Hexadecimal conversion into octal-CodePudding

I want to ask, why the devc++ operation no problem, but a uploaded to the blue cup practice system is displayed "error", thank ^_^

#include
11 # define N
# define M 100001
using namespace std;
Int a [N].
Int b [M];
Int main ()
{int n;
Cin> n;
for(int i=0; ICin> Hex> B [I];
for(int i=0; I{[I] a [I]=b;
Coutreturn 0;
}

CodePudding user response:

Can provide a little like data range, the result processing requirements, sample input and output of what? So hang any information directly, we don't even know what your problem is

CodePudding user response:

reference 1st floor User_Ghost response:

can provide a little like data range, the processing requirements, sample input and output of what? So hang any information directly, we don't know where your problem is

Resource limitations
Time limit: 1.0 s memory limit: 512.0 MB
Problem description
Given n hexadecimal positive integer, output their corresponding octal number,

Input format
Enter the first act of a positive integer n (1 & lt;=n<=10),
The next n lines, each line by A 0 ~ 9, uppercase string composed of A ~ F, said to convert hexadecimal positive integers, each hexadecimal number length is not more than 100000,

The output format
Output n lines, each behavior input corresponding octal positive integer,

[note]
Input hexadecimal number don't have a leading zero, such as 012 a,
Output octal number cannot have a leading 0,

The sample input
2
39
ABC 123

Sample output
71
4435274

[tip]
First to convert hexadecimal number to a hexadecimal number, again by a hexadecimal number into octal,
Sorry

CodePudding user response:

Each hexadecimal number length can't more than 1000000 how to understand, is the numerical size

CodePudding user response:

 
/* * 
C 
 * @ file hex2oct.* @ brief, 
*/

# include & lt; Assert. H> 
# include & lt; Ctype. H> 
# include & lt; stdio.h> 
# include & lt; Stdlib. H> 
# include & lt; String. H> 

# define M 100000 

The static 
Char * 
Chop (char * s) 
{
int n; 

N=strlen (s); 
If (n & gt; 0 & amp; & (=='\ r' s [n - 1) | |=='\ n' s/n - 1)) 
S [] - n='\ 0'; 
If (n & gt; 0 & amp; & (=='\ r' s [n - 1) | |=='\ n' s/n - 1)) 
S [] - n='\ 0'; 
return s; 
} 


The static 
Void 
Hex2bin (char * hex, int len, char * bin) 
{
int i; 
Int b0=1=2 b1, b2=4, b3=8; 

For (I=0; I & lt; Len. I++) {
Bin [I * 4 + 0]=(b3 & amp; Hex [I])? 1:0; 
Bin [I * 4 + 1)=(b2 & amp; Hex [I])? 1:0; 
Bin [I * 4 + 2]=(b1 & amp; Hex [I])? 1:0; 
Bin [I * 4 + 3]=(b0 & amp; Hex [I])? 1:0; 
} 
} 


The static 
Void 
Bin2oct (char * bin, int len) 
{
Int I, n; 
Char * b; 

For (I=0; I & lt; Len - 1; I++) {
If (bin [I]) 
break; 
} 
B=bin + I; 
N=len - I; 

The switch (n % 3) {
Case 1: 
B -=2; 
N +=2; 
break; 
Case 2: 
B -=1; 
N +=1; 
break; 
} 

For (I=0; I & lt; n; I +=3) {
Printf (" % d ", [I + 0] * 4 b + b * 2 + b [I + 1] [I] + 2); 
} 
printf("\n"); 
} 

Int 
The main (void) 
{
Int I, n; 
Int j, m; 
Char * s; 
Int c; 
Hex, char * * bin; 

Assert (1==the scanf (" % d \ n ", & amp; N)); 

S=malloc (M + 1); 

For (I=0; I & lt; n; I++) {
The fgets (s, M, stdin); 
Chop (s); 
M=strlen (s); 
Hex=malloc (m); 
Bin=malloc (m * 4); 
For (j=0; J & lt; m; J++) {
C=s [j]; 
If (isdigit (c)) {
C=c - '0'; 
{} else if (isxdigit (c)) 
C=toupper (c) - 'A' + 10; 
} 
Hex [j]=c; 
} 
Hex2bin (hex, m, bin); 
Bin2oct (bin, m * 4); 
Free (hex); 
Free (bin); 
} 

Free (s); 

return 0; 
} 

/* 
Input: 
18 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
A 
B 
C 
D 
E 
F 
0123456789 abcdef 
FEDCBA9876543210 
Output: 
0 
1 
2 
3 
4 
5 
6 
7 
10 
11. 
12 
13 
14 
15 
16 
17 
4432126361152746757 
1773345651416625031020 
*/

CodePudding user response:

Three hexadecimal corresponding to the four octal
Playing table + alignment
finished

CodePudding user response:

Hex
If using integer storage
Two requires 1 byte memory
A 100000 - bit memory is 50000 bytes of memory
Your int only 4 bytes sent away

CodePudding user response:

Even can't save the input
At the back of the conversion is meaningless