Emag
eht htiw Em Pleh
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 2578 | Accepted: 1731 |
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.|
+---+---+---+---+---+---+---+---+
【题目来源】
【题目大意】
这题和上一题是相关联的，上一题是给棋盘让你输出棋子的坐标，这题时给坐标让你输出棋盘。
【题目分析】
先将整个棋盘的初始状态打表存放起来，然后每次都初始化，输入坐标后更新数组的值，最后输出。
思路清晰就能1A。
#include<cstdio>
#include<cstring>
char Map[17][33];
char Graph[17][33]=
{
‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,
‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,
‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,
‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,
‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,
‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,
‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,
‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,
‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,
‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,
‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,
‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,
‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,
‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,
‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,
‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,‘:‘,‘:‘,‘:‘,‘|‘,‘.‘,‘.‘,‘.‘,‘|‘,
‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,‘-‘,‘-‘,‘-‘,‘+‘,
};void make_table()
{
for(int i=0;i<17;i++)
for(int j=0;j<33;j++)
Map[i][j]=Graph[i][j];
}char w[500];
char b[500];void update1(char c,char x1,char y1)
{
int x=(8-(y1-‘0‘)+1)*2-1;
int y=(x1-‘a‘+1)*4-2;
Map[x][y]=c;
}void update2(char c,char x1,char y1)
{
int x=(8-(y1-‘0‘)+1)*2-1;
int y=(x1-‘a‘+1)*4-2;
Map[x][y]=c+32;
}void update3(char x1,char y1)
{
int x=(8-(y1-‘0‘)+1)*2-1;
int y=(x1-‘a‘+1)*4-2;
Map[x][y]=‘P‘;
}void update4(char x1,char y1)
{
int x=(8-(y1-‘0‘)+1)*2-1;
int y=(x1-‘a‘+1)*4-2;
Map[x][y]=‘p‘;
}int main()
{
while(scanf("White: %s",w)!=EOF)
{
getchar();
scanf("Black: %s",b);
getchar();
make_table();
int len1=strlen(w);
int len2=strlen(b);
for(int i=0;i<len1;)
{
if(w[i]>=‘A‘&&w[i]<=‘Z‘)
{
update1(w[i],w[i+1],w[i+2]);
i+=4;
}
else if(w[i]>=‘a‘&&w[i]<=‘z‘)
{
update3(w[i],w[i+1]);
i+=3;
}
}
for(int i=0;i<len2;)
{
if(b[i]>=‘A‘&&b[i]<=‘Z‘)
{
update2(b[i],b[i+1],b[i+2]);
i+=4;
}
else if(b[i]>=‘a‘&&b[i]<=‘z‘)
{
update4(b[i],b[i+1]);
i+=3;
}
}
for(int i=0;i<17;i++)
{
for(int j=0;j<33;j++)
printf("%c",Map[i][j]);
puts("");
}
}
return 0;
}
模拟 + 打表 --- Emag eht htiw Em Pleh