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61 lines
1.5 KiB
61 lines
1.5 KiB
3 years ago
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#include <stdio.h>
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#include <stdbool.h>
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#include <string.h>
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#include <stdlib.h>
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#define POSSIBLE_SPAWN_AGES 9
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unsigned long countFish(unsigned long *fishCount){
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unsigned long count = 0;
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for( unsigned int i = 0; i < POSSIBLE_SPAWN_AGES; i++ ){
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count += fishCount[i];
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}
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return count;
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}
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int main( int argc, char *argv[] ){
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if( argc == 2 ) {
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FILE *fp=fopen(argv[1], "r");
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unsigned int no;
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char ch;
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// I realised that the ordering of the fish doesn't matter. What matters is how many of them are of each age.
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// This array will contain the number of fish of age 1 in fishCount[1] and so on.
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unsigned long fishCount[POSSIBLE_SPAWN_AGES] = {0};
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while ( fscanf( fp, "%i%c", &no, &ch ) ){
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fishCount[no]++;
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if ( ch == '\n' ) break;
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}
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for ( unsigned int i = 0; i < 256; i++ ){
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// "backup" the number of fish of age 0 as the memmove will loose this
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unsigned long age0 = fishCount[0];
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memmove( &fishCount[0], &fishCount[1], sizeof(unsigned long) * (POSSIBLE_SPAWN_AGES - 1) );
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// All the fish that had a spawn_age of 0 will now have a spawn age of 6
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fishCount[6] += age0;
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// They will also all have a new fish with age of 8
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fishCount[8] = age0;
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//printf("\nDay %i:\n", i+1);
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//for( unsigned int j = 0; j < POSSIBLE_SPAWN_AGES; j++){
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// printf("Fish with spawn age %i: %lu\n", j, fishCount[j]);
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//}
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}
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printf("There are %lu Fish.\n",countFish(fishCount));
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fclose(fp);
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return 0;
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} else {
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printf("You need to provide a file\n");
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return 1;
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}
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}
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