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#include <ctype.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "aocdailies.h"
#include "aocutils.h"
/* === aoc202408 =======================================================
Oh! another square of text!
Idea: for all points p with an antenna find all points q>p with
a corresponding frequency. For each such pair calculate the 2 antinodes
and add the resulting points (if not there already) to an array.
The answer to Part One is the number of elements in the array
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#include <ctype.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "aocdailies.h"
#include "aocutils.h"
/* === aoc202409 =======================================================
WOW! input consists of a 20,000 long string of numbers!
===================================================================== */
void aoc202409(char *data, size_t len) {
(void)len; // unused argument
int *disk = malloc(512 * sizeof *disk);
size_t rdisk = 512;
size_t ndisk = 0;
int fileid = 0;
while (*data) {
for (int k = 0; k < *data - '0'; k++) {
if (ndisk == rdisk) {
// assume it works
rdisk = (13*rdisk)/8;
disk = realloc(disk, rdisk * sizeof *disk);
}
disk[ndisk++] = fileid;
}
fileid++;
data++;
if (*data) {
for (int k = 0; k < *data - '0'; k++) {
if (ndisk == rdisk) {
// assume it works
rdisk = (13*rdisk)/8;
disk = realloc(disk, rdisk * sizeof *disk);
}
disk[ndisk++] = -1;
}
data++;
}
}
int *left = disk, *right = disk + ndisk;
for (;;) {
while (*left != -1) left++;
if (left >= right) break;
while (right[-1] == -1) right--;
// swap *left and right[-1]
int tmp = *--right;
*right = *left;
*left++ = tmp;
}
unsigned long long chksum = 0;
size_t index = 0;
while (disk[index] >= 0) {
chksum += (size_t)disk[index] * index;
index++;
}
printf("%llu\n", chksum);
free(disk);
}
/* === aoc202408 =======================================================
Oh! another square of text!
Idea: for all points p with an antenna find all points q>p with
a corresponding frequency. For each such pair calculate the 2 antinodes
and add the resulting points (if not there already) to an array.
The answer to Part One is the number of elements in the array
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