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Exercicio para casa semana 4 #23

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1b64087
exercicio casa logica semana 4
elianeferreira Jun 27, 2024
c7508c4
alteração exercicio semana 4
elianeferreira Jun 27, 2024
03c00e8
resolução exercicio para casa - professora
elianeferreira Jun 29, 2024
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255 changes: 255 additions & 0 deletions exercicios/para-casa/eliane-ferreira/exercicio_casa.ipynb
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Desenvolva um programa que faça a tabuada de um número qualquer inteiro que será digitado pelo usuário, mas a tabuada não deve necessariamente iniciar em 1 e terminar em 10, o valor inicial e final devem ser informados também pelo usuário. "
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"5 x 0 = 0\n",
"5 x 1 = 5\n",
"5 x 2 = 10\n",
"5 x 3 = 15\n",
"5 x 4 = 20\n",
"5 x 5 = 25\n",
"5 x 6 = 30\n",
"5 x 7 = 35\n",
"5 x 8 = 40\n",
"5 x 9 = 45\n",
"5 x 10 = 50\n",
" \n"
]
}
],
"source": [
"multiplicador = int(input('Digite um número'))\n",
"inicial = int(input('Digite o número para tabuada'))\n",
"final= int(input('Digite o número final'))\n",
"\n",
"for i in range(inicial,final+1):\n",
" resultado = multiplicador * i\n",
" print(f\"{multiplicador} x {i} = {resultado}\")\n",
" #i += 3\n",
" \n",
"\n",
"#num = int(input(\"Digite um número:\"))\n",
"#num2 = int(input('Digite o segundo número'))\n",
"#contador = 2\n",
"\n",
"#while contador <= 9:\n",
" # resultado = int(num) * int(num2)\n",
" # print(f\"{num} x {num2}= {resultado}\")\n",
" ##contador += 1\n",
" \n",
"#print(\"condição aceita\")"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"numero = int(input('Digite um número para ver sua tabuada'))\n",
"\n",
"inicio = int(input(\"Digite o valor inicial da tabuada\"))\n",
"\n",
"fim = int(input(\"Digite o valor final da tabuada\"))\n",
"\n",
"for i in range (inicio, fim +1):\n",
" print(f\"{numero} x {i} = {numero * i}\")"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"try:\n",
" numero = int(input('Digite um número para ver sua tabuada'))\n",
"\n",
" inicio = int(input(\"Digite o valor inicial da tabuada\"))\n",
"\n",
" fim = int(input(\"Digite o valor final da tabuada\"))\n",
"\n",
" for i in range (inicio, fim +1):\n",
" print(f\"{numero} x {i} = {numero * i}\")\n",
"\n",
" if inicio > fim:\n",
" raise ValueError(\" o valor inicial deve ser menor ou igual ao valor final.\")\n",
" \n",
" for i in range(inicio, fim + 1):\n",
" print(f\"{numero} x {i} = {numero * 1}\")\n",
"except ValueError as ve:\n",
" print(\"f Entrada invalida: {ve}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Faça um programa que calcule o fatorial de um número inteiro fornecido pelo usuário. Ex.: 5!=5.4.3.2.1=120. A saída deve ser conforme o exemplo abaixo:\n",
"Fatorial de: 5\n",
"5! = 5 . 4 . 3 . 2 . 1 = 120"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"5! = 5.4.3.2.1=120\n"
]
}
],
"source": [
"num = int(input('Calcule o fatorial de :'))\n",
"fatorial = 1\n",
"fatores = []\n",
"\n",
"\n",
"for i in range (num,0,-1):\n",
" fatorial = fatorial * i \n",
" fatores.append(str(i)) \n",
"\n",
"fatorial_str = f\"{num}! = {'.'.join(fatores)} = {fatorial}\"\n",
"print(fatorial_str)\n",
"\n",
"\n",
"\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"num = int(input('Calcule o fatorial de :'))\n",
"\n",
"resultado = 1\n",
"\n",
"for i in range(1, num +1 ):\n",
" resultado *= 1\n",
"\n",
" print(f\"O fatorial é {num}\")"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"120\n"
]
}
],
"source": [
"numero = int(input(\"Fatorial de: \") )\n",
"\n",
"resultado=1\n",
"for n in range(1,numero+1):\n",
" resultado *= n\n",
"\n",
"print(resultado)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Um funcionário de uma empresa recebe aumento salarial anualmente: Sabe-se que:\n",
"Esse funcionário foi contratado em 1995, com salário inicial de R$ 1.000,00;\n",
"Em 1996 recebeu aumento de 1,5% sobre seu salário inicial;\n",
"A partir de 1997 (inclusive), os aumentos salariais sempre correspondem ao dobro do percentual do ano anterior. Faça um programa que determine o salário atual desse funcionário. Após concluir isto, altere o programa permitindo que o usuário digite o salário inicial do funcionário."
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Salario atual do funcionario acrescido de 15% é: 1015.0\n",
"O salario com aumento dobrado é : 1015.03\n"
]
}
],
"source": [
"salario_inicial = float(input(\"digite seu salario inicial\"))\n",
"salario_dobro = float(input(\"Digite o salário atual\"))\n",
"novo_salario = salario_inicial + ( salario_inicial * 0.015)\n",
"\n",
"print(f\"Salario atual do funcionario acrescido de 15% é: {novo_salario}\" )\n",
"\n",
"resultado = novo_salario + 2 * 0.015\n",
"print(f\"O salario com aumento dobrado é : {resultado}\" )\n",
"\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"salario_inicial = float(input(\"Digite o salário inicial do funcionário: \"))\n",
"ano_contratacao = 1995\n",
"ano_atual = 2024\n",
"percentual_aumento = 1.5/100\n",
"salario_atual = salario_inicial\n",
"for ano in range(ano_contratacao + 1, ano_atual + 1):\n",
" if ano == 1996:\n",
" salario_atual += salario_inicial * percentual_aumento\n",
" else:\n",
" percentual_aumento *= 2\n",
" salario_atual += salario_inicial * percentual_aumento\n",
" \n",
"print(f\"O salário atual do funcionario em {ano_atual} é R$ {salario_atual:.2f}\")"
]
}
],
"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": "3.12.4"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
29 changes: 28 additions & 1 deletion exercicios/para-sala/exercicios_for.ipynb
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Expand Up @@ -238,6 +238,33 @@
" \n",
"print(f\"A soma dos primeiros 5 números naturais é: {soma}\")"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A média dos números é: 46.57\n"
]
}
],
"source": [
"numeros = [23, 45, 12, 67 , 89 , 34 , 56]\n",
"\n",
"soma = 0\n",
"\n",
"for num in numeros : \n",
" soma = soma + num \n",
"\n",
" media = soma / len (numeros)\n",
"\n",
"print(f\"A média dos números é: {media:.2f}\")\n",
"\n"
]
}
],
"metadata": {
Expand All @@ -256,7 +283,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.11.4"
"version": "3.12.4"
}
},
"nbformat": 4,
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26 changes: 13 additions & 13 deletions exercicios/para-sala/exercicios_while.ipynb
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Original file line number Diff line number Diff line change
Expand Up @@ -102,24 +102,24 @@
},
{
"cell_type": "code",
"execution_count": 8,
"execution_count": 1,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"1 x 0 = 0\n",
"1 x 1 = 1\n",
"1 x 2 = 2\n",
"1 x 3 = 3\n",
"1 x 4 = 4\n",
"1 x 5 = 5\n",
"1 x 6 = 6\n",
"1 x 7 = 7\n",
"1 x 8 = 8\n",
"1 x 9 = 9\n",
"1 x 10 = 10\n",
"4 x 0 = 0\n",
"4 x 1 = 4\n",
"4 x 2 = 8\n",
"4 x 3 = 12\n",
"4 x 4 = 16\n",
"4 x 5 = 20\n",
"4 x 6 = 24\n",
"4 x 7 = 28\n",
"4 x 8 = 32\n",
"4 x 9 = 36\n",
"4 x 10 = 40\n",
"Final\n"
]
}
Expand Down Expand Up @@ -153,7 +153,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.11.4"
"version": "3.12.4"
}
},
"nbformat": 4,
Expand Down