72 lines
2.1 KiB
C
72 lines
2.1 KiB
C
/* Three distinct points are plotted at random on a Cartesian plane, for which -1000 ≤ x, y ≤ 1000, such that a triangle is formed.
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*
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* Consider the following two triangles:
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*
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* A(-340,495), B(-153,-910), C(835,-947)
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* X(-175,41), Y(-421,-714), Z(574,-645)
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*
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* It can be verified that triangle ABC contains the origin, whereas triangle XYZ does not.
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*
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* Using triangles.txt, a 27K text file containing the co-ordinates of one thousand "random" triangles, find the number of triangles
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* for which the interior contains the origin.
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*
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* NOTE: The first two examples in the file represent the triangles in the example given above.*/
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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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typedef struct point
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{
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int x;
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int y;
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} point;
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int main(int argc, char **argv)
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{
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int count = 0;
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FILE *fp;
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point p1, p2, p3;
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double a, b, c, elapsed;
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struct timespec start, end;
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clock_gettime(CLOCK_MONOTONIC, &start);
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if((fp = fopen("triangles.txt", "r")) == NULL)
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{
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fprintf(stderr, "Error while opening file %s\n", "triangles.txt");
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return 1;
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}
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/* Using the barycentric coordinates method to determine if (0, 0) is inside the triangles.*/
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while(fscanf(fp, "%d,%d,%d,%d,%d,%d", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y) != EOF)
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{
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a = (double)((p2.y - p3.y) * (-p3.x) + (p3.x - p2.x) * (-p3.y)) /
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((p2.y - p3.y) * (p1.x - p3.x) + (p3.x - p2.x) * (p1.y - p3.y));
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b = (double)((p3.y - p1.y) * (-p3.x) + (p1.x - p3.x) * (-p3.y)) /
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((p2.y - p3.y) * (p1.x - p3.x) + (p3.x - p2.x) * (p1.y - p3.y));
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c = 1 - a - b;
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if(a >= 0 && a <= 1 && b >= 0 && b <= 1 && c >= 0 && c <= 1)
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{
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count++;
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}
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}
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fclose(fp);
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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 102\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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