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Emag eht htiw Em Pleh

Description

This problem is a reverse case of the . You are given the output of the problem H and your task is to find the corresponding input.

Input

according to output of .

Output

according to input of .

Sample Input

White: Ke1,Qd1,Ra1,Rh1,Bc1,Bf1,Nb1,a2,c2,d2,f2,g2,h2,a3,e4Black: 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

代码：

//poj2993 poj2996 的逆向题目 #include 
   
    #include 
    
     using namespace std;char c[17][33];bool first(int k){	if(k == 1 || k == 5 || k == 9 || k == 13)		return true;	return false;}void init(int myRow, int myCol){	for(int i = 0; i < myRow; i++)	{		for(int j = 0; j < myCol; j++)		{			if(i % 2 == 0)			{				for(int k = 0; k < myCol; k++)				{					if(k % 4 == 0)						c[i][k] = '+';					else 						c[i][k] = '-';				}				break;			}			else			{				if(first(i)) //白黑 				{					for(int k = 0; k < myCol; k++)					{						if(k % 4 == 0)							c[i][k] = '|';						else						{							if(k % 8 < 4)							{								c[i][k] = '.';							}							else							{								c[i][k] = ':';							}						}					}					break;				}				else //黑白 				{					for(int k = 0; k < myCol; k++)					{						if(k % 4 == 0)							c[i][k] = '|';						else						{							if(k % 8 > 4)							{								c[i][k] = '.';							}							else							{								c[i][k] = ':';							}						}					}					break;				}			}		}	}}bool role(char cc){	if(cc == 'K' || cc == 'Q' || cc == 'R' || cc == 'B' || cc == 'N')		return true;	return false;}void deal(int myRow, int myCol){	string b;	getline(cin, b); //白 	string a;	getline(cin, a); //黑 	int k = 0;	while(b[k++] != 'K');	int m = k - 1;	while(m < b.length())	{		if(role(b[m]))		{			int col = (b[m + 1] - 'a') * 4 + 2;			int row = 15 - (int(b[m + 2] - '0') - 1 ) * 2; 			switch(b[m])			{					case 'K':					c[row][col] = 'K';				break;			case 'Q':				c[row][col] = 'Q'; 				break;			case 'R':				c[row][col] = 'R'; 				break;			case 'B':				c[row][col] = 'B'; 				break;			case 'N':				c[row][col] = 'N'; 				break;			}			m += 4;		}		else		{			int col = col = (b[m] - 'a') * 4 + 2;			int row = 15 - (int(b[m + 1] - '0') - 1 ) * 2; 			c[row][col] = 'P';			m += 3;			}	}	//	getline(cin, b);	k = 0;	while(a[k++] != 'K');	m  = k - 1;	while(m < a.length())	{		if(role(a[m]))		{			int col = (a[m + 1] - 'a') * 4 + 2;			int row = 15 - (int(a[m + 2] - '0') - 1 ) * 2; 			switch(a[m])			{					case 'K':					c[row][col] = 'k';				break;			case 'Q':				c[row][col] = 'q'; 				break;			case 'R':				c[row][col] = 'r'; 				break;			case 'B':				c[row][col] = 'b'; 				break;			case 'N':				c[row][col] = 'n'; 				break;			}			m += 4;		}		else		{			int col = col = (a[m] - 'a') * 4 + 2;			int row = 15 - (int(a[m + 1] - '0') - 1 ) * 2; 			c[row][col] = 'p';			m += 3;			}	}}void print(int myRow, int myCol){	for(int i = 0; i < myRow; i++)	{		for(int j = 0; j < myCol; j++)		{			cout<
    
   

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