Addo solution for problem 112 in C

C/p112.c

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+/* Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.
+ *
+ * Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.
+ *
+ * We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.
+ *
+ * Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy.
+ * In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.
+ *
+ * Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.
+ *
+ * Find the least number for which the proportion of bouncy numbers is exactly 99%.*/
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <time.h>
+
+int is_bouncy(int n);
+
+int main(int argc, char **argv)
+{
+   int i = 100, n_bouncy = 0;
+   double ratio = 0.0, elapsed;
+   struct timespec start, end;
+
+   clock_gettime(CLOCK_MONOTONIC, &start);
+
+   while(1)
+   {
+      if(is_bouncy(i))
+      {
+         n_bouncy++;
+      }
+
+      ratio = (double)n_bouncy / i;
+      
+      if(ratio == 0.99)
+      {
+         break;
+      }
+     
+      i++;
+   }
+
+   clock_gettime(CLOCK_MONOTONIC, &end);
+
+   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+
+   printf("Project Euler, Problem 112\n");
+   printf("Answer: %d\n", i);
+
+   printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+   return 0;
+}
+
+/* Check if the number is bouncy.*/
+int is_bouncy(int n)
+{
+   int i = 0, l;
+   char n_string[100];
+
+   sprintf(n_string, "%d", n);
+
+   l = strlen(n_string);
+
+   /* Ignore consecutive equal digits.*/
+   while(i < l - 1 && n_string[i] == n_string[i+1])
+   {
+      i++;
+   }
+
+   /* If all the digits are the same, the number is not bouncy.*/
+   if(i == l - 1)
+   {
+      return 0;
+   }
+
+   /* If the first two different digits are increasing, if a successive
+    * digit smaller than the previous digit is found, the number is bouncy.*/
+   if(n_string[i] < n_string[i+1])
+   {
+      for(; i < l - 1; i++)
+      {
+         if(n_string[i] > n_string[i+1])
+         {
+            return 1;
+         }
+      }
+   }
+
+   /* If the first two different digits are decreasing, if a successive
+    * digit larger than the previous digit is found, the number is bouncy.*/
+   if(n_string[i] > n_string[i+1])
+   {
+      for(; i < l - 1; i++)
+      {
+         if(n_string[i] < n_string[i+1])
+         {
+            return 1;
+         }
+      }
+   }
+
+   return 0;
+}