euler/ipython/EulerProblem016.ipynb

143 lines
3.0 KiB
Plaintext

{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 16\n",
"\n",
"$2^{15}$ = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.\n",
"\n",
"What is the sum of the digits of the number $2^{1000}$?"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Seriously? Okay, probably more difficult in C."
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"1366\n"
]
}
],
"source": [
"s = sum(map(int, str(2**1000)))\n",
"assert(s == 1366)\n",
"print(s)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Okay, let's do it without big ints."
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"def double_number_string(number_string):\n",
" number_string = number_string[::-1] # reverse\n",
" result = []\n",
" carriage = 0\n",
" for c in number_string:\n",
" s = carriage + int(c) + int(c)\n",
" carriage = s // 10\n",
" next_digit = s % 10\n",
" result.append(str(next_digit))\n",
" if carriage > 0:\n",
" result.append(str(carriage))\n",
" result = \"\".join(result)[::-1]\n",
" return result\n",
"\n",
"assert(double_number_string(\"132\") == \"264\")\n",
"assert(double_number_string(\"965\") == \"1930\")"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"1366\n"
]
}
],
"source": [
"def powers_of_two(n):\n",
" \"\"\" Retuns nth power of two as a string. \"\"\"\n",
" assert(n >= 0)\n",
" if n == 0:\n",
" return \"1\"\n",
" s = \"2\"\n",
" for _ in range(n - 1):\n",
" s = double_number_string(s)\n",
" return s\n",
"\n",
"assert(powers_of_two(3) == \"8\")\n",
"assert(powers_of_two(10) == \"1024\")\n",
"\n",
"number_string = powers_of_two(1000)\n",
"print(sum(map(int, number_string)))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Here we go. The conversion to integer and the sum would be easy in C."
]
}
],
"metadata": {
"completion_date": "Sun, 31 Aug 2014, 20:49",
"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": "3.5.4"
},
"tags": [
"c",
"manual"
]
},
"nbformat": 4,
"nbformat_minor": 0
}