x1-a+1)*4-2;
    Map[x][y]=p;
}


int main()
{
   while(scanf("White: %s",w)!=EOF)
    {
        getchar();
        scanf("Black: %s",b);
        getchar();
        make_table();
        int len1=strlen(w);
        int len2=strlen(b);
        for(int i=0;i<len1;)
        {
            if(w[i]>=A&&w[i]<=Z)
            {
                update1(w[i],w[i+1],w[i+2]);
                i+=4;
            }
            else if(w[i]>=a&&w[i]<=z)
            {
                update3(w[i],w[i+1]);
                i+=3;
            }
        }
        for(int i=0;i<len2;)
        {
            if(b[i]>=A&&b[i]<=Z)
            {
                update2(b[i],b[i+1],b[i+2]);
                i+=4;
            }
            else if(b[i]>=a&&b[i]<=z)
            {
                update4(b[i],b[i+1]);
                i+=3;
            }
        }
        for(int i=0;i<17;i++)
        {
          for(int j=0;j<33;j++)
             printf("%c",Map[i][j]);
          puts("");
        }
    }
    return 0;
}
bubuko.com,布布扣

 

 

 

 

 

模拟 + 打表 --- Emag eht htiw Em Pleh,布布扣,bubuko.com

模拟 + 打表 --- Emag eht htiw Em Pleh

标签:des   style   class   blog   c   code   

原文地址:http://www.cnblogs.com/acmer-jsb/p/3747979.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!