144 lines
3.8 KiB
C
144 lines
3.8 KiB
C
/* By replacing the 1st digit of the 2-digit number *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.
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*
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* By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes among the ten
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* generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this
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* family, is the smallest prime with this property.
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*
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* Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime
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* value family.*/
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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 <string.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 count_digit(int n, int d);
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int replace(int n);
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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;
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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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/* N set to 1000000 as a reasonable limit, which turns out to be enough.*/
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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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/* Checking only odd numbers with at least 4 digits.*/
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for(i = 1001; i < N; i += 2)
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{
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/* The family of numbers needs to have at least one of 0, 1 or 2 as
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* repeated digits, otherwise we can't get a 8 number family (there
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* are only 7 other digits). Also, the number of repeated digits must
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* be 3, otherwise at least 3 resulting numbers will be divisible by 3.
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* So the smallest number of this family must have three 0s, three 1s or
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* three 2s.*/
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if(count_digit(i, 0) >= 3 || count_digit(i, 1) >= 3 ||
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count_digit(i, 2) >= 3)
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{
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/* If i is prime, try replacing digits, if obtaining 8 primes, then
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* this is the result.*/
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if(primes[i] && replace(i) == 8)
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{
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break;
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}
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}
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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 51\n");
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printf("Answer: %d\n", i);
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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 count_digit(int n, int d)
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{
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int count = 0;
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while(n > 0)
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{
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if(n % 10 == d)
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{
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count++;
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}
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n /= 10;
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}
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return count;
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}
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int replace(int n)
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{
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int i, j, k, w, l, count, max = 0, s_to_n;
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char n_to_s[7];
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sprintf(n_to_s, "%d", n);
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l = strlen(n_to_s);
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for(i = 0; i < l - 3; i++)
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{
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for(j = i + 1; j < l - 2; j++)
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{
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/* Replacing the last digit can't give 8 primes, because at least
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* six of the numbers obtained will be divisible by 2 and/or 5.*/
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for(k = j + 1; k < l - 1; k++)
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{
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count = 0;
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for(w = 0; w < 10; w++)
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{
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if(i == 0 && w == 0)
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{
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continue;
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}
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sprintf(n_to_s, "%d", n);
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n_to_s[i] = w + '0';
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n_to_s[j] = w + '0';
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n_to_s[k] = w + '0';
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s_to_n = atoi(n_to_s);
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if(primes[s_to_n])
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{
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if(count == 0 && s_to_n != n)
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{
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continue;
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}
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count++;
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}
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}
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if(count > max)
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{
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max = count;
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}
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}
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}
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}
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return max;
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}
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