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