93 lines
2.5 KiB
C
93 lines
2.5 KiB
C
/* If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
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*
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* {20,48,52}, {24,45,51}, {30,40,50}
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*
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* For which value of p ≤ 1000, is the number of solutions maximised?*/
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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 <time.h>
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int main(int argc, char **argv)
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{
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int p, m, n, a, b, c, count, max = 0, res = 0, tmpa, tmpb, tmpc, i;
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int savedc[1000] = {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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/* Start with p=12 (the smallest pythagorean triplet is (3,4,5) and 3+4+5=12.*/
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for(p = 12; p <= 1000; p++)
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{
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count = 0;
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a = 0;
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b = 0;
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c = 0;
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/* Generate pythagorean triplets.*/
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for(m = 2; m * m < p; m++)
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{
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for(n = 1; n < m; n++)
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{
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a = m * m - n * n;
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b = 2 * m * n;
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c = m * m + n * n;
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/* Increase counter if a+b+c=p and the triplet is new,
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* then save the value of c to avoid counting the same
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* triplet more than once.*/
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if(a + b + c == p && !savedc[c])
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{
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savedc[c] = 1;
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count++;
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}
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i = 2;
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tmpa = a;
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tmpb = b;
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tmpc = c;
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/* Check all the triplets obtained multiplying a, b and c
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* for integer numbers, until the perimeters exceeds p.*/
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while(tmpa + tmpb + tmpc < p)
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{
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tmpa = a * i;
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tmpb = b * i;
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tmpc = c * i;
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/* Increase counter if the new a, b and c give a perimeter=p.*/
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if(tmpa + tmpb + tmpc == p && !savedc[tmpc])
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{
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savedc[tmpc] = 1;
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count++;
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}
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i++;
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}
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}
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}
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/* If the current value is greater than the maximum,
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* save the new maximum and the value of p.*/
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if(count > max)
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{
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max = count;
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res = p;
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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 39\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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