108 lines
2.6 KiB
C
108 lines
2.6 KiB
C
/* The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime,
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* and, (ii) each of the 4-digit numbers are permutations of one another.
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*
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* There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit
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* increasing sequence.
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*
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* What 12-digit number do you form by concatenating the three terms in this sequence?*/
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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 10000
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int is_permutation(int a, int b);
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int *primes;
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int main(int argc, char **argv)
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{
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int i = 1489, j, found = 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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/* Starting from i=1489 (bigger than the first number in the sequence given in the problem),
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* check odd numbers. If they're prime, loop on even numbers j (odd+even=odd, odd+odd=even and
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* we need odd numbers because we're looking for primes) up to 4254 (1489+2*4256=10001 which has
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* 5 digits.*/
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while(i < N)
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{
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if(primes[i])
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{
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for(j = 2; j < 4255; j += 2)
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{
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/* If i, i+j and i+2*j are all primes and they have
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* all the same digits, the result has been found.*/
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if(i + 2 * j < N && primes[i+j] && primes[i+2*j] &&
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is_permutation(i, i+j) && is_permutation(i, i+2*j))
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{
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found = 1;
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break;
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}
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}
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}
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if(found)
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{
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break;
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}
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i += 2;
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}
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free(primes);
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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 49\n");
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printf("Answer: %d%d%d\n", i, i+j, i+2*j);
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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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