594 - One Little, Two Little, Three Little Endians

[globi]

Thanks a lot ( I got accepted ).

[szymcio2001]

Some additional test cases for everyone who has a problem with negative input:
0 -> 0
-1 -> -1
-2 -> -16777217
-65536 -> 65535

[Ronald29]

Thx, I already have AC

[Ronald29]

Thx, I've already got AC

[aakash_mandhar]

Try this code. this is very easy problem
[cpp]
Code removed.
Thx Shamim for the suggestion
[/cpp]

Hope u like the code

[aakash_mandhar]

Removed.. Thx Shamim for the suggestion

[RustB]

This is giving me correct answers with my trial inputs, but the judge returns W.A. Why is that?
[c]
#include <stdio.h>

union inp
{
long num;
char a[4];
} in,out;

int main(void)
{
while(!feof(stdin))
{
int i;
in.num=out.num=0;
scanf("%d",&in.num);
for(i=0;i<4;i++)
out.a=in.a[3-i];
printf("%d converts to %d\n",in.num,out.num);
}
}
[/c]

[Ryan Pai]

There's probably an end-of-line character at the end of the data. You should always make your solution not depend on there being/not being one.

So after it reads the last number, there's still that last character, the end-of-line. So the feof() returns false. But the next iteration you don't read a number, so it outputs 0.

Also, the endianness of your system might be different than the judges (unless you checked what they use and compared it to what you use).

[acmforacm]

i got WA~
who can help me ?

Code: Select all

#include <stdio.h>
#include <math.h>

main()
{
        long long n , num;
        int i , j , k;
        int binary[4][8];
        while ( scanf ( " %lld" , &n ) == 1 ){
                printf ( "%lld" , n );
                for ( i=0; i<4; ++i )
                        for ( j=0; j<8; ++j )
                                binary[i][j] = 0;
                i = j = num = 0;
                while ( n >= 1 ){
                        binary[i][j++] = n % 2;
                        n /= 2;
                        if ( j == 8 ){
                                j = 0;
                                i++;
                        }
                }
                for ( i=3 , k=0; i>=0; --i , ++k ){
                        for ( j=0; j<8; ++j ){
                                num += binary[i][j] * pow( 2 , ( ( k * 8 ) + j ) );
                        }
                }
                printf ( " converts to %lld\n" , num );
        }
}

[Ghust_omega]

Hi acmforacm i found one bug in your code not handlle negative inputs fine see what is the ouput for this
-7 , -5 is 0 you have to check your code
Hope its Helps
Keep posting !

[linux]

Hey, RustB! Don't be worried about endians they are same as you did. But about end-of-line character ryan pai is right. I think you better use scanf by which I got AC from you program.

[B.E]

I would just like to say that the problem is very easy(did it in under a minute). The code I used was (which the CPU time was 0.004 seconds) The following. It works by using the 80386 assembly instruction to byte swap an integer (the instruction is often used to convert big endian numbers to little endian numbers and vice versa).

Code: Select all

#include <stdio.h>
#include <stdlib.h>

int main(){
	char buff[13];
	register long l;
	while (gets(buff)){
		l = atol(buff);
		__asm__ __volatile__ ("bswap %0" : "+r" (l));
		printf("%s converts to %d\n", buff, l);
	}
	return 0;
}

[lovemagic]

can somebody explain when the output should be negetive? & how can i handle negetive number?

[Jan]

1. can somebody explain when the output should be negetive?
2. how can i handle negetive number?

1. What is the value of binary(1111111...111 -> 32 1's) for a signed 32-bit integer?
2. No special condition I think.

[lovemagic]

Thanx for ur reply.Now i understood.
Got AC......