100 lines
2.5 KiB
C
100 lines
2.5 KiB
C
/* Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers
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* less than or equal to n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and
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* relatively prime to nine, φ(9)=6.
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* The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.
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*
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* Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.
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*
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* Find the value of n, 1 < n < 10^7, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.*/
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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 <float.h>
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#include "projecteuler.h"
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#define N 10000000
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int is_permutation(int a, int b);
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int main(int argc, char **argv)
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{
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int i, a, b, p, n = -1;
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int *primes;
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double elapsed, min = DBL_MAX;
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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(i = 2; i < N; i++)
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{
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/* When n is prime, phi(n)=(n-1), so to minimize n/phi(n) we should
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* use n prime. But n-1 can't be a permutation of n. The second best
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* bet is to use semiprimes. For a semiprime n=p*q, phi(n)=(p-1)(q-1).
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* So we check if a number is semiprime, if yes calculate phi, finally
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* check if phi(n) is a permutation of n and update the minimum if it's
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* smaller.*/
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if(is_semiprime(i, &a, &b, primes))
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{
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p = phi_semiprime(i, a, b);
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if(is_permutation(p, i) && (double)i / p < min)
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{
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n = i;
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min = (double)i / p;
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}
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}
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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 70\n");
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printf("Answer: %d\n", n);
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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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int is_permutation(int a, int b)
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{
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int i;
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int digits1[10] = {0}, digits2[10] = {0};
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/* Get digits of a.*/
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while(a > 0)
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{
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digits1[a%10]++;
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a /= 10;
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}
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/* Get digits of b.*/
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while(b > 0)
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{
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digits2[b%10]++;
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b /= 10;
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}
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/* If they're not the same, return 0.*/
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for(i = 0; i < 10; i++)
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{
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if(digits1[i] != digits2[i])
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{
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return 0;
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}
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}
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return 1;
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}
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