Problem 37

37.c

@@ -0,0 +1,43 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <stdbool.h>
+#include <math.h>
+#include "primes.h"
+
+bool* primesTable;
+
+bool isLeftTrunc (int n) {
+    int digits = floor (log10 (n)) + 1;
+    for (int i = 1; i <= digits; i++) {
+        if (!primesTable[n % (int) pow (10, i)])
+            return false;
+    }
+    return true;
+}
+
+bool isRightTrunc (int n) {
+    while (n > 0) {
+        if (!primesTable[n])
+            return false;
+        n /= 10;
+    }
+    return true;
+}
+
+int main () {
+    int* primes;
+    int numOfPrimes = listOfPrimes (1000000, &primes, &primesTable);
+    // unfortunately I couldn't find a deterministic way to find
+    // the upper-bound of all truncatable primes... 1 mil will do
+    
+    int sum = 0;
+    for (int i = 0; i < numOfPrimes; i++) {
+        if (isLeftTrunc (primes[i]) &&
+            isRightTrunc (primes[i]) &&
+            primes[i] > 10) {
+            printf ("%d\n", primes[i]);
+            sum += primes[i];
+        }
+    }
+    printf ("sum: %d\n", sum);
+}