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Correct function count_divisors

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This commit is contained in:
daniele
2019-09-22 21:12:58 +02:00
parent 95611e4401
commit dd628af770
2 changed files with 28 additions and 24 deletions
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18
C/projecteuler.c
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@ -143,7 +143,7 @@ int *sieve(int n)
* there is no need check i>sqrt(n).*/
limit = floor(sqrt(n));
for(i = 3; i < limit; i += 2)
for(i = 3; i <= limit; i += 2)
{
/* Find the next number not crossed out, which is prime.*/
if(primes[i])
@ -166,19 +166,23 @@ int count_divisors(int n)
{
int i, limit, count = 0;
/* For every divisor below the square root of n, there is a corresponding one
* above the square root, so it's sufficient to check up to the square root of n
* and count every divisor twice. If n is a perfect square, the last divisor is
* wrongly counted twice and must be corrected.*/
limit = floor(sqrt(n));
for(i = 1; i < limit; i++)
for(i = 1; i <= limit; i++)
{
if(n % i == 0)
{
count += 2;
}
if(n == limit * limit)
{
count--;
}
}
if(n == limit * limit)
{
count--;
}
return count;