模拟 + 打表 --- Emag eht htiw Em Pleh

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。

 

#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;
}