Inverting the bits of an even number in C++ [closed] - c++

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A girl writes N numbers on the board (odd and even numbers) and then, she modifies only even numbers and inverts its binary representation (from the left to the right ) and replaces each even number. Write a code for the same.
0 <= n <= 10^7
I made a code for this, where my partial code looks like this:
int a[100000];
while ( t != 0 ) // t is the number in which input is taken
{
k = t & 1;
if ( k == 1 )
a[i] = 0; // a is the array in which bits of new number will be stored
else
a[i] = 1;
i++;
t = t >> 1;
}
for ( j = i; j >= 0; j-- )
{
if (a[j] == 1)
{
num = num + pow(2,j)*a[j]; // num is the number to be formed
}
}
cout<<num<<"\n";
But my answer comes out to be wrong for some values, for example for 8, it outputs 7. What is wrong with this? Thanks!
Problem link:
http://www.spoj.com/problems/EC_CONB/
Edit: ( In response to Pete's answer )
while ( t != 0 )
{
k = t & 1;
if ( k == 1 )
a[i] = 0;
else
{
a[i] = 1;
num = num + pow(2,i);
}
i++;
t = t >> 1;
}
cout<<num<<"\n";
}
This still shows the same problem, outputs the value of 8 as 7.

Answer 1

The problem actually wants you to reverse the bits not invert them. So the part where you check if k == 1, and putting 0 instead is not correct, because that inverts the bits.
What you need to do is reverse the ordering of the bits as follows:
1000 (8) -> 0001 (1)
1010 (10) -> 0101 (5)
Sample code based on your code:
while (t != 0)
{
k = t & 1;
// push the output array to the left
num <<= 1;
// add the read bit to the beginning of the output array
num += k;
t = t >> 1;
}
Explanation:
The basic idea is that we read bits from the input number one-by-one, and push them to the beginning of the output number.
Below is a trace of reversing a number (1011) at each iteration:
iterations 0 1 2 3 4
input 1011 101 10 1 0
output 0 1 11 110 1101

Answer 2

You appear to be reading a[i] after incrementing i, so the behaviour is undefined.
Instead of creating an array, accumulate the result in the first loop.

Answer 3

Arrays? Loops? Not necessary.
Reversing the number like that is equivalent to completely reversing the integer and then right-justifying it, like this (not tested)
// reverse from http://graphics.stanford.edu/~seander/bithacks.html#ReverseParallel
// swap odd and even bits
v = ((v >> 1) & 0x55555555) | ((v & 0x55555555) << 1);
// swap consecutive pairs
v = ((v >> 2) & 0x33333333) | ((v & 0x33333333) << 2);
// swap nibbles ...
v = ((v >> 4) & 0x0F0F0F0F) | ((v & 0x0F0F0F0F) << 4);
// swap bytes
v = ((v >> 8) & 0x00FF00FF) | ((v & 0x00FF00FF) << 8);
// swap 2-byte long pairs
v = ( v >> 16 ) | ( v << 16);
// right-justify
v /= v & -v;
v should be an uint32_t.