Decoding
Time Limit: 2 Seconds Memory Limit: 65536 KB
Chip and Dale have devised an encryption method to hide their (written) text messages. They first agree secretly on two numbers that will be used as the number of rows (R) and columns (C) in a matrix. The sender encodes an intermediate format using the following
rules:
1.
The text is formed with uppercase letters [A-Z] and <space>.
2.
Each text character will be represented by decimal values as follows:
<space> = 0, A = 1, B = 2, C = 3, ..., Y = 25, Z = 26 The sender enters the 5 digit binary representation of the characters‘ values in a spiral pattern along the matrix as shown below. The matrix is padded out with zeroes (0) to fill the matrix completely.
For example, if the text to encode is: "ACM" and R=4 and C=4, the matrix would be filled in as follows:
A = 00001, C = 00011, M = 01101(one extra 0) The bits in the matrix are then concatenated together in row major order and sent to the receiver. The example above would be encoded as: 0000110100101100
Input
The first line of input contains a single integer N, (1 <= N <= 1000) which is the number of datasets that follow.
Each dataset consists of a single line of input containing R (1 <= R <= 21), a space, C (1 <= C <= 21), a space, and a string of binary digits that represents the contents of the matrix (R * Cbinary digits).
The binary digits are in row major order.
Output
For each dataset, you should generate one line of output with the following values: The dataset number as a decimal integer (start counting at one), a space, and the decoded text message. You should throw away any trailing spaces and/or partial characters found
while decoding.
Sample Input
4
4 4 0000110100101100
5 2 0110000010
2 6 010000001001
5 5 0100001000011010110000010
Sample Output
1 ACM
2 HI
3 HI
4 HI HO
题意： 输入r c str。把str字符串 按行填入r*c。再照图中的顺序，转换出来，成一串字符。再把每五个二进制数转换为一个A-Z字母，连续五个0为空格。把字母或空格输出。但是尾部的空格要去除，并且如果没有字母全都是空格，那么也是不输出的。
#include<stdio.h>
#include<string>
#include<iostream>
#include<map>
#include<string.h>
using namespace std;
int mp[30][30];
char bol[1000];
int dir[4][2]={0,1,1,0,0,-1,-1,0};
int main()
{
int t;
string str;
int cas=1;
int r,c;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&r,&c);
cin>>str;
int pos=0;
for(int i=0;i<r;i++)
{
for(int j=0;j<c;j++)
{
mp[i][j]=str[pos++]-'0';
}
}
int x=0,y=0;
int xx,yy;
int ff=0;
int tem=0;
int bin=0;
int ling=0;
printf("%d ",cas++);
for(int i=0;i<r*c;i++)
{
tem*=2;
tem+=mp[x][y];
mp[x][y]=-1;
if((i+1)%5==0)
{
if(tem==0)
ling++;
else
{
for(int j=0;j<ling;j++)
{
printf(" ");
}
ling=0;
printf("%c",'A'+tem-1);
tem=0;
}
}
xx=x+dir[ff][0];
yy=y+dir[ff][1];
if(xx>=0&&xx<r&&yy>=0&&yy<c&&mp[xx][yy]!=-1)
{
int a=1;
}
else
{
ff=(ff+1)%4;
xx=x+dir[ff][0];
yy=y+dir[ff][1];
}
x=xx;
y=yy;
}
puts("");
}
return 0;
}