day 23 support more inputs
[aoc_eblake.git] / 2018 / day11.c
blob7ac17288672145e840c53a7499fb41903743b1e6
1 #define _GNU_SOURCE 1
2 #include <stdio.h>
3 #include <string.h>
4 #include <stdlib.h>
5 #include <stdarg.h>
6 #include <stdbool.h>
7 #include <assert.h>
9 void debug(const char *fmt, ...) {
10 va_list ap;
11 if (getenv("DEBUG")) {
12 va_start(ap, fmt);
13 vfprintf(stderr, fmt, ap);
14 va_end(ap);
18 #define LIMIT 300
20 static struct point {
21 int8_t value;
22 int row[LIMIT];
23 } grid[LIMIT][LIMIT];
24 static struct best {
25 int x;
26 int y;
27 int sum;
28 } best[LIMIT];
30 static int8_t level(int x, int y, int s) {
31 int id = x + 10;
32 int p = (id * y + s) * id;
33 return (p % 1000) / 100 - 5;
36 static void rows(int x, int y) {
37 grid[x][y].row[0] = grid[x][y].value;
38 for (int i = 1; i < LIMIT - y; i++)
39 grid[x][y].row[i] = grid[x][y].row[i - 1] + grid[x][y + i].value;
42 static void find(int size) {
43 int i, j;
44 bool dup = false;
45 best[size - 1].sum = -5 * size * size;
46 for (j = 0; j < LIMIT - size + 1; j++) {
47 int sum = 0;
48 for (i = 0; i < size; i++)
49 sum += grid[i][j].row[size - 1];
50 for (i = 0; i < LIMIT - size + 1; i++) {
51 if (i)
52 sum += grid[i + size - 1][j].row[size - 1] -
53 grid[i - 1][j].row[size - 1];
54 if (sum > best[size - 1].sum) {
55 best[size - 1].sum = sum;
56 best[size - 1].x = i + 1;
57 best[size - 1].y = j + 1;
58 dup = false;
59 } else if (sum == best[size - 1].sum)
60 dup = true;
63 if (dup)
64 debug("ignoring duplicate best candidates for size %d\n", size);
67 int main(int argc, char **argv) {
68 int s = 1955;
69 int i, j;
70 int max = 0;
72 if (argc > 1)
73 s = atoi(argv[1]);
74 for (j = 0; j < LIMIT; j++)
75 for (i = 0; i < LIMIT; i++)
76 grid[i][j].value = level(i + 1, j + 1, s);
77 for (j = 0; j < LIMIT; j++)
78 for (i = 0; i < LIMIT; i++)
79 rows(i, j);
80 for (i = 0; i < LIMIT; i++) {
81 debug("%d\n", i);
82 find(i + 1);
83 if (best[i].sum > best[max].sum)
84 max = i;
85 else if (best[i].sum == best[max].sum)
86 debug("size %d has same best as %d\n", i, max);
88 printf("best at %d,%d,%d with %d\n", best[max].x, best[max].y, max + 1,
89 best[max].sum);
90 return 0;