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2019 day 7, part 1

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This commit is contained in:
Bruno Raoult
2022-09-30 19:22:12 +02:00
parent 626bd1df41
commit 155e066ed2
3 changed files with 143 additions and 13 deletions
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81
2019/day07/aoc-c.c
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@@ -41,10 +41,13 @@ typedef struct {
char *mnemo;
} ops_t;
#define MAXOPS 1024
#define MAXINPUT 4
#define MAXOPS 1024
typedef struct {
int length; /* total program length */
int cur; /* current position */
int curinput, lastinput;
int input[MAXINPUT]; /* input */
int mem [MAXOPS]; /* should really be dynamic */
} program_t;
@@ -72,7 +75,45 @@ static int _flag_pow10[] = {1, 100, 1000, 10000};
INDIRECT(p, n + i) = val; } \
while (0)
static int run(program_t *p, int in)
/**
* permute - get next permutation of an array of integers
* @len: length of array
* @array: address of array
*
* Algorithm: lexicographic permutations
* https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
* Before the initial call, the array must be sorted (e.g. 0 2 3 5)
*
* Return: 1 if next permutation was found, 0 if no more permutation.
*
*/
static int permute_next(int len, int *array)
{
int k, l;
/* 1. Find the largest index k such that a[k] < a[k + 1] */
for (k = len - 2; k >= 0 && array[k] >= array[k + 1]; k--)
;
/* No more permutations */
if (k < 0)
return 0;
/* 2. Find the largest index l greater than k such that a[k] < a[l] */
for (l = len - 1; array[l] <= array[k]; l--)
;
/* 3. Swap the value of a[k] with that of a[l] */
swap(array[k], array[l]);
/* 4. Reverse sequence from a[k + 1] up to the final element */
for (l = len - 1, k++; k < l; k++, l--)
swap(array[k], array[l]);
return 1;
}
static void dup_program(program_t *from, program_t *to)
{
*to = *from;
}
static int run(program_t *p)
{
int out = -1;
while (1) {
@@ -90,7 +131,7 @@ static int run(program_t *p, int in)
poke(p, p->cur, 3, peek(p, p->cur, 1) * peek(p, p->cur, 2));
break;
case INP:
poke(p, p->cur, 1, in);
poke(p, p->cur, 1, p->input[p->curinput++]);
break;
case OUT:
out = peek(p, p->cur, 1);
@@ -132,15 +173,20 @@ static int usage(char *prg)
int main(int ac, char **av)
{
int opt, part = 1, in = -1;
program_t p = { 0 };
program_t p = { 0 }, p1;
int phase[] = {0, 1, 2, 3, 4};
while ((opt = getopt(ac, av, "d:p:i:")) != -1) {
while ((opt = getopt(ac, av, "d:p:i:o:")) != -1) {
switch (opt) {
case 'd':
debug_level_set(atoi(optarg));
break;
case 'i':
in = atoi(optarg);
p.input[p.lastinput++] = atoi(optarg);
break;
case 'o':
for (ulong i = 0; i < strlen(optarg); ++i)
phase[i] = optarg[i] - '0';
break;
case 'p': /* 1 or 2 */
part = atoi(optarg);
@@ -152,11 +198,32 @@ int main(int ac, char **av)
}
}
for (int i = 0; i < 5; ++i)
printf("phase[%d]=%d\n", i, phase[i]);
parse(&p);
int out, max = 0;
do {
out = 0;
for (int i = 0; i < 5; ++i) {
dup_program(&p, &p1);
p1.input[p1.lastinput++] = phase[i];
p1.input[p1.lastinput++] = out;
out = run(&p1);
}
if (out > max) {
max = out;
printf("new max: %c%c%c%c%c out=%d max=%d\n", phase[0] + '0',
phase[1] + '0', phase[2] + '0', phase[3] + '0', phase[4] + '0',
out, max);
}
} while (permute_next(5, phase));
exit(0);
if (optind < ac)
return usage(*av);
if (in == -1)
in = part == 1? 1: 5;
parse(&p);
printf("%s : res=%d\n", *av, run(&p, in));
printf("%s : res=%d\n", *av, run(&p));
exit (0);
}