81 lines
2.3 KiB
C
81 lines
2.3 KiB
C
/* Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
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*
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* 37 36 35 34 33 32 31
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* 38 17 16 15 14 13 30
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* 39 18 5 4 3 12 29
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* 40 19 6 1 2 11 28
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* 41 20 7 8 9 10 27
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* 42 21 22 23 24 25 26
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* 43 44 45 46 47 48 49
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*
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* It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers
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* lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
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*
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* If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued,
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* what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?*/
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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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int main(int argc, char **argv)
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{
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int i = 1, l = 1, step = 2, count = 0, diag = 5;
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double ratio, elapsed;
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struct timespec start, end;
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clock_gettime(CLOCK_MONOTONIC, &start);
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/* Starting with 1, the next four numbers in the diagonal are 3 (1+2), 5 (1+2+2), 7 (1+2+2+2)
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* and 9 (1+2+2+2+2). Check which are prime, increment the counter every time a new prime is
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* found, and divide by the number of elements of the diagonal, which are increase by 4 at
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* every cycle. The next four number added to the diagonal are 13 (9+4), 17 (9+4+4), 21 and 25.
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* Then 25+6 etc., at every cycle the step is increased by 2. Continue until the ratio goes below 0.1.*/
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do
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{
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i += step;
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if(is_prime(i))
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{
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count++;
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}
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i += step;
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if(is_prime(i))
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{
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count++;
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}
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i += step;
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if(is_prime(i))
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{
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count++;
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}
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i += step;
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ratio = (double)count / diag;
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step += 2;
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diag += 4;
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l += 2;
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} while(ratio >= 0.1);
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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 58\n");
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printf("Answer: %d\n", l);
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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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