Emag eht htiw Em Pleh
| Time Limit: 1000MS |
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Memory Limit: 65536K |
| Total Submissions: 2994 |
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Accepted: 1979 |
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
#include
#include
#include
using namespace std; int main() { char s1[260],s2[260],s[260][260]; while(gets(s1)!=NULL) { //getchar(); gets(s2); for(int i=0;i<17;) { for(int j=0;j<33;j++) if(j%4==0) s[i][j]='+'; else s[i][j]='-'; i+=2; } int kkk=0,num=0; for(int i=1;i<17;) { num++; kkk=3*num-1; for(int j=0;j<33;j++) if(j%4==0) s[i][j]='|'; else {kkk++; if((kkk/3)%2) s[i][j]='.'; else s[i][j]=':'; } i+=2; } for(int i=7;i
='a'&&s1[i]<='z') { if(s1[i-1]>='A'&&s1[i-1]<='Z') { int a=17-(int)((s1[i+1]-'0')*2); int b=(int)((s1[i]-'a')*4)+2; s[a][b]=s1[i-1]; } else { int a=17-(int)((s1[i+1]-'0')*2); int b=(int)((s1[i]-'a')*4)+2; s[a][b]='P'; } } } for(int i=7;i
='a'&&s2[i]<='z') { if(s2[i-1]>='A'&&s2[i-1]<='Z') { int a=17-(int)((s2[i+1]-'0')*2); int b=(int)((s2[i]-'a')*4)+2; s[a][b]=s2[i-1]-'A'+'a'; //格式转换 } elsewww.2cto.com { int a=17-(int)((s2[i+1]-'0')*2); int b=(int)((s2[i]-'a')*4)+2; s[a][b]='p'; } } } for(int i=0;i<17;i++) { for(int j=0;j<33;j++) { cout<
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