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Andre2070Andre2070
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'primer_comming'
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‎03 - Flujos de Control/Mi_TareaResuelta.ipynb

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{
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"cells": [
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"1) Crear una variable que contenga un elemento del conjunto de números enteros y luego imprimir por pantalla si es mayor o menor a cero"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"a = -2\n",
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"if a > 0:\n",
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" print(\"a es mayor a cero\")\n",
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"else:\n",
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" print(\"es menor a cero\")"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"2) Crear dos variables y un condicional que informe si son del mismo tipo de dato\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"a = 10\n",
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"b = \"10\"\n",
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"if type(a)==type(b):\n",
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" print(\"a y b son del mismo tipo\")\n",
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"else:\n",
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" print(\"Son de diferente tipo\")\n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"3) Para los valores enteros del 1 al 20, imprimir por pantalla si es par o impar\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"for i in range(1,21):\n",
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" if i%2==0:\n",
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" print(\"El \",i,\" es par\")\n",
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" else:\n",
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" print(\"El \",i,\" es impar\")\n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"4) En un ciclo for mostrar para los valores entre 0 y 5 el resultado de elevarlo a la potencia igual a 3\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"for num in range(0,6):\n",
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" print(num,\" elevado a la tercera potencia es: \",num**3)\n",
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"\n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"5) Crear una variable que contenga un número entero y realizar un ciclo for la misma cantidad de ciclos\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"x = 6\n",
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"for rep in range(x):\n",
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" print(\"Esta es la repetición: \", rep)"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"6) Utilizar un ciclo while para realizar el factorial de un número guardado en una variable, sólo si la variable contiene un número entero mayor a 0\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 27,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"120\n"
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]
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}
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],
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"source": [
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"num = 5\n",
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"factorial = 1\n",
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"if (num > 0):\n",
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" factorial = num\n",
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" while (num > 2):\n",
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" num -=1\n",
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" factorial = num*factorial\n",
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" print(factorial)\n",
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"else:\n",
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" print(\"La variable no es mayor a cero\")\n",
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" \n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"7) Crear un ciclo for dentro de un ciclo while\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"y = 0\n",
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"while y < 5:\n",
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" for i in range(1,y):\n",
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" print(\"For\", i)\n",
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" print(\"While\", y)\n",
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" y+=1\n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"8) Crear un ciclo while dentro de un ciclo for\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 31,
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"metadata": {},
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"outputs": [],
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"source": [
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"n = 4\n",
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"for i in range(1,n):\n",
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" while (n < 4):\n",
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" n -= 1\n",
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" print(\"while\",n)\n",
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" print(\"for\",i)"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"9) Imprimir los números primos existentes entre 0 y 30\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"el numero 2 es primo\n",
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"el numero 3 es primo\n",
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"el numero 5 es primo\n",
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"el numero 7 es primo\n",
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"el numero 11 es primo\n",
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"el numero 13 es primo\n",
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"el numero 17 es primo\n",
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"el numero 19 es primo\n",
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"el numero 23 es primo\n",
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"el numero 29 es primo\n"
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]
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}
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],
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"source": [
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"n=2\n",
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"fin = 30\n",
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"\n",
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"while n <= fin:\n",
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" primo = True\n",
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" for i in range(2,n):\n",
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" modulo = n%i\n",
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" if modulo == 0:\n",
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" primo= False\n",
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" \n",
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" if primo:\n",
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" print(f\"el numero {n} es primo\")\n",
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" n+=1\n",
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" \n",
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"\n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"10) ¿Se puede mejorar el proceso del punto 9? Utilizar las sentencias break y/ó continue para tal fin\n",
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"11) mirar qué tanto se optimizó"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"n=2\n",
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"fin = 30\n",
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"ciclos_sin = 0 #contar el número de ciclos\n",
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"while n <= fin:\n",
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" primo = True\n",
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" for i in range(2,n):\n",
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" ciclos_sin += 1\n",
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" modulo = n%i\n",
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" if modulo == 0:\n",
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" primo= False\n",
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" break \n",
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" if primo:\n",
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" print(f\"el numero {n} es primo\")\n",
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" n+=1\n",
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"print(f\"Los ciclos sin break fueron: {ciclos_sin}\")"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"13) Aplicando continue, armar un ciclo while que solo imprima los valores divisibles por 12, dentro del rango de números de 100 a 300\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"n=99\n",
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"f=300\n",
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"while (n<=300):\n",
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" n+=1\n",
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" if n%12 != 0:\n",
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" continue\n",
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" else:\n",
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" print(n)\n",
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"\n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"14) Utilizar la función **input()** que permite hacer ingresos por teclado, para encontrar números primos y dar la opción al usario de buscar el siguiente"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"n=2\n",
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"primo = True\n",
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"buscar_primo = True\n",
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"\n",
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"while buscar_primo: \n",
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" for i in range(2,n):\n",
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" modulo = n%i\n",
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" if modulo == 0:\n",
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" primo= False\n",
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" break\n",
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" if primo:\n",
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" print(f\"el numero {n} es primo\")\n",
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" print(\"Desea encontrar el siguien? \")\n",
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" continuar = input(\"ingrese si o no: \")\n",
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" if continuar != 'si':\n",
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" print(\"se finaliza la búsqueda\")\n",
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" buscar_primo = False\n",
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" break\n",
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" else:\n",
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" primo = True\n",
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" n+=1"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"15) Crear un ciclo while que encuentre dentro del rango de 100 a 300 el primer número divisible por 3 y además múltiplo de 6"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 8,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"102\n"
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]
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}
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],
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"source": [
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"n = 99\n",
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"fin = 300\n",
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"while(n<=fin):\n",
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" n+=1\n",
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" if n%3 == 0 and n%6 == 0:\n",
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" print(n)\n",
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" break\n",
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" else:\n",
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" continue"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.4"
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},
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"orig_nbformat": 4
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

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