Recent projects

calculadora

Assembly · an anonymous user
18 hours ago
section .data menu db "Menu Calculadora", 10, \ "1. Suma (50+10)", 10 \ "2- REsta (50-10)", 10 \ "3. Multiplicacion (50*10)", 10 ;La diagonal invertida sirve para poder ordenar el codigo en diferentes line

12981번

Python · an anonymous user
18 hours ago
s, p = map(int, input().split()) a = input() D = [] for c in a : D.append(c) A, C, G, T = map(int, input().split()) cnt = 0

clase_anm_13nov_arq_y_progr_compu

Assembly · an anonymous user
18 hours ago
section .data menu db "Menu de Calculadora", 10, \ "1. Suma (50+10)", 10, \ "2. Resta (50-10)", 10, \ "3. Multiplicacion (50*10)", 10 \ ;La diagonal invertida sirve para poder ordenar el codigo en diferentes lineas ;sin perder la perten

4장[문제5].c

C · pinxxose
18 hours ago
// N(100이하)개 정수 입력 받기 -> 가장 큰 값, 두 번째로 큰 값 순서대로 출력 (줄바꿈, 둘이 같을 수 있음) // - max1 보다 num 이 크다고 판명 // 1. max1 을 업데이트 // 2. 기존 max1 을 max2 로 옮김 // 2 // 4 1 // - 기존 max2 보다는 큰데 max1 보다 작은경우 // max2 = 1 #include <stdio.h>

위상정렬.... 인데 너무 어렵다

Python · chanheeko0709
18 hours ago
from collections import deque v, n = map(int, input().split()) r = [] g = [[] for _ in range(v + 1)] for _ in range(n): r.append(list(map(int,input().split())))

untied

Python · an anonymous user
18 hours ago
import numpy as np import IPython rate = 48000 duration = 1.0 t = np.linspace(0., duration, int(rate*duration)) x = np.sin(2.0*np.pi*440.0*t) IPython.display.Audio(x, rate=rate, autoplay=True)

dql_dml.sql

MySQL · an anonymous user
18 hours ago
-- INSERTAR 10 REGISTROS EN CADA TABLA INSERT INTO usuario (id_usuario, nombre, apellido, correo, telefono) VALUES (1,'Cristopher','Mena','laura1@gmail.com','0000000000'), (2,'Valery','Martinez','andres2@gmail.com','0000000000'), (3,'Nicolle','Lope

Quiz11_202606017.py

Python · na324
18 hours ago
print("*"*18) print("202606017나경문") print("*"*18) # 좌석 상태 출력 함수 정의 def show(seat_dict): for seat in seat_dict: print(seat, end="\t") print()

진자어이업느각도구하기

C# · 1skdssae
19 hours ago
using System; namespace MyCompiler { class Program { public static int innerSum(int n) { return (n-2)*180; } public static int inner(int n) { return (n-2)*180/n; }

EJERCICIO SISTEMA DE GESTION DE INSCRIPCIONES

SQL · an anonymous user
19 hours ago
/* Proyecto de Aplicación Práctica - Fundamentos de Base de Datos */ /* Sistema de Gestión de Inscripciones Académicas */ /* Motor: SQLite (ejecutado en myCompiler - www.mycompiler.io/es/new/sql) */ /* 1. CONFIGURACIÓN */ /* En SQLite la base de da

10번

C++ · sum479
19 hours ago
#include <iostream> #include <vector> #include <algorithm> #include <string> using namespace std; int main(){ ios_base::sync_with_stdio(false); cin.tie(NULL); int n;

ssss

Assembly · an anonymous user
19 hours ago
;============================================================================== ; PROYECTO: Contador Multiplexado de 2 Dígitos ; Placa SimuPIC - PIC16F628A ; Objetivo: Contar de 50 hasta 99 de a 3, luego volver a 50 y de a 1 bajar hasta 0. ;========

:)

C · an anonymous user
19 hours ago
#include <stdio.h> int main() { printf("Hello world!\n"); }

Interpolación Polinomica (Verificación)

Python · an anonymous user
19 hours ago
import numpy as np x0=0 x1=25000 x2=50000 x3=75000 x4=100000 x5=125000 y0=9.8100 y1=9.7335

Interpolación Newton (Verificación)

Python · an anonymous user
19 hours ago
from sympy import * x0=0 x1=25000 x2=50000 x3=75000 x4=100000 x5=125000 y0=9.8100 y1=9.7335

Interpolación Lagrange (Verificación)

Python · an anonymous user
19 hours ago
from sympy import * x0=0 x1=25000 x2=50000 x3=75000 x4=100000 x5=125000 y0=9.8100 y1=9.7335

ejercicio 1 examen

TypeScript · jorgerosales
19 hours ago
function verificarStock( cantidad : number) : string { if( cantidad=== 0){ return "Alerta: Producto agotado!"; } else if(cantidad <= 5 ) { return "Alerta!,quedan pocas unidades"; } else { return "stock suficiente,disponible"; }

Inter.Lagrange (Verificación)

Python · an anonymous user
19 hours ago
from sympy import * x0=0 x1=40000 x2=80000 x3=120000 x4=160000 y0=9.8100 y1=9.6878 y2=9.5682

Inter.Lagrange

Python · an anonymous user
19 hours ago
from sympy import * x0=0 x1=40000 x2=80000 x3=120000 x4=160000 y0=9.8100 y1=9.6878 y2=9.5682

dp2 à rendre

C++ · an anonymous user
19 hours ago
#include<iostream> #include<time.h> #include<string.h> using namespace std; //question 4 and 5 int uniquePathsWithObstacles(int** obstacleGrid, int m, int n){ // Cas extrêmes : départ ou arrivée bloqués if(obstacleGrid[0][0] == 1 || obsta