POJ 2993 Emag eht htiw Em Pleh &&POJ 2996的反序

 

http://poj.org/problem?id=2993

学会了用strtok()函数分割字符串

Emag eht htiw Em Pleh

Time Limit: 1000MS		Memory Limit: 65536K

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

/* Author : yan * Question : Emag eht htiw Em Pleh * Data && Time : Thursday, December 30 2010 12:02 AM */ #include<stdio.h> char map[10][10]; char cache[16][4]; void solve_w(int cnt) { int i; for(i=0;i<cnt;i++) { if(cache[i][2]!='/0')//字符串长度为3 { map[8+'0'-cache[i][2]][cache[i][1]-'a']=cache[i][0]; } else map[8+'0'-cache[i][1]][cache[i][0]-'a']='P'; } } void solve_b(int cnt) { int i; for(i=0;i<cnt;i++) { if(cache[i][2]!='/0')//字符串长度为3 { map[7-(cache[i][2]-1-'0')][cache[i][1]-'a']=cache[i][0]+32; } else { map[7-(cache[i][1]-1-'0')][cache[i][0]-'a']='p'; } } } int main() { //freopen("input","r",stdin); int i,j; int cnt; char flag[10]; char input[100]; char step[]=","; char *token; for(i=0;i<8;i++) { for(j=0;j<8;j++) { if((i%2==0&&j%2!=0)||(i%2!=0&&j%2==0)) map[i][j]=':'; else map[i][j]='.'; } } scanf("%s",flag);//无用 scanf("%s",input); cnt=0; token=strtok(input,step); while(token) { strcpy(cache[cnt++],token); token=strtok(NULL,step); } if(flag[0]=='W') solve_w(cnt); else solve_b(cnt); ////////// scanf("%s",flag);//无用 scanf("%s",input); cnt=0; token=strtok(input,step); while(token) { strcpy(cache[cnt++],token); token=strtok(NULL,step); } if(flag[0]=='B') solve_b(cnt); else solve_w(cnt); for(i=0;i<8;i++) { printf("+---+---+---+---+---+---+---+---+/n"); printf("|"); for(j=0;j<8;j++) { if((i%2==0&&j%2!=0)||(i%2!=0&&j%2==0)) printf("%c%c%c%c",':',map[i][j],':','|'); else printf("%c%c%c%c",'.',map[i][j],'.','|'); } printf("/n"); } printf("+---+---+---+---+---+---+---+---+"); return 0; }