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daniele
2024-02-20 19:46:29 +01:00
parent adca45e24f
commit 48503b8e27
16 changed files with 72 additions and 64 deletions
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68
C/p069.c
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@ -2,15 +2,15 @@
* relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
*
* n Relatively Prime φ(n) n/φ(n)
* 2 1 1 2
* 3 1,2 2 1.5
* 4 1,3 2 2
* 5 1,2,3,4 4 1.25
* 6 1,5 2 3
* 7 1,2,3,4,5,6 6 1.1666...
* 8 1,3,5,7 4 2
* 9 1,2,4,5,7,8 6 1.5
* 10 1,3,7,9 4 2.5
* 2 1 1 2
* 3 1,2 2 1.5
* 4 1,3 2 2
* 5 1,2,3,4 4 1.25
* 6 1,5 2 3
* 7 1,2,3,4,5,6 6 1.1666...
* 8 1,3,5,7 4 2
* 9 1,2,4,5,7,8 6 1.5
* 10 1,3,7,9 4 2.5
*
* It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
*
@ -26,41 +26,41 @@
int main(int argc, char **argv)
{
int i, res = 1;
int i, res = 1;
double elapsed;
struct timespec start, end;
struct timespec start, end;
clock_gettime(CLOCK_MONOTONIC, &start);
clock_gettime(CLOCK_MONOTONIC, &start);
i = 1;
i = 1;
/* Using Euler's formula, phi(n)=n*prod(1-1/p), where p are the distinct
* primes that divide n. So n/phi(n)=1/prod(1-1/p). To find the maximum
* value of this function, the denominator must be minimized. This happens
* when n has the most distinct small prime factor, i.e. to find the solution
* we need to multiply the smallest consecutive primes until the result is
* larger than 1000000.*/
while(res < N)
{
i++;
if(is_prime(i))
{
res *= i;
}
}
/* Using Euler's formula, phi(n)=n*prod(1-1/p), where p are the distinct
* primes that divide n. So n/phi(n)=1/prod(1-1/p). To find the maximum
* value of this function, the denominator must be minimized. This happens
* when n has the most distinct small prime factor, i.e. to find the solution
* we need to multiply the smallest consecutive primes until the result is
* larger than 1000000.*/
while(res < N)
{
i++;
if(is_prime(i))
{
res *= i;
}
}
/* We need the previous value, because we want i<1000000.*/
res /= i;
/* We need the previous value, because we want i<1000000.*/
res /= i;
clock_gettime(CLOCK_MONOTONIC, &end);
clock_gettime(CLOCK_MONOTONIC, &end);
elapsed = (end.tv_sec-start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
elapsed = (end.tv_sec-start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
printf("Project Euler, Problem 69\n");
printf("Project Euler, Problem 69\n");
printf("Answer: %d\n", res);
printf("Elapsed time: %.9lf seconds\n", elapsed);
printf("Elapsed time: %.9lf seconds\n", elapsed);
return 0;
}