69 lines
1.9 KiB
C
69 lines
1.9 KiB
C
/* Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are
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* 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.
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*
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* n Relatively Prime φ(n) n/φ(n)
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* 2 1 1 2
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* 3 1,2 2 1.5
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* 4 1,3 2 2
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* 5 1,2,3,4 4 1.25
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* 6 1,5 2 3
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* 7 1,2,3,4,5,6 6 1.1666...
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* 8 1,3,5,7 4 2
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* 9 1,2,4,5,7,8 6 1.5
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* 10 1,3,7,9 4 2.5
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*
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* It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
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*
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* Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.*/
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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 1000000
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int main(int argc, char **argv)
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{
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int i, res = 1;
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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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i = 1;
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/* Using Euler's formula, phi(n)=n*prod(1-1/p), where p are the distinct
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* primes that divide n. So n/phi(n)=1/prod(1-1/p). To find the maximum
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* value of this function, the denominator must be minimized. This happens
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* when n has the most distinct small prime factor, i.e. to find the solution
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* we need to multiply the smallest consecutive primes until the result is
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* larger than 1000000.*/
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while(res < N)
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{
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i++;
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if(is_prime(i))
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{
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res *= i;
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}
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}
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/* We need the previous value, because we want i<1000000.*/
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res /= i;
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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 69\n");
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printf("Answer: %d\n", res);
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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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