project-euler-solutions/C/projecteuler.c
Daniele Fucini 6b29655330
Add more solutions
Added solutions for problems 21, 22, 23, 24 and 25 both in C and
in python
2019-09-20 11:12:51 +02:00

158 lines
2.2 KiB
C

#include <stdlib.h>
#include <math.h>
#include "projecteuler.h"
int is_prime(long int num)
{
int i, limit;
if(num <= 3)
{
return num == 2 || num == 3;
}
if(num % 2 == 0 || num % 3 == 0)
{
return 0;
}
limit = floor(sqrt(num));
for(i = 5; i <= limit; i += 6)
{
if(num % i == 0 || num % (i + 2) == 0)
{
return 0;
}
}
return 1;
}
long int gcd(long int a, long int b)
{
long int tmp;
while(b != 0)
{
tmp = b;
b = a % b;
a = tmp;
}
return a;
}
long int lcm(long int a, long int b)
{
return a * b / gcd(a, b);
}
long int lcmm(long int *values, int n)
{
int i;
long int value;
if(n == 2)
{
return lcm(values[0], values[1]);
}
else
{
value = values[0];
for(i = 0; i < n - 1; i++)
{
values[i] = values[i+1];
}
return lcm(value, lcmm(values, n-1));
}
}
int *sieve(int n)
{
int i, j, limit;
int *primes;
if((primes = (int *)malloc(n*sizeof(int))) == NULL)
{
return NULL;
}
primes[0] = 0;
primes[1] = 0;
primes[2] = 1;
primes[3] = 1;
for(i = 4; i < n - 1; i += 2)
{
primes[i] = 0;
primes[i+1] = 1;
}
limit = floor(sqrt(n));
for(i = 3; i < limit; i += 2)
{
if(primes[i])
{
for(j = i * i; j < n; j += 2 * i)
{
primes[j] = 0;
}
}
}
return primes;
}
int count_divisors(int n)
{
int i, limit, count = 0;
limit = floor(sqrt(n));
for(i = 1; i < limit; i++)
{
if(n % i == 0)
{
count += 2;
}
if(n == limit * limit)
{
count--;
}
}
return count;
}
void quick_sort(void **array, int l, int r, int (*cmp)(void *lv, void *rv))
{
int i, j;
void *pivot, *tmp;
if (l >= r) return;
pivot = array[l];
i = l - 1;
j = r + 1;
while (i < j) {
while (cmp(array[++i], pivot) < 0);
while (cmp(array[--j], pivot) > 0);
if (i < j) {
tmp = array[i];
array[i] = array[j];
array[j] = tmp;
}
}
quick_sort(array, l, j, cmp);
quick_sort(array, j+1, r, cmp);
}