Add solutions for problems 97 and 99 in C

C/p097.c

@@ -0,0 +1,48 @@
+/* The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form 26972593−1;
+ * it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form 2p−1, have been found which contain more digits.
+ *
+ * However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: 28433×2^7830457+1.
+ *
+ * Find the last ten digits of this prime number.*/
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <time.h>
+
+int main(int argc, char **argv)
+{
+   int b = 2, e = 7830457, e_first;
+   long int  m = 10000000000, c, result;
+   double elapsed;
+   struct timespec start, end;
+
+   clock_gettime(CLOCK_MONOTONIC, &start);
+
+   /* Using modular exponentiation algorithm to calculate 2^7830457. The algorithm is:
+    * Set c=1, e'=0
+    * Increment e' by 1
+    * Set c=b*c%m, where b is the base (2) and m is the modulo (10000000000 since
+    * we want the last 10 digits.
+    * If e'<e (where e is the exponent), repeat step 3, otherwise c contains the result.*/
+   c = 1;
+   e_first = 0;
+
+   while(e_first < e)
+   {
+      e_first++;
+      c = (b * c) % m;
+   }
+
+   result = (c * 28433 + 1) % m;
+
+   clock_gettime(CLOCK_MONOTONIC, &end);
+
+   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+
+   printf("Project Euler, Problem 97\n");
+   printf("Answer: %ld\n", result);
+
+   printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+   return 0;
+}

C/p099.c

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+/* Comparing two numbers written in index form like 2^11 and 3^7 is not difficult, as any calculator would confirm that 
+ * 2^11 = 2048 < 3^7 = 2187.
+ *
+ * However, confirming that 632382^518061 > 519432^525806 would be much more difficult, as both numbers contain over three million digits.
+ *
+ * Using base_exp.txt, a 22K text file containing one thousand lines with a base/exponent pair on each line, 
+ * determine which line number has the greatest numerical value.
+ *
+ * NOTE: The first two lines in the file represent the numbers in the example given above.*/
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <math.h>
+#include <time.h>
+
+int main(int argc, char **argv)
+{
+   int i, max_i = -1, base, exp;
+   double curr, max = 0, elapsed;
+   struct timespec start, end;
+   FILE *fp;
+
+   clock_gettime(CLOCK_MONOTONIC, &start);
+
+   if((fp = fopen("base_exp.txt", "r")) == NULL)
+   {
+      fprintf(stderr, "Error while opening file %s\n", "base_exp.txt");
+      return 1;
+   }
+
+   for(i = 1; i <= 1000; i++)
+   {
+      fscanf(fp, "%d,%d", &base, &exp);
+      /* If 
+       * a^x > b^y
+       * log(a^x) > log(b^y)
+       * x*log(a) > y*log(b).
+       * So for each b^e, calculate e*log(b) and compare these numbers instead.*/
+      curr = exp * log(base);
+
+      if(curr > max)
+      {
+         max = curr;
+         max_i = i;
+      }
+   }
+
+   fclose(fp);
+   
+   clock_gettime(CLOCK_MONOTONIC, &end);
+
+   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+
+   printf("Project Euler, Problem 99\n");
+   printf("Answer: %d\n", max_i);
+
+   printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+   return 0;
+}