POJ 2993：Emag eht htiw Em Pleh：模拟题

Emag eht htiw Em Pleh

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2555		Accepted: 1716

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

Source

CTU Open 2005

上一个模拟题的逆过程

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
char tem[35];
char tem1[35]="+---+---+---+---+---+---+---+---+";
char tem2[35];
int a[20][35];
int p;
void chuli()
{
	int i,j;
	int bianhao;
	for(i=0;i<strlen(tem);)
	{
		while(tem[i]==',')
			i++;
		if(tem[i]<91)
		{
			switch(tem[i])
			{
			case 'K':bianhao=1;break;
			case 'Q':bianhao=2;break;
			case 'R':bianhao=3;break;
			case 'B':bianhao=4;break;
			case 'N':bianhao=5;break;
			default:break;
			};
			a[9-(tem[i+2]-'0')][tem[i+1]-'a'+1]=p+bianhao;
			i+=3;
		}
		else
		{
			a[9-(tem[i+1]-'0')][tem[i]-'a'+1]=p+6;
			i+=2;
		}
	}
}
void fuzhi(int hang)
{
	int i,j;
	if(hang%2)
		strcpy(tem2,"|...|:::|...|:::|...|:::|...|:::|");
	else
		strcpy(tem2,"|:::|...|:::|...|:::|...|:::|...|");
	for(i=1;i<=8;i++)
	{
		switch(a[hang][i])
		{
		case 1:
			tem2[(i-1)*4+2]='K';
			break;
		case 2:
			tem2[(i-1)*4+2]='Q';
			break;
		case 3:
			tem2[(i-1)*4+2]='R';
			break;
		case 4:
			tem2[(i-1)*4+2]='B';
			break;
		case 5:
			tem2[(i-1)*4+2]='N';
			break;
		case 6:
			tem2[(i-1)*4+2]='P';
			break;
		case 7:
			tem2[(i-1)*4+2]='k';
			break;
		case 8:
			tem2[(i-1)*4+2]='q';
			break;
		case 9:
			tem2[(i-1)*4+2]='r';
			break;
		case 10:
			tem2[(i-1)*4+2]='b';
			break;
		case 11:
			tem2[(i-1)*4+2]='n';
			break;
		case 12:
			tem2[(i-1)*4+2]='p';
			break;
		default:break;
		};
	}
}
void output()
{
	int i,j;
	for(i=1;i<=8;i++)
	{
		printf("%s\n",tem1);
		fuzhi(i);
		printf("%s\n",tem2);
	}
	printf("%s\n",tem1);
}
int main()
{
	int i,j;
	memset(a,0,sizeof(a));
	scanf("%s",tem);
	scanf("%s",tem);
	p=0;
	chuli();
	scanf("%s",tem);
	scanf("%s",tem);
	p=6;
	chuli();
	output();
	return 0;
}