Week 148, part 2: C solution

1 files changed, 145 insertions, 0 deletions

diff --git a/challenge-148/abigail/c/ch-2.c b/challenge-148/abigail/c/ch-2.c
new file mode 100644
index 0000000000..63e2db46e1
--- /dev/null
+++ b/challenge-148/abigail/c/ch-2.c
@@ -0,0 +1,145 @@
+# include <stdlib.h>
+# include <stdio.h>
+# include <string.h>
+# include <stdbool.h>
+# include <math.h>
+
+/*
+ * See https://theweeklychallenge.org/blog/perl-weekly-challenge-148
+ */
+
+/*
+ * Run as: cc -o ch-2.o ch-2.c; ./ch-2.o
+ */
+
+# define COUNT          5
+# define A              0
+# define B              1
+# define C              2
+# define SUM            3
+/* Overkill, but this will do */
+# define MAX_DIVISOR  100
+
+int main (void) {
+    int ** out;
+    int    k = 0;
+
+    if ((out = (int **) malloc (COUNT * sizeof (int *))) == NULL) {
+        perror ("Malloc failed");
+        return (1);
+    }
+    for (size_t i = 0; i < COUNT; i ++) {
+        if ((out [i] = (int *) malloc (SUM * sizeof (int))) == NULL) {
+            perror ("Malloc failed");
+            return (1);
+        }
+        out [i] [A]   = 999999;
+        out [i] [B]   = 999999;
+        out [i] [C]   = 999999;
+        out [i] [SUM] = out [i] [A] + out [i] [B] + out [i] [C];
+    }
+
+    size_t max_index = 0;
+
+
+    while (3 * k + 2 <= out [max_index] [SUM]) { 
+        int a  = 3 * k + 2;
+        int f1 =     k + 1;
+        int f2 = 8 * k + 5;
+
+        int * d1;
+        int * d2;
+
+        if ((d1 = (int *) malloc (MAX_DIVISOR * sizeof (int))) == NULL) {
+            perror ("Malloc failed");
+            return (1);
+        }
+        if ((d2 = (int *) malloc (MAX_DIVISOR * sizeof (int))) == NULL) {
+            perror ("Malloc failed");
+            return (1);
+        }
+
+        size_t d1c = 0;
+        size_t d2c = 0;
+
+        /*
+         * Get the divisors for f1
+         */
+        for (int i = 1; i * i <= f1; i ++) {
+            if (f1 % i == 0) {
+                d1 [d1c ++] = i;
+                if (i != f1 / i) {
+                    d1 [d1c ++] = f1 / i;
+                }
+            }
+        }
+
+        /*
+         * Get the square divisors of f2
+         */
+        for (int i = 1; i * i <= f2; i ++) {
+            if (f2 % i == 0) {
+                int s1 = (int) sqrt (i);
+                int s2 = (int) sqrt (f2 / i);
+                if (s1 * s1 == i) {
+                    d2 [d2c ++] = s1;
+                }
+                if (s2 * s2 == f2 / i && s2 != s1) {
+                    d2 [d2c ++] = s2;
+                }
+            }
+        }
+
+        for (size_t i = 0; i < d1c; i ++) {
+            for (size_t j = 0; j < d2c; j ++) {
+                int b = d1 [i] * d2 [j];
+                int c = f1 * f1 * f2 / (b * b);
+                if (a + b + c < out [max_index] [SUM]) {
+                    /*
+                     * Avoid duplicates
+                     */
+                    bool seen = false;
+                    for (size_t i = 0; i < COUNT; i ++) {
+                        if (out [i] [A] == a && out [i] [B] == b) {
+                            seen = true;
+                            break;
+                        }
+                    }
+                    if (seen) {
+                        break;
+                    }
+
+                    out [max_index] [A]   = a;
+                    out [max_index] [B]   = b;
+                    out [max_index] [C]   = c;
+                    out [max_index] [SUM] = a + b + c;
+
+                    /*
+                     * Find the next max index
+                     */
+                    max_index = 0;
+                    int max_sum = out [max_index] [SUM];
+                    for (size_t l = 1; l < COUNT; l ++) {
+                        if (out [l] [SUM] > max_sum) {
+                            max_sum   = out [l] [SUM];
+                            max_index = l;
+                        }
+                    }
+                }
+            }
+        }
+        free (d1);
+        free (d2);
+        k ++;
+    }
+
+    /*
+     * Print the array
+     */
+    for (size_t i = 0; i < COUNT; i ++) {
+        printf ("%d %d %d\n", out [i] [A], out [i] [B], out [i] [C]);
+        free (out [i]);
+    }
+
+    free (out);
+}