/* 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.*/

#define _POSIX_C_SOURCE 199309L

#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;
}