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382 changes: 382 additions & 0 deletions 03 - Flujos de Control/Mi_TareaResuelta.ipynb
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{
"cells": [
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"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"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"a = -2\n",
"if a > 0:\n",
" print(\"a es mayor a cero\")\n",
"else:\n",
" print(\"es menor a cero\")"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"2) Crear dos variables y un condicional que informe si son del mismo tipo de dato\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"a = 10\n",
"b = \"10\"\n",
"if type(a)==type(b):\n",
" print(\"a y b son del mismo tipo\")\n",
"else:\n",
" print(\"Son de diferente tipo\")\n"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"3) Para los valores enteros del 1 al 20, imprimir por pantalla si es par o impar\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"for i in range(1,21):\n",
" if i%2==0:\n",
" print(\"El \",i,\" es par\")\n",
" else:\n",
" print(\"El \",i,\" es impar\")\n"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"4) En un ciclo for mostrar para los valores entre 0 y 5 el resultado de elevarlo a la potencia igual a 3\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"for num in range(0,6):\n",
" print(num,\" elevado a la tercera potencia es: \",num**3)\n",
"\n"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"5) Crear una variable que contenga un número entero y realizar un ciclo for la misma cantidad de ciclos\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"x = 6\n",
"for rep in range(x):\n",
" print(\"Esta es la repetición: \", rep)"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"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"
]
},
{
"cell_type": "code",
"execution_count": 27,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"120\n"
]
}
],
"source": [
"num = 5\n",
"factorial = 1\n",
"if (num > 0):\n",
" factorial = num\n",
" while (num > 2):\n",
" num -=1\n",
" factorial = num*factorial\n",
" print(factorial)\n",
"else:\n",
" print(\"La variable no es mayor a cero\")\n",
" \n"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"7) Crear un ciclo for dentro de un ciclo while\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"y = 0\n",
"while y < 5:\n",
" for i in range(1,y):\n",
" print(\"For\", i)\n",
" print(\"While\", y)\n",
" y+=1\n"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"8) Crear un ciclo while dentro de un ciclo for\n"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {},
"outputs": [],
"source": [
"n = 4\n",
"for i in range(1,n):\n",
" while (n < 4):\n",
" n -= 1\n",
" print(\"while\",n)\n",
" print(\"for\",i)"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"9) Imprimir los números primos existentes entre 0 y 30\n"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"el numero 2 es primo\n",
"el numero 3 es primo\n",
"el numero 5 es primo\n",
"el numero 7 es primo\n",
"el numero 11 es primo\n",
"el numero 13 es primo\n",
"el numero 17 es primo\n",
"el numero 19 es primo\n",
"el numero 23 es primo\n",
"el numero 29 es primo\n"
]
}
],
"source": [
"n=2\n",
"fin = 30\n",
"\n",
"while n <= fin:\n",
" primo = True\n",
" for i in range(2,n):\n",
" modulo = n%i\n",
" if modulo == 0:\n",
" primo= False\n",
" \n",
" if primo:\n",
" print(f\"el numero {n} es primo\")\n",
" n+=1\n",
" \n",
"\n"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"10) ¿Se puede mejorar el proceso del punto 9? Utilizar las sentencias break y/ó continue para tal fin\n",
"11) mirar qué tanto se optimizó"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"n=2\n",
"fin = 30\n",
"ciclos_sin = 0 #contar el número de ciclos\n",
"while n <= fin:\n",
" primo = True\n",
" for i in range(2,n):\n",
" ciclos_sin += 1\n",
" modulo = n%i\n",
" if modulo == 0:\n",
" primo= False\n",
" break \n",
" if primo:\n",
" print(f\"el numero {n} es primo\")\n",
" n+=1\n",
"print(f\"Los ciclos sin break fueron: {ciclos_sin}\")"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"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"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"n=99\n",
"f=300\n",
"while (n<=300):\n",
" n+=1\n",
" if n%12 != 0:\n",
" continue\n",
" else:\n",
" print(n)\n",
"\n"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"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"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"n=2\n",
"primo = True\n",
"buscar_primo = True\n",
"\n",
"while buscar_primo: \n",
" for i in range(2,n):\n",
" modulo = n%i\n",
" if modulo == 0:\n",
" primo= False\n",
" break\n",
" if primo:\n",
" print(f\"el numero {n} es primo\")\n",
" print(\"Desea encontrar el siguien? \")\n",
" continuar = input(\"ingrese si o no: \")\n",
" if continuar != 'si':\n",
" print(\"se finaliza la búsqueda\")\n",
" buscar_primo = False\n",
" break\n",
" else:\n",
" primo = True\n",
" n+=1"
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
"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"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"102\n"
]
}
],
"source": [
"n = 99\n",
"fin = 300\n",
"while(n<=fin):\n",
" n+=1\n",
" if n%3 == 0 and n%6 == 0:\n",
" print(n)\n",
" break\n",
" else:\n",
" continue"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.11.4"
},
"orig_nbformat": 4
},
"nbformat": 4,
"nbformat_minor": 2
}
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