. Our encrypted message is composed of eight basic characters, respectively is A B C D # 1 2 3, for long-distance transmission, now need to transfer encoding of the eight characters, transfer encoding with binary encoding (long-distance transmission, digital code anti-interference ability strong), the binary coding for three long code (all characters coding length), coding rules are as follows:
A: 000 B: 001 C, 010 D, 011 1:100 2:10 3:11 0 # 1:111
Now enter a line (up to 10) to encrypt the message code, please translate it for the corresponding transfer encoding and output,
For example: input ABC#, output 000001010111,
Tip: encoding rules can be stored in an array, become the code table,
Format remind:
Input: input from A B C D 1 2 3 # of A line of characters, # can only appear in the end, after the carriage returns,
Output: a line of binary code, the last carriage returns, pay attention to the 0 s and 1 s of binary encoding is character, not Numbers,

CodePudding user response:

The teacher the assignment.

 
//A: 000 B: 001 C, 010 D, 011 1:100 2:10 3:11 0 # 1:111 
#include  
#include  

Int main (int arg c, char * argv []) 
{
Char mess [12]. 

The scanf (" % s ", mess); 

for (int i=0; i {
The switch (mess [I]) 
{
Case 'A' : 
Printf (" 000 "); 
break; 

Case 'B' : 
Printf (" 001 "); 
break; 

Case 'C' : 
Printf (" 010 "); 

Case 'D' : 
Printf (" 011 "); 
break; 

Case '1' : 
Printf (" 100 "); 
break; 

Case '2' : 
Printf (" 101 "); 
break; 

Case '3' : 
Printf (" 110 "); 
break; 

Case '#' : 
Printf (" 111 "); 
return 0; 

Default: 
break; 
} 
} 

return 0; 
} 

CodePudding user response:

When # under a new line

CodePudding user response:

Reference:

 # include & lt; stdio.h> 
#include  

Int main () 
{//A B C D 1 2 3 # 
Char [s] [4]={" 000 ", "001", "010", "011", "100", "101", "110", "111"}; 
Char STR, PTR [64]; 

Do {
The scanf (" % c ", & amp; STR); 
The switch (STR) {
Case 'A' : strcat (PTR, s [0]). 
break; 
Case 'B' : strcat (PTR, s [1]). 
break; 
Case 'C' : strcat (PTR, s [2]). 
break; 
Case 'D' : strcat (PTR, s [3]). 
break; 
Case '1' : strcat (PTR, s [4]); 
break; 
Case '2' : strcat (PTR, s [5]); 
break; 
Case: '3' strcat (PTR, s [6]); 
break; 
Case '#' : strcat (PTR, s [7]); 
break; 
Default: break; 
} 
} while (STR!='#'); 
Printf (" % s \ n ", PTR); 

return 0; 
} 

CodePudding user response:

reference 1/f, Kira Skyler response:

teacher assigned homework.

 
//A: 000 B: 001 C, 010 D, 011 1:100 2:10 3:11 0 # 1:111 
#include  
#include  

Int main (int arg c, char * argv []) 
{
Char mess [12]. 

The scanf (" % s ", mess); 

for (int i=0; i {
The switch (mess [I]) 
{
Case 'A' : 
Printf (" 000 "); 
break; 

Case 'B' : 
Printf (" 001 "); 
break; 

Case 'C' : 
Printf (" 010 "); 

Case 'D' : 
Printf (" 011 "); 
break; 

Case '1' : 
Printf (" 100 "); 
break; 

Case '2' : 
Printf (" 101 "); 
break; 

Case '3' : 
Printf (" 110 "); 
break; 

Case '#' : 
Printf (" 111 "); 
return 0; 

Default: 
break; 
} 
} 

return 0; 
} 

In c + + how to write, we don't learn c language

CodePudding user response:

The

reference 3 floor QZJHJXJ response:

for reference:

 # include & lt; stdio.h> 
#include  

Int main () 
{//A B C D 1 2 3 # 
Char [s] [4]={" 000 ", "001", "010", "011", "100", "101", "110", "111"}; 
Char STR, PTR [64]; 

Do {
The scanf (" % c ", & amp; STR); 
The switch (STR) {
Case 'A' : strcat (PTR, s [0]). 
break; 
Case 'B' : strcat (PTR, s [1]). 
break; 
Case 'C' : strcat (PTR, s [2]). 
break; 
Case 'D' : strcat (PTR, s [3]). 
break; 
Case '1' : strcat (PTR, s [4]); 
break; 
Case '2' : strcat (PTR, s [5]); 
break; 
Case: '3' strcat (PTR, s [6]); 
break; 
Case '#' : strcat (PTR, s [7]); 
break; 
Default: break; 
} 
} while (STR!='#'); 
Printf (" % s \ n ", PTR); 

return 0; 
} 

In c + + how to write, we don't learn c language

CodePudding user response:

Header file modification: # include & lt; Iostream. H>
#include
using namespace std;

Then type: cin> STR;//the scanf (" % c ", & amp; STR);
Output: coutOther are all the same, need not be changed,

CodePudding user response:

reference 5 floor _ClutchFactor reply:

Quote: refer to the third floor QZJHJXJ response:

for reference:

 # include & lt; stdio.h> 
#include  

Int main () 
{//A B C D 1 2 3 # 
Char [s] [4]={" 000 ", "001", "010", "011", "100", "101", "110", "111"}; 
Char STR, PTR [64]; 

Do {
The scanf (" % c ", & amp; STR); 
The switch (STR) {
Case 'A' : strcat (PTR, s [0]). 
break; 
Case 'B' : strcat (PTR, s [1]). 
break; 
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