Advent-of-Code/2021/day6/solution-part2.c

#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#include <stdlib.h>

#define POSSIBLE_SPAWN_AGES 9

unsigned long countFish(unsigned long *fishCount){
	unsigned long count = 0;
	for( unsigned int i = 0; i < POSSIBLE_SPAWN_AGES; i++ ){
		count += fishCount[i];
	}
	return count;
}

int main( int argc, char *argv[] ){
	if( argc == 2 ) {


		FILE *fp=fopen(argv[1], "r");

		unsigned int no;
		char ch;
		// I realised that the ordering of the fish doesn't matter. What matters is how many of them are of each age.
		// This array will contain the number of fish of age 1 in fishCount[1] and so on.
		unsigned long fishCount[POSSIBLE_SPAWN_AGES] = {0};
		while ( fscanf( fp, "%i%c", &no, &ch ) ){
			fishCount[no]++;
			if ( ch == '\n' ) break;
		}

		for ( unsigned int i = 0; i < 256; i++ ){
			// "backup" the number of fish of age 0 as the memmove will loose this
			unsigned long age0 = fishCount[0];
			memmove( &fishCount[0], &fishCount[1], sizeof(unsigned long) * (POSSIBLE_SPAWN_AGES - 1) );

			// All the fish that had a spawn_age of 0 will now have a spawn age of 6
			fishCount[6] += age0;

			// They will also all have a new fish with age of 8
			fishCount[8] = age0;

			//printf("\nDay %i:\n", i+1);
			//for( unsigned int j = 0; j < POSSIBLE_SPAWN_AGES; j++){
			//	printf("Fish with spawn age %i: %lu\n", j, fishCount[j]);
			//}
		}
		printf("There are %lu Fish.\n",countFish(fishCount));


		fclose(fp);



		return 0;
	} else {
		printf("You need to provide a file\n");
		return 1;
	}
}