poj 2993 Emag eht htiw Em Pleh（模拟）_this problem is a reverse case of the problem 2996-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_41156122/article/details/80034812

本文介绍了一种根据已知棋盘输出进行逆向解析输入的算法实现。该算法通过读取黑白双方的棋子位置信息，转换为棋盘布局形式输出。文中详细展示了如何使用C++语言进行具体实现。

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Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4063		Accepted: 2601

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.|

+---+---+---+---+---+---+---+---+

就是把2996那道题倒着来一遍；

但是一直wa，过不了w过不了OTZ

希望有dalao指出哪里有bug；

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn=40;
char mp[maxn][maxn];
char str1[40];
char str2[40];
int main()
{
    for(int i=0;i<8;i++)
    {
        for(int j=0;j<8;j++)
        {
            mp[2*i][4*j]='+';
            mp[2*i+1][4*j]='|';
            mp[2*i][4*j+1]='-';
            mp[2*i][4*j+2]='-';
            mp[2*i][4*j+3]='-';
        }
        mp[2*i][32]='+';
        mp[2*i+1][32]='|';
        mp[2*i][33]='\0';
    if(i==7)
    {
        for(int j=0;j<8;j++)
        {
            mp[2*i+2][4*j]='+';
            mp[2*i+2][4*j+1]='-';
            mp[2*i+2][4*j+2]='-';
            mp[2*i+2][4*j+3]='-';
        }
        mp[2*i+2][32]='+';
        mp[2*i+2][33]='\0';
    }
    for(int j=0;j<8;j++)
    {
        if((i+j)%2==0)
        {
            mp[2*i+1][4*j+1]='.';
            mp[2*i+1][4*j+2]='.';
            mp[2*i+1][4*j+3]='.';
        }
        else
        {
            mp[2*i+1][4*j+1]=':';
            mp[2*i+1][4*j+2]=':';
            mp[2*i+1][4*j+3]=':';
        }
     }
     mp[2*i+1][33]='\0';
    }
    scanf("%s%s",&str1,&str2);
    if(strcmp(str1,"White:")==0)
    {
        for(int i=0;str2[i]!=0;)
        {
            if(str2[i]==','){i++;continue;}
            if(str2[i]>='A'&&str2[i]<='Z')
            {
                int x=8-(str2[i+2]-'0');
                int y=str2[i+1]-'a';
                mp[2*x+1][4*y+2]=str2[i];
                i+=3;
            }
            else
            {
                int x=8-(str2[i+1]-'0');
                int y=str2[i]-'a';
                mp[2*x+1][4*y+2]='P';
                i+=2;
            }
        }
    }
    else
    {
        for(int i=0;str2[i]!=0;)
        {
            if(str2[i]==','){i++;continue;}
            if(str2[i]>='A'&&str2[i]<='Z')
            {
                int x=8-(str2[i+2]-'0');
                int y=str2[i+1]-'a';
                mp[2*x+1][4*y+2]=str2[i]-'A'+'a';
                i+=3;
            }
            else
            {
                int x=8-(str2[i+1]-'0');
                int y=str2[i]-'a';
                mp[2*x+1][4*y+2]='p';
                i+=2;
            }
        }

    }
  scanf("%s%s",&str1,&str2);
    if(strcmp(str1,"White:")==0)
    {
        for(int i=0;str2[i]!=0;)
        {
            if(str2[i]==','){i++;continue;}
            if(str2[i]>='A'&&str2[i]<='Z')
            {
                int x=8-(str2[i+2]-'0');
                int y=str2[i+1]-'a';
                mp[2*x+1][4*y+2]=str2[i];
                i+=3;
            }
            else
            {
                int x=8-(str2[i+1]-'0');
                int y=str2[i]-'a';
                mp[2*x+1][4*y+2]='P';
                i+=2;
            }
        }
    }
    else
    {
        for(int i=0;str2[i]!=0;)
        {
            if(str2[i]==','){i++;continue;}
            if(str2[i]>='A'&&str2[i]<='Z')
            {
                int x=8-(str2[i+2]-'0');
                int y=str2[i+1]-'a';
                mp[2*x+1][4*y+2]=str2[i]-'A'+'a';
                i+=3;
            }
            else
            {
                int x=8-(str2[i+1]-'0');
                int y=str2[i]-'a';
                mp[2*x+1][4*y+2]='p';
                i+=2;
            }
        }

    }
    for(int i=0;i<17;i++)
        printf("%s\n",mp[i]);
    return 0;
}