57 lines
1.6 KiB
C
57 lines
1.6 KiB
C
/* Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.
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*
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* If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:
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*
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* 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
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*
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* It can be seen that there are 21 elements in this set.
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*
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* How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?*/
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#define _POSIX_C_SOURCE 199309L
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#include <stdio.h>
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#include <stdlib.h>
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#include <math.h>
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#include <time.h>
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#include "projecteuler.h"
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#define N 1000001
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int main(int argc, char **argv)
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{
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int i;
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int *primes;
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long int count = 0;
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double elapsed;
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struct timespec start, end;
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clock_gettime(CLOCK_MONOTONIC, &start);
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if((primes = sieve(N)) == NULL)
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{
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fprintf(stderr, "Error! Sieve function returned NULL\n");
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return 1;
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}
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/* For any denominator d, the number of reduced proper fractions is
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* the number of fractions n/d where gcd(n, d)=1, which is the definition
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* of Euler's Totient Function phi. It's sufficient to calculate phi for each
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* denominator and sum the value.*/
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for(i = 2; i < N; i++)
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{
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count += phi(i, primes);
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}
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clock_gettime(CLOCK_MONOTONIC, &end);
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elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
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printf("Project Euler, Problem 72\n");
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printf("Answer: %ld\n", count);
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printf("Elapsed time: %.9lf seconds\n", elapsed);
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return 0;
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}
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