105 lines
2.9 KiB
C
105 lines
2.9 KiB
C
/* It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way,
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* but there are many more examples.
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*
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* 12 cm: (3,4,5)
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* 24 cm: (6,8,10)
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* 30 cm: (5,12,13)
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* 36 cm: (9,12,15)
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* 40 cm: (8,15,17)
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* 48 cm: (12,16,20)
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*
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* In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow
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* more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.
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*
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* 120 cm: (30,40,50), (20,48,52), (24,45,51)
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*
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* Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?*/
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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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#include "projecteuler.h"
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#define N 1500000
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int main(int argc, char **argv)
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{
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int i, a, b, c, tmpa, tmpb, tmpc, m, n, count = 0;
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int *l;
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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((l = (int *)calloc(N + 1, sizeof(int))) == NULL)
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{
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fprintf(stderr, "Error while allocating memory\n");
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return 1;
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}
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/* Generate all Pythagorean triplets using Euclid's algorithm:
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* For m>=2 and n<m:
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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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* This gives a primitive triple if gcd(m, n)=1 and exactly one
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* of m and n is odd. To generate all the triples, generate all
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* the primitive one and multiply them by i=2,3, ..., n until the
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* perimeter is larger than the limit. The limit for m is 865, because
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* when m=866 even with the smaller n (i.e. 1) the perimeter is greater
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* than the given limit.*/
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for(m = 2; m < 866; m++)
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{
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for(n = 1; n < m; n++)
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{
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if(gcd(m, n) == 1 && ((m % 2 == 0 && n % 2 != 0) || (m % 2 != 0 && n % 2 == 0)))
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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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if(a + b + c <= N)
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{
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l[a+b+c]++;
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}
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i = 2;
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tmpa = i * a;
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tmpb = i * b;
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tmpc = i * c;
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while(tmpa + tmpb + tmpc <= N)
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{
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l[tmpa+tmpb+tmpc]++;
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i++;
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tmpa = i * a;
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tmpb = i * b;
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tmpc = i * c;
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}
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}
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}
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}
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for(i = 0; i <= N; i++)
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{
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if(l[i] == 1)
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count++;
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}
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free(l);
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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 75\n");
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printf("Answer: %d\n", count);
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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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