poj2993 翻转2996

Emag eht htiw Em Pleh

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

#include<stdio.h>
#include<string.h>
char map[10][10],str1[100],str2[100];
int main(){
while(gets(str1)){
gets(str2);
int len1=strlen(str1);
int len2=strlen(str2);
for(int i=0;i<=len1-1;i++){
if(str1[i]>=‘1‘&&str1[i]<=‘9‘){
if(str1[i-2]!=‘,‘)
map[str1[i]-‘1‘+1][str1[i-1]-‘a‘+1]=str1[i-2];
else
map[str1[i]-‘1‘+1][str1[i-1]-‘a‘+1]=‘P‘;
}
}
for(int i=0;i<=len2-1;i++){
if(str2[i]>=‘1‘&&str2[i]<=‘9‘){
if(str2[i-2]!=‘,‘)
map[str2[i]-‘1‘+1][str2[i-1]-‘a‘+1]=str2[i-2]+32;
else
map[str2[i]-‘1‘+1][str2[i-1]-‘a‘+1]=‘p‘;
}
}

printf("+---+---+---+---+---+---+---+---+\n");
for(int i=8;i>=1;i--){
printf("|");
for(int j=1;j<=8;j++){
if((i+j)%2==0){
printf(":");
if(map[i][j]) printf("%c",map[i][j]);
else printf(":");
printf(":");
}
else {
printf(".");
if(map[i][j]) printf("%c",map[i][j]);
else printf(".");
printf(".");
}
printf("|");
}
printf("\n");
printf("+---+---+---+---+---+---+---+---+\n");
}
memset(str1,0,sizeof(str1));
memset(str2,0,sizeof(str2));
memset(map,0,sizeof(map));
}
return 0;
}