| Time Limit: 1000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
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.
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.|
+---+---+---+---+---+---+---+---+
题意就是把2996的那道题的输出变输入，输入变输出。我觉得一下子变简单了很多，起码没那么多的麻烦事了。
思路就是现在是一个空棋盘，然后题目给了各个棋子的位置，拿这些棋子去占相应的位置，之后再输出占好各个位置的棋盘。
代码：
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std; const string temp="+---+---+---+---+---+---+---+---+";
string test[20];
string cal[5]; int i,len;
char row,col; int main()
{
test[1]="|...|:::|...|:::|...|:::|...|:::|";
test[2]="|:::|...|:::|...|:::|...|:::|...|";
test[3]="|...|:::|...|:::|...|:::|...|:::|";
test[4]="|:::|...|:::|...|:::|...|:::|...|";
test[5]="|...|:::|...|:::|...|:::|...|:::|";
test[6]="|:::|...|:::|...|:::|...|:::|...|";
test[7]="|...|:::|...|:::|...|:::|...|:::|";
test[8]="|:::|...|:::|...|:::|...|:::|...|"; cin>>cal[0];
cin>>cal[1];
cin>>cal[3];
cin>>cal[2]; len=cal[1].length();
for(i=0;i<len;i++)
{
if(cal[1][i]==',')continue;
else if(cal[1][i]=='K')
{
row=cal[1][i+1];
col=cal[1][i+2]; test[9-(col-'0')][(row-'a')*4+2]='K';
i=i+2;
}
else if(cal[1][i]=='Q')
{
row=cal[1][i+1];
col=cal[1][i+2]; test[9-(col-'0')][(row-'a')*4+2]='Q';
i=i+2;
}
else if(cal[1][i]=='R')
{
row=cal[1][i+1];
col=cal[1][i+2]; test[9-(col-'0')][(row-'a')*4+2]='R';
i=i+2;
}
else if(cal[1][i]=='B')
{
row=cal[1][i+1];
col=cal[1][i+2]; test[9-(col-'0')][(row-'a')*4+2]='B';
i=i+2;
}
else if(cal[1][i]=='N')
{
row=cal[1][i+1];
col=cal[1][i+2]; test[9-(col-'0')][(row-'a')*4+2]='N';
i=i+2;
}
else if(cal[1][i]>='a'&&cal[1][i]<='h')
{
row=cal[1][i];
col=cal[1][i+1]; test[9-(col-'0')][(row-'a')*4+2]='P';
i=i+1;
}
} len=cal[2].length();
for(i=0;i<len;i++)
{
if(cal[2][i]==',')continue;
else if(cal[2][i]=='K')
{
row=cal[2][i+1];
col=cal[2][i+2]; test[9-(col-'0')][(row-'a')*4+2]='k';
i=i+2;
}
else if(cal[2][i]=='Q')
{
row=cal[2][i+1];
col=cal[2][i+2]; test[9-(col-'0')][(row-'a')*4+2]='q';
i=i+2;
}
else if(cal[2][i]=='R')
{
row=cal[2][i+1];
col=cal[2][i+2]; test[9-(col-'0')][(row-'a')*4+2]='r';
i=i+2;
}
else if(cal[2][i]=='B')
{
row=cal[2][i+1];
col=cal[2][i+2]; test[9-(col-'0')][(row-'a')*4+2]='b';
i=i+2;
}
else if(cal[2][i]=='N')
{
row=cal[2][i+1];
col=cal[2][i+2]; test[9-(col-'0')][(row-'a')*4+2]='n';
i=i+2;
}
else if(cal[2][i]>='a'&&cal[2][i]<='h')
{
row=cal[2][i];
col=cal[2][i+1]; test[9-(col-'0')][(row-'a')*4+2]='p';
i=i+1;
}
} cout<<temp<<endl;
for(i=1;i<=8;i++)
{
cout<<test[i]<<endl;
cout<<temp<<endl;
} return 0;
}
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