Fırat Salgür neuro-sys

## textmode.asm
; nasm mode13.asm -fbin -omode13.com
SECTION         .text
                org     100h
start:          mov     dx, filenamez
                mov     al, 0
                call    openfile
                jnc     fileopened
                mov     dx, filenotopeneds
                call    printstr
                jmp     filedone

## mode13.asm

start:          call    mode13          ; Set mode 13
repeat_row:     mov     ax, [posx]
                cmp     ax, 320         ; Check if posx >= 320
                jz      end_xloop       ; Stop
                mov     al, [color]
                mov     cx, [posx]
                mov     dx, [posy]
                call    setpixel        ; Set pixel at posx, posy with color
                inc     byte [color]    ; color += 1

## spider.bas
10 mode 1
15 midx=320:midy=200:s=10
20 for x=0 to midy step s
30 move midx+x, midy
40 draw midx, midy+(midy-x)
50 draw midx-x, midy
60 draw midx, midy-(midy-x)
70 draw midx+x, midy
80 next

## rdaily-20190211-hard.asm
BDOS            EQU     &5
C_WRITESTR      EQU     &9

; Structs and the member sizes

edge_status     equ     1               ; 0 negative, 1 positive
edge_node       equ     2               ; pointer to node this edge connects to
edge_next       equ     2               ; pointer to next edge
edge_s_offset   equ     edge_status + edge_node + edge_next

## rdaily-20190211-int.asm
; This challenge is about a simple card flipping solitaire game. You're
; presented with a sequence of cards, some face up, some face down. You
; can remove any face up card, but you must then flip the adjacent cards
; (if any). The goal is to successfully remove every card. Making the
; wrong move can get you stuck.
;
; In this challenge, a 1 signifies a face up card and a 0 signifies a
; face down card. We will also use zero-based indexing, starting from
; the left, to indicate specific cards. So, to illustrate a game,
; consider this starting card set.

## rdaily-20190211-easy.asm
; Description
; A number is input in computer then a new no should get
; printed by adding one to each of its digit. If you encounter a 9,
; insert a 10 (don't carry over, just shift things around).
;
; For example, 998 becomes 10109.
;
; Bonus
; This challenge is trivial to do if you map it to a string to
; iterate over the input, operate, and then cast it back. Instead, try

## euler-2.asm
; Each new term in the Fibonacci sequence is generated by adding the
;  previous two terms. By starting with 1 and 2, the first 10 terms
;  will be:
; 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
; By considering the terms in the Fibonacci sequence whose values do
;  not exceed four million, find the sum of the even-valued terms.

                ORG     &8000

                DI

## euler-1.asm
; If we list all the natural numbers below 10 that
; are multiples of 3 or 5, we get 3, 5, 6 and
; 9. The sum of these multiples is 23.  Find the
; sum of all the multiples of 3 or 5 below 1000.

                ORG     &8000
                RUN     &8000,break

                DI
                LD      DE, 999         ; COUNT INTEGERS BELOW 1000

## sprite.asm
  org &4000
  run &4000

  di
  im 1

  ; set ei, ret to &38 interrupt
  ld a, &fb
  ld (&38), a
  ld a, &c9

## crtc.c
#include <stdio.h>

typedef unsigned char u8;
typedef unsigned short u16;
typedef unsigned int u32;

#define VIDEO_ADDR(MA, RA) \
    ((MA & 0x3FF) << 1) | ((RA & 7) << 11) | ((MA & 0x3000) << 2)

#define PRINT_CHAR_ADDR