From 8302da73a4ef0c3d0c44c689aba9161ae8eb8d17 Mon Sep 17 00:00:00 2001
From: sijanec <anton@sijanec.eu>
Date: Mon, 4 Jan 2021 22:24:14 +0100
Subject: =?UTF-8?q?sloven=C5=A1=C4=8Dina=20in=20nekaj=20zajebancije=20z=20?=
 =?UTF-8?q?rtk=20laNSKO=20offline=20nalogo?=
MIME-Version: 1.0
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---
 inf/rtk/offline/resi.c | 67 ++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 67 insertions(+)
 create mode 100644 inf/rtk/offline/resi.c

(limited to 'inf/rtk/offline/resi.c')

diff --git a/inf/rtk/offline/resi.c b/inf/rtk/offline/resi.c
new file mode 100644
index 0000000..43f97cc
--- /dev/null
+++ b/inf/rtk/offline/resi.c
@@ -0,0 +1,67 @@
+#if __INCLUDE_LEVEL__ != 0
+#pragma once
+#endif
+/* najdaljši ponavljajoči kos niza */
+#include <stdio.h>
+#include <stdlib.h>
+#include <offline.h>
+#include <string.h>
+#define RTK_NAENKRAT 60000
+/* lahko bi šli čez niz znakov for(strlen)for(strlen)for(strlen), ampak to
+ * bi trajalo zelo dolgo, čas*1000000^3 in bi imelo O(n^3) kompleksnost.
+ * ta implementacija najde najdaljši string in ima kompleksnost
+ * PRIBLIŽNO OKOLI O(n^2), kar je bistveno hitreje, PRIBLIŽNO čas*1000000^2. */
+struct rtk_kos npk (const char * s, const size_t l) {
+	struct rtk_kos k;
+	k.l = 0; k.o = 0; 
+	unsigned char ** z = calloc(l+1, sizeof(unsigned char *));
+	size_t i = 0;
+	size_t j = 0;
+	for (i = 0; i < l+1; i++)
+		z[i] = calloc(l+1, sizeof(unsigned char));
+	for (i = 1; i <= l; i++)
+		for (j = i+1; j <= l; j++)
+			if (s[i-1] == s[j-1] && z[i-1][j-1] < (j-1)) {
+				z[i][j] = z[i-1][j-1] + 1;
+				if (z[i][j] > k.l) {
+					k.l = z[i][j];
+					k.o = k.o > i ? k.o : i;
+				}
+			} else
+				z[i-1][j-1] = 0;
+	k.o = k.o-k.l;
+	for (i = 0; i < l+1; i++) {
+		free(z[i]); z[i] = NULL;
+	}
+	free(z);
+	z = NULL;
+	return k;
+}
+#if __INCLUDE_LEVEL__ == 0
+int main (int argc, char ** argv) {
+	char c = getchar();
+	size_t v = 0;
+	char * s = malloc(sizeof(char)*1);
+	struct rtk_kos k1;
+	struct rtk_kos k2;
+	size_t l = 0;
+	while (!feof(stdin)) {
+		s = realloc(s, sizeof(char)*v+2);
+		s[v++] = c;
+		s[v] = '\0';
+		c = getchar();
+	}
+	l = v;
+	for (v = 0; v*RTK_NAENKRAT <= l;) {
+		k1 = npk(s+(v*(RTK_NAENKRAT/2)), RTK_NAENKRAT);
+		k2 = npk(s+(++v*(RTK_NAENKRAT/2)), RTK_NAENKRAT);
+	}
+	
+	for (v = 0; v < k.l; v++)
+		putchar(s[v+k.o]);
+	putchar('\n'); /* za dobro mero */
+	free(s);
+	s = NULL;
+	return 0;
+}
+#endif
-- 
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