Emag eht htiw Em Pleh 分类： POJ 2015-06-29 18:54 10人阅读 评论(0) 收藏

Emag eht htiw Em Pleh

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2937		Accepted: 1944

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.|
+---+---+---+---+---+---+---+---+
还是模拟题
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;

const int Max=1100000;

char str[]="+---+---+---+---+---+---+---+---+";
char s1[]="|:::|...|:::|...|:::|...|:::|...|";
char s2[]="|...|:::|...|:::|...|:::|...|:::|";
char Map[55][55];
void Trans(char s[])
{
    bool flag=false;
    if(s[0]=='B')
    {
        flag=true;
    }
    for(int i=7; s[i];)
    {
        if(s[i]>='A'&&s[i]<='Z')
        {
            int u=s[i+1]-'a'+1;
            int v=s[i+2]-'0';
            if(flag)
            {
                Map[v][u]=s[i]+32;
            }
            else
            {
                Map[v][u]=s[i];
            }
            i+=4;
        }
        else if(s[i]>='a'&&s[i]<='z')
        {
            int u=s[i]-'a'+1;
            int v=s[i+1]-'0';
            if(flag)
            {
                Map[v][u]='p';
            }
            else
            {
                Map[v][u]='P';
            }
            i+=3;
        }
    }
}
int main()
{
    char white[110];
    char black[110];
    memset(white,'\0',sizeof(white));
    memset(black,'\0',sizeof(black));
    gets(white);
    gets(black);
    memset(Map,0,sizeof(Map));
    Trans(white);
    Trans(black);
    int u=8;
    for(int i=17; i>=1; i--)
    {
        if(i%2)
        {
            printf("%s\n",str);
        }
        else
        {
            int ans=2;
            int v=1;
            if((i/2)%2)
            {
                for(int j=0; j<33; j++)
                {
                    if(j!=ans)
                    {
                        printf("%c",s1[j]);
                    }
                    else
                    {
                        if(Map[u][v])
                        {
                            printf("%c",Map[u][v]);
                        }
                        else
                        {
                            printf("%c",s1[j]);
                        }
                        v++;
                        ans+=4;
                    }
                }
            }
            else
            {
                for(int j=0; j<33; j++)
                {
                    if(j!=ans)
                    {
                        printf("%c",s2[j]);
                    }
                    else
                    {
                        if(Map[u][v])
                        {
                            printf("%c",Map[u][v]);
                        }
                        else
                        {
                            printf("%c",s2[j]);
                        }
                        v++;
                        ans+=4;
                    }
                }
            }
            printf("\n");
            u--;
        }
    }
    return 0;
}

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