HDU 1987 Decoding (模拟)

Decoding

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 810    Accepted Submission(s): 253

[align=left]Problem Description[/align]
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:



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

[align=left]Input[/align]
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<=20), a space, C (1<=C<=20), a space, and a string of binary digits that represents the contents of the matrix (R * C binary digits).

The binary digits are in row major order.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]

4
4 4 0000110100101100
5 2 0110000010
2 6 010000001001
5 5 0100001000011010110000010

[align=left]Sample Output[/align]

1 ACM
2 HI
3 HI
4 HI HO

[align=left]Source[/align]
2008
“Shun Yu Cup” Zhejiang Collegiate Programming Contest - Warm Up（1）

[align=left]Recommend[/align]
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题解：好麻烦的模拟...

AC代码：

#include<bits/stdc++.h>
using namespace std;

char a[28]={" ABCDEFGHIJKLMNOPQRSTUVWXYZ"};

int convert(char *s)
{
int sum=0;
for(int i=0; i<5; i++)
{
sum *=2;
sum += s[i]-'0';
}
return sum;
}

void convert_again(char *sss,int n,int m)
{
char temp[1000]="";
char space[2]=" ";
char s[6];
s[5] = '\0';
char ss[22][22];
char *p=sss;
for(int i=0; i<n; i++)
{
strncpy(ss[i],p,m);
ss[i][m]='\0';
p +=m;
}

int b1,b2,t,tt; //b1低行，b2低列，t,高行，tt 高列；
b1 = b2 = 0;
t = n;
tt = m;
int k=0;
int g=0;
int tst=0;
int begin=1;
while(1)
{
if(b1>=t||b2>=tt) break;
for(g=b2; g<tt; g++)
{
s[k++] = ss[b1][g];
if(k == 5)
{
if(a[convert(s)]!=' ')
{
printf("%s",temp);
strcpy(temp,"");
printf("%c",a[convert(s)]);
tst+=5;
begin=0;
}
else
{
if(!begin)
{
strcat(temp,space);
}
}
k=0;
}
}
b1++;
if(b1>=t||b2>=tt) break;
for(g=b1; g<t; g++)
{
s[k++] = ss[g][tt-1];
if(k == 5)
{
if(a[convert(s)]!=' ')
{
printf("%s",temp);
strcpy(temp,"");
printf("%c",a[convert(s)]);
tst+=5;
begin=0;
}
else
{
if(!begin)
{
strcat(temp,space);
}
}
k=0;
}
}
tt--;
if(b1>=t||b2>=tt) break;
for(g=tt-1; g>=b2; g--)
{
s[k++] = ss[t-1][g];
if(k == 5)
{
if(a[convert(s)]!=' ')
{
printf("%s",temp);
strcpy(temp,"");
printf("%c",a[convert(s)]);
tst+=5;
begin=0;
}
else
{
if(!begin)
{
strcat(temp,space);
}
}
k=0;
}
}
t--;
if(b1>=t||b2>=tt) break;
for(g=t-1; g>=b1; g--)
{
s[k++] = ss[g][b2];
if(k == 5)
{
if(a[convert(s)]!=' ')
{
printf("%s",temp);
strcpy(temp,"");
printf("%c",a[convert(s)]);
tst+=5;
begin=0;
}
else
{
if(!begin)
{
strcat(temp,space);
}
}
k=0;
}
}
b2++;
if(b1>=t||b2>=tt) break;
}
}

int main()
{
int t,r,c;
scanf("%d",&t);
int cas=1;
while(t--)
{

char s[410];
scanf("%d%d%s",&r,&c,s);
printf("%d ",cas++);
convert_again(s,r,c);
puts("");
}
return 0;
}