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day1 converted to ipython notebook
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‎notebooks/Day_1.ipynb

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Question 1\n",
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"\n",
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"### **Question:**\n",
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"\n",
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"> **_Write a program which will find all such numbers which are divisible by 7 but are not a multiple of 5,\n",
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"> between 2000 and 3200 (both included).The numbers obtained should be printed in a comma-separated sequence on a single line._**\n",
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"\n",
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"---\n",
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"\n",
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"### Hints:\n",
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"\n",
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"> **_Consider use range(#begin, #end) method._**\n",
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"\n",
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"---\n",
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"\n",
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"**Main author's Solution: Python 2**"
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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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"l=[]\n",
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"for i in range(2000, 3201):\n",
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" if (i%7==0) and (i%5!=0):\n",
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" l.append(str(i))\n",
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"\n",
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"print ','.join(l)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"---\n",
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"\n",
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"**My Solution: Python 3**\n",
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"- **Using for loops**\n",
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" > > > > > > > 100-plus-Python-programming-exercises-extended/master"
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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(2000,3201):\n",
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" if i%7 == 0 and i%5!=0:\n",
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" print(i,end=',')\n",
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"print(\"\\b\")"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"- **Using generators and list comprehension**"
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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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"print(*(i for i in range(2000, 3201) if i%7 == 0 and i%5 != 0), sep=\",\")"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"> > > > > > > 100-plus-Python-programming-exercises-extended/master\n",
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"\n",
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"# Question 2\n",
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"\n",
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"### **Question:**\n",
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"\n",
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"> **_Write a program which can compute the factorial of a given numbers.The results should be printed in a comma-separated sequence on a single line.Suppose the following input is supplied to the program: 8\n",
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"> Then, the output should be:40320_**\n",
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"\n",
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"---\n",
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"\n",
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"### Hints:\n",
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"\n",
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"> **_In case of input data being supplied to the question, it should be assumed to be a console input._**\n",
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"\n",
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"---\n",
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"\n",
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"**Main author's Solution: Python 2**"
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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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"def fact(x):\n",
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" if x == 0:\n",
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" return 1\n",
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" return x * fact(x - 1)\n",
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"\n",
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"x = int(raw_input())\n",
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"print fact(x)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"---\n",
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"\n",
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"**My Solution: Python 3**\n",
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"\n",
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"- **Using While Loop**\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 = int(input()) #input() function takes input as string type\n",
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" #int() converts it to integer type\n",
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"fact = 1\n",
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"i = 1\n",
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"while i <= n:\n",
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" fact = fact * i;\n",
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" i = i + 1\n",
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"print(fact)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"- **Using For Loop**\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 = int(input()) #input() function takes input as string type\n",
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" #int() converts it to integer type\n",
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"fact = 1\n",
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"for i in range(1,n+1):\n",
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" fact = fact * i\n",
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"print(fact)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"- **Using Lambda Function**\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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"# Solution by: harshraj22\n",
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"\n",
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"n = int(input())\n",
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"def short_fact(x): return 1 if x <= 1 else x*short_fact(x-1)\n",
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"print(short_fact(n))\n"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"---\n",
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"\n",
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"# Question 3\n",
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"\n",
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"### **Question:**\n",
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"\n",
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"> **_With a given integral number n, write a program to generate a dictionary that contains (i, i x i) such that is an integral number between 1 and n (both included). and then the program should print the dictionary.Suppose the following input is supplied to the program: 8_**\n",
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"\n",
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"> **_Then, the output should be:_**"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"{1: 1, 2: 4, 3: 9, 4: 16, 5: 25, 6: 36, 7: 49, 8: 64}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"---\n",
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"\n",
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"### Hints:\n",
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"\n",
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"> **_In case of input data being supplied to the question, it should be assumed to be a console input.Consider use dict()_**\n",
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"\n",
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"---\n",
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"\n",
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"**Main author's Solution: Python 2**"
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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 = int(raw_input())\n",
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"d = dict()\n",
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"for i in range(1,n+1):\n",
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" d[i] = i * i\n",
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"print d"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**My Solution: Python 3:**\n",
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"\n",
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"- **Using for loop**"
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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 = int(input())\n",
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"ans = {}\n",
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"for i in range (1,n+1):\n",
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" ans[i] = i * i\n",
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"print(ans)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"- **Using dictionary comprehension**"
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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 = int(input())\n",
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"ans={i : i*i for i in range(1,n+1)}\n",
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"print(ans)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"---\n",
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"\n",
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"## Conclusion\n",
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"\n",
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"**_These was the solved problems of day 1. The above problems are very easy for the basic syntex learners.I have shown some easy ways of coding in my solutions. Lets see how to face and attack new problems in the next day._**\n",
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"\n",
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"[**_go to next day_**](https://github.com/darkprinx/100-plus-Python-programming-exercises-extended/blob/master/Status/Day%202.md \"Next Day\")\n",
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"\n",
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"[**_Discussion_**](https://github.com/darkprinx/100-plus-Python-programming-exercises-extended/issues/3)"
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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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"name": "52918bd4-40ea-43d9-b560-0e0992d7c326",
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"display_name": "'Python Interactive'"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 4
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}

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