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This will be optimized using the following logic. The operation num%8 will have a value greater than 0 only if the least significant 3 bits of a number are set. This is because of the fact that least significant 3 bits represent 2^0=1, 2^1=2 and 2^2=4. Any other bit that is set will be a multiple of 8 (8,16,32...) and will be useless for num%8 operation as the result will be always 0.

So we can just add the least significant 3 bits to the sum.

sum = 0;
while (!EOF)
{
sum += num&0x7;
}
if (sum & 0x07)
// not divisible by 8
else
// divisible by 8.

So basically we have removed the % operator and substituted it with the much faster bitwise operator &. Also since we are using only the lower order 3 bits for sum calculations the possibility of overflow is also less (until it's told that we are dealing with thousands of numbers in each file)

int isdiv8(FILE*fp)
{

int total 0;

while(fscanf(fp, "%d", &num))
{
total += num % 8;
total %= 8;
}

if (!total)
return 1;
else
return 0;

}

Find sum of the numbers in each file keeping in mind that at any time the sum < 8. If sum >= 8 then sum = sum %8. Now the final remainder if 0 will mean the sum of numbers is divisible by 8.

You need to check for overflow as well, reset it at 65528.....

We need to just check that in binary representation of the number last 3 digits should be 0. This can be checked by the condition (no && 7 == 0).