Question

write a c program that http://gyazo.com/614ad16944c56d97947b2bba6c77e8e3

Answer 1

im not good at programming, not sure where to start.

Answer 2

I can code this in C++,you can code it in C similarly,I'll explain the logic and everything.
#include<iostream.h>
#include<conio.h>
int main()
{
clrscr();
int Input,NumberOfTimes=0;
cout<<"Enter the Number : ";
cin>>Input;
for(int i=0;i<Input;i++)
{
if(Input%2!=0)
{
Input=Input/2;
NumberOfTimes++;
}
}
cout<<"The number of times the Number can be halved = > " <<NumberOfTimes;
getch();
}

Do you get what I did?
Till the number doesn't become odd,i.e the remainder is NON 0 then we will keep haling the number using a for loop.
Any doubts?

Answer 3

i cant use for loops, only while loops...

Answer 4

can you please help me with this using while loops?

Answer 5

Replace
-----------------------------
for(int i=0;i<Input;i++)
{
if(Input%2!=0)
{
Input=Input/2;
NumberOfTimes++;
}
}

------------------------------
with
------------------------------
while(Input%2!=0)
{
Input=Input/2;
NumberOfTimes++;
}

Answer 6

@Omniscience ?

Answer 7

not working...how do i count how many times it has been havled?

Answer 8

If its an even number then on dividing by 2 the remainder will be 0,for odd it will be non 0.
Then the number will be halved i.e Input=Input/2;
And If number is halved then we increment the variable NumberOfTimes++;
then finally print it.This is the logic.I don't know why it isn't working with C maybe some syntax thing but you know the logic try to code yourself though i don't think there is a fault in my program it should work well :/

Answer 9

```
int halfCount = 0;
while (n % 2 == 0) { // Check that n is even
n = n / 2; // Haves the number
halfCount++; // Adds to half count
}

```

Answer 10

Also, the input and output are very different in C from C++. If you are trying to learn C, then use of the C++ libraries will hurt you rather than help you.

Answer 11

How about:

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

int evenHalf( int number )
{
number /= 2;
if( number % 2 )
return 0;
else
return( evenHalf( number ) + 1 );
}

int main( int argc, char* argv[] )
{
if( argc > 1 )
printf( "%d\n", evenHalf( atoi( argv[ 1 ] ) ) );
else
printf( "Usage: %s <integer>\n", argv[ 0 ] );

return( 0 );
}

Answer 12

Not to demean this site, but if you're looking for help with C programming this is the site I always use:

It's filled with a lot of really experienced programmers, but they expect you to do the work. They will most def help you solve your problem.

http://cboard.cprogramming.com/

Answer 13

worked, thanks guys! http://gyazo.com/d2f11c5c30d4dd6888d53f80d75cf4eb

Answer 14

but the logic remains the same